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Trigonometry Study Materials PDF With Practice Questions Worksheet

Trigonometry Study Materials PDF With Practice Questions Worksheet: Trignometry is one of the major section of Advance Mathematics for different exams including competitive exams. Trignometry Study Materials PDF With Practice Questions Worksheet is available here to download in English and Hindi language.

Trigonometry is the branch of mathematics dealing with the relations of the sides and angles of triangles and with the relevant functions of any angles. Throughout history, trigonometry has been applied in areas such as geodesy, surveying, celestial mechanics, and navigation.

Table of Contents

Trigonometry Study Materials PDF For Competitive Exams

Trigonometry basics.

The concept of Trigonometry is given by a Greek mathematician Hipparchus. Trigonometry is all about a right-angled triangle.

It is one of those divisions in mathematics that helps in finding the angles and missing sides of a triangle by the help of trigonometric ratios.

The angles are either measured in radians or degrees. The usual trigonometry angles are 0°, 30°, 45°, 60° and 90°, which are commonly used.

Trigonometry Basics

Six Important Trigonometric Functions

The six important trigonometric functions (trigonometric ratios) are calculated by the below formulas using above figure. It is necessary to get knowledge regarding the sides of the right angle triangle, because it defines the set of important trigonometric functions.

Trigonometry Ratios Table

The standard angles of trigonometrical ratios are 0°, 30°, 45°, 60° and 90° . The values of trigonometrical ratios of standard angles are very important to solve the trigonometrical problems.

The values of trigonometrical ratios of standard angles are very important to solve the trigonometrical problems. Therefore, it is necessary to remember the value of the trigonometrical ratios of these standard angles. The sine, cosine and tangent of the standard angles are given below in the table.

Trigonometry Ratios Table

Trigonometry Formula

The Trigonometric formulas or Identities are the equations which are true in the case of Right-Angled Triangles. Some of the special trigonometric identities are as given below –

1. Pythagorean Identities

2. Sum and Difference identities-

For angles A and B, we have the following relationships:

3. If A, B and C are angles and a, b and c are the sides of a triangle, then,

Cosine Laws

Trigonometry Questions & Answers For Competitive Exams

Here we have attached some Trigonometry questions and their solutions for competitive exams like SSC, Railway, UPSC & other exams.

Question 1: In a ΔABC right angled at B if AB = 12, and BC = 5 find sin A and tan A, cos C and cot C

AC=√((AB)^2+(BC)^2 ) =√(〖12〗^2+5^2 ) =√(144+25) =√169=13

When we consider t-ratios of∠A we have Base AB = 12 Perpendicular = BC = 5 Hypotenuse = AC = 13 sinA=Perpendicular/Hypotenuse=5/13 tanA=Perpendicular/Base=5/12

When we consider t-ratios of ∠C, we have Base = BC = 5 Perpendicular = AB = 12 Hypotenuse = AC = 13

cosC = Base/Hypotenuse = 5/13 cotC = Base/Perpendicular = 5/12

Question 2 : Find the value of 2 sin2 30° tan 60° – 3 cos2 60° sec2 30°

Solution: 2(1/2)^2×√3-3(1/2)^2×(2/√3)^2 =2×1/4×√3-3×1/4×4/3=√3/2-1=(√3-2)/2

Question 3 : In a right triangle ABC right angle at B the six trigonometric ratios of ∠C

Solution: sinA=Perpendicular/Hypotenuse=3/5

Base=√((Hypotenuse)^2-(Perpendicualr)^2 ) =√(5^2-3^2 ) =√(25-9)=√16=4

Now sinC=BC/AC=4/5,cosecC=5/4 cosC=3/5=AB/AC,secC=5/3 tanC=AB/AC=4/3,cotC=3/4

Question 4 : Find the value of 2 sin2 30° tan 60° – 3 cos2 60° sec2 30°

Question 5 : bFind the value θ sin2θ=√3

Solution: sin2θ= √3/2 2θ = 60 θ = 30°

Question 6 : Find the value of x. Tan 3x = sin 45° cos 45° + sin 30°

Solution: tan3x=1/√2×1/√2+1/2 =1/2+1/2=1 ⇒tan3x=1 ⇒ tan3x = tan45° 3x = 45° X = 15°

Trigonometry Problems & Solutions

Practice these questions given here to get a deep knowledge of Trigonometry. Use the formulas and table given in this article wherever necessary.

Q.1: In △ABC, right-angled at B, AB=22cm and BC=17cm. Find:

(a) sin A Cos B

(b) tan A tan B

Q.2: If 12cot θ= 15, then find sec θ?

Q.3: In Δ PQR, right-angled at Q, PR + QR = 30 cm and PQ = 10 cm. Determine the values of sin P, cos P and tan P.

Q.4: If sec 4θ = cosec (θ- 300), where 4θ is an acute angle, find the value of A.

Height And Distance

Sometimes, we have to find the height of a tower, building, tree, distance of a ship, width of a river, etc.

Though we cannot measure them easily, we can determine these by using trigonometric ratios.

Line of Sight

The line of sight or the line of vision is a straight line to the object we are viewing. If the object is above the horizontal from the eye, we have to lift up our head to view the object. In this process, our eye move, through an angle. This angle is called the angle of elevation of the object.

If the object is below the horizontal from the eye, then we have no turn our head downwards no view the object. In this process, our eye move through an angle. This angle is called the angle of depression of the object.

Example: A 25 m long ladder is placed against a vertical wall of a building. The foot of the ladder is 7m from base of the building. If the top of the ladder slips 4m, then the foot of the Ladder will slide by how much distance.

Sol: Let the height of the wall be h.

Now, h = √(〖25〗^2-7^2 ) = √(576 ) = 24m QS = √(625-400) = √(225 )=15m Required distance, X = (15-7) = 8m

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Trigonometry Practice Questions Worksheet PDF

Here we have attached the trigonometry practice questions pdf alongwith trigonometry worksheet pdf. You can also download Complete Trigonometry Study Materials PDF from the table given below:

Trigonometry Frequently Asked Questions

What do you Mean by Trigonometry? Ans. –   Trigonometry is one of the branches of mathematics which deals with the relationship between the sides of a triangle (right triangle) with its angles. There are 6 trigonometric functions to define It.

What are the Different Trigonometric Functions? Ans. –   The 6 trigonometric functions are: Sine function, Cosine function, Tan function, Sec function, Cot function, Cosec function

Who is the Father of Trigonometry? Ans. –   Hipparchus was a Greek astronomer who lived between 190-120 B.C. He is considered the father of trigonometry.

What are the Applications of Trigonometry in Real Life? Ans. –   The real life applications of trigonometry is in the calculation of height and distance. Some of the sectors where the concepts of trigonometry is extensively used are aviation department, navigation, criminology, marine biology, etc.

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MATH Worksheets 4 Kids

Trigonometry Charts

Quadrants and Angles

Degrees, Minutes and Seconds

Reference and Coterminal Angles

Trigonometric Identities

Trigonometry Worksheets for High School

Explore the surplus collection of trigonometry worksheets that cover key skills in quadrants and angles, measuring angles in degrees and radians, conversion between degrees, minutes and radians, understanding the six trigonometric ratios, unit circles, frequently used trigonometric identities, evaluating, proving and verifying trigonometric expressions and the list go on...

List of Trigonometry Worksheets

Explore the trigonometry worksheets in detail.

Grasp and retain trigonometric concepts with ease employing these visually appealing charts for quadrants and angles, right triangle trigonometric ratio chart, trigonometric ratio tables, allied angles and unit circle charts to mention a few.

Identify the quadrant encompassing the terminal side of the angle with this set of quadrants and angles worksheets. Draw the indicated angle on the coordinate plane, measure the angles in the quadrant and represent as degrees and radians and a lot more.

Conversion of Degrees and Radians

Introduce the two ways to measure an angle, namely degrees and radians with this set of worksheets. Adequate worksheets are provided to assist in practicing prompt conversions of degrees to radians and vice-versa.

To specifically and accurately measure the size of an angle in degrees, it is further broken down into degrees, minutes and seconds. This worksheet stack consists of ample exercises to practice conversion between degrees, minutes and seconds.

Determine the reference angles in degrees and radians, find the coterminal angles for the indicated angles, and positive and negative coterminal angles with this assemblage of reference and coterminal angles worksheets.

Trigonometric Ratios | Right Triangle Trigonometry

Kick start your learning with these trig ratio worksheets. Identify the legs, side and angles, introduce the six trigonometric ratios both primary trig ratios and reciprocal trig ratios and much more with these trigonometric ratio worksheets.

Included here are fundamental identities like quotient, reciprocal, cofunction and Pythagorean identities, sum and difference identities, sum-to-product, product-to-sum, double angle and half angle identities and ample trig expression to be simplified, proved and verified using the trigonometric formulas.

Unit Circle Worksheets

Packed in these unit circle worksheets are exercises to find the coordinates of a point on the unit circle, determine the corresponding angle measure, use the unit circle to find the six trigonometric ratios and a lot more.

Trigonometric Ratios of Allied Angles

Allied angle worksheets here enclose exercises like finding the exact value of the trigonometric ratio offering angle measures in degrees or radians, evaluating trig ratios of allied angles and proving the trigonometric statements to mention just a few.

Evaluating Trigonometric Expressions

These worksheets outline the concept of evaluating trigonometric expressions involving primary, reciprocal and fundamental trigonometric ratios, evaluating expressions using a calculator, evaluate using allied angles and more!

Evaluating Trigonometric Functions Worksheets

With this set of evaluating trigonometric functions worksheets at your disposal, you have no dearth of practice exercises. Begin with substituting the specified x-values in trigonometric functions and solve for f(x).

Inverse Trigonometric Function Worksheets

Utilize this adequate supply of inverse trigonometric ratio worksheets to find the exact value of inverse trig ratios using charts and calculators, find the measure of angles, solve the equations, learn to evaluate inverse and the composition of trigonometric functions and a lot more.

Law of Sines Worksheets

Navigate through this law of sines worksheets that encompass an array of topics like finding the missing side and the unknown angles, solving triangles, an ambiguous case in a triangle, finding the area of SAS triangle and more.

Law of Cosines Worksheets

Incorporate the law of cosines worksheets to elevate your understanding of the concept and practice to find the missing sides of a triangle, finding the unknown angles (SAS & SSS), solving triangles and much more.

Solving Triangles Worksheets

Access this huge collection of solving triangles worksheets to comprehend the topics like solving triangles, finding the area of the triangle, solving the triangle using the given area and much more worksheets are included.

Principal Solutions of Trig Equations Worksheets

Reinforce the concept of principal solutions of trigonometric equations with this adequate supply of worksheets like solving linear trigonometric equations, solving trigonometric equations in quadratic form and much more.

General Solutions of Trig Equations Worksheets

Employ this assortment of general solutions of trigonometric equations worksheets that feature ample of exercises to hone your skills in solving different types of trigonometric equations to obtain the general solutions.

Sample Worksheets

Trigonometry Charts

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Unit: Trigonometry

Unit circle introduction.

The Pythagorean identity

Trigonometric values of special angles

Graphs of sin(x), cos(x), and tan(x)

Amplitude, midline and period

Transforming sinusoidal graphs

Graphing sinusoidal functions

Sinusoidal models

About this unit

trigonometry quadrant questions pdf

1000+ Trigonometry PDF (Questions & Solution with Shortcut Tricks) – Download Now

Trigonometry pdf  questions & solution with shortcut tricks.

Trigonometry PDF  : Trigonometry  is one of the most important topic that comes under Banking (IBPS, SBI, RBI, SEBI, NABARD, LIC), SSC (CGL, CHSL, MTS, CPO, SI, JE), Railway (RRB NTPC, Grade D, ALP, JE, TC), Defence (UPSC CDS/NDA/NA, Police, Army, Navy, Airforce) & Teaching Exams. If you know different types & patterns of Trigonometry   then you can easily get 4-5 marks in just 3-4 minutes.

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3.3: Solving Trigonometric Equations

Learning Objectives

Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles , which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles.

Photo of the Egyptian pyramids near a modern city.

In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids.

Solving Linear Trigonometric Equations in Sine and Cosine

Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all. Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The period of both the sine function and the cosine function is \(2\pi\). In other words, every \(2\pi\) units, the y- values repeat. If we need to find all possible solutions, then we must add \(2\pi k\),where \(k\) is an integer, to the initial solution. Recall the rule that gives the format for stating all possible solutions for a function where the period is \(2\pi\):

\[\sin \theta=\sin(\theta \pm 2k\pi)\]

There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.

Example \(\PageIndex{1A}\): Solving a Linear Trigonometric Equation Involving the Cosine Function

Find all possible exact solutions for the equation \(\cos \theta=\dfrac{1}{2}\).

From the unit circle, we know that

\[ \begin{align*} \cos \theta &=\dfrac{1}{2} \\[4pt] \theta &=\dfrac{\pi}{3},\space \dfrac{5\pi}{3} \end{align*}\]

These are the solutions in the interval \([ 0,2\pi ]\). All possible solutions are given by

\[\theta=\dfrac{\pi}{3} \pm 2k\pi \quad \text{and} \quad \theta=\dfrac{5\pi}{3} \pm 2k\pi \nonumber\]

where \(k\) is an integer.

Example \(\PageIndex{1B}\): Solving a Linear Equation Involving the Sine Function

Find all possible exact solutions for the equation \(\sin t=\dfrac{1}{2}\).

Solving for all possible values of \(t\) means that solutions include angles beyond the period of \(2\pi\). From the section on Sum and Difference Identities, we can see that the solutions are \(t=\dfrac{\pi}{6}\) and \(t=\dfrac{5\pi}{6}\). But the problem is asking for all possible values that solve the equation. Therefore, the answer is

\[t=\dfrac{\pi}{6}\pm 2\pi k \quad \text{and} \quad t=\dfrac{5\pi}{6}\pm 2\pi k \nonumber\]

How to: Given a trigonometric equation, solve using algebra

Example \(\PageIndex{2}\): Solve the Linear Trigonometric Equation

Solve the equation exactly: \(2 \cos \theta−3=−5\), \(0≤\theta<2\pi\).

Use algebraic techniques to solve the equation.

\[\begin{align*} 2 \cos \theta-3&= -5\\ 2 \cos \theta&= -2\\ \cos \theta&= -1\\ \theta&= \pi \end{align*}\]

Exercise \(\PageIndex{1}\)

Solve exactly the following linear equation on the interval \([0,2\pi)\): \(2 \sin x+1=0\).

\(x=\dfrac{7\pi}{6},\space \dfrac{11\pi}{6}\)

Solving Equations Involving a Single Trigonometric Function

When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see [link]). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is \(\pi\),not \(2\pi\). Further, the domain of tangent is all real numbers with the exception of odd integer multiples of \(\dfrac{\pi}{2}\),unless, of course, a problem places its own restrictions on the domain.

Example \(\PageIndex{3A}\): Solving a Trignometric Equation Involving Sine

Solve the problem exactly: \(2 {\sin}^2 \theta−1=0\), \(0≤\theta<2\pi\).

As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate \(\sin \theta\). Then we will find the angles.

\[\begin{align*} 2 {\sin}^2 \theta-1&= 0\\ 2 {\sin}^2 \theta&= 1\\ {\sin}^2 \theta&= \dfrac{1}{2}\\ \sqrt{ {\sin}^2 \theta }&= \pm \sqrt{ \dfrac{1}{2} }\\ \sin \theta&= \pm \dfrac{1}{\sqrt{2}}\\ &= \pm \dfrac{\sqrt{2}}{2}\\ \theta&= \dfrac{\pi}{4}, \space \dfrac{3\pi}{4},\space \dfrac{5\pi}{4}, \space \dfrac{7\pi}{4} \end{align*}\]

As \(\sin \theta=−\dfrac{1}{2}\), notice that all four solutions are in the third and fourth quadrants.

Example \(\PageIndex{3B}\): Solving a Trigonometric Equation Involving Cosecant

Solve the following equation exactly: \(\csc \theta=−2\), \(0≤\theta<4\pi\).

We want all values of \(\theta\) for which \(\csc \theta=−2\) over the interval \(0≤\theta<4\pi\).

\[\begin{align*} \csc \theta&= -2\\ \dfrac{1}{\sin \theta}&= -2\\ \sin \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{7\pi}{6},\space \dfrac{11\pi}{6},\space \dfrac{19\pi}{6}, \space \dfrac{23\pi}{6} \end{align*}\]

Example \(\PageIndex{3C}\): Solving an Equation Involving Tangent

Solve the equation exactly: \(\tan\left(\theta−\dfrac{\pi}{2}\right)=1\), \(0≤\theta<2\pi\).

Recall that the tangent function has a period of \(\pi\). On the interval \([ 0,\pi )\),and at the angle of \(\dfrac{\pi}{4}\),the tangent has a value of \(1\). However, the angle we want is \(\left(\theta−\dfrac{\pi}{2}\right)\). Thus, if \(\tan\left(\dfrac{\pi}{4}\right)=1\),then

\[\begin{align*} \theta-\dfrac{\pi}{2}&= \dfrac{\pi}{4}\\ \theta&= \dfrac{3\pi}{4} \pm k\pi \end{align*}\]

Over the interval \([ 0,2\pi )\),we have two solutions:

\(\theta=\dfrac{3\pi}{4}\) and \(\theta=\dfrac{3\pi}{4}+\pi=\dfrac{7\pi}{4}\)

Exercise \(\PageIndex{2}\)

Find all solutions for \(\tan x=\sqrt{3}\).

\(\dfrac{\pi}{3}\pm \pi k\)

Example \(\PageIndex{4}\): Identify all Solutions to the Equation Involving Tangent

Identify all exact solutions to the equation \(2(\tan x+3)=5+\tan x\), \(0≤x<2\pi\).

We can solve this equation using only algebra. Isolate the expression \(\tan x\) on the left side of the equals sign.

There are two angles on the unit circle that have a tangent value of \(−1\): \(\theta=\dfrac{3\pi}{4}\) and \(\theta=\dfrac{7\pi}{4}\).

Solve Trigonometric Equations Using a Calculator

Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.

Example \(\PageIndex{5A}\): Using a Calculator to Solve a Trigonometric Equation Involving Sine

Use a calculator to solve the equation \(\sin \theta=0.8\),where \(\theta\) is in radians.

Make sure mode is set to radians. To find \(\theta\), use the inverse sine function. On most calculators, you will need to push the 2 ND button and then the SIN button to bring up the \({\sin}^{−1}\) function. What is shown on the screen is \({\sin}^{−1}\).The calculator is ready for the input within the parentheses. For this problem, we enter \({\sin}^{−1}(0.8)\), and press ENTER. Thus, to four decimals places,

\({\sin}^{−1}(0.8)≈0.9273\)

The solution is

\(\theta≈0.9273\pm 2\pi k\)

The angle measurement in degrees is

\[\begin{align*} \theta&\approx 53.1^{\circ}\\ \theta&\approx 180^{\circ}-53.1^{\circ}\\ &\approx 126.9^{\circ} \end{align*}\]

Note that a calculator will only return an angle in quadrants I or IV for the sine function since that is the range of the inverse sine. The other angle is obtained by using \(\pi−\theta\).

Example \(\PageIndex{5B}\): Using a Calculator to Solve a Trigonometric Equation Involving Secant

Use a calculator to solve the equation \( \sec θ=−4, \) giving your answer in radians.

We can begin with some algebra.

\[\begin{align*} \sec \theta&= -4\\ \dfrac{1}{\cos \theta}&= -4\\ \cos \theta&= -\dfrac{1}{4} \end{align*}\]

Check that the MODE is in radians. Now use the inverse cosine function

\[\begin{align*}{\cos}^{-1}\left(-\dfrac{1}{4}\right)&\approx 1.8235\\ \theta&\approx 1.8235+2\pi k \end{align*}\]

Since \(\dfrac{\pi}{2}≈1.57\) and \(\pi≈3.14\),\(1.8235\) is between these two numbers, thus \(\theta≈1.8235\) is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See Figure \(\PageIndex{2}\).

Graph of angles theta =approx 1.8235, theta prime =approx pi - 1.8235 = approx 1.3181, and then theta prime = pi + 1.3181 = approx 4.4597

So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is \(\theta '≈\pi−1.8235≈1.3181\). The other solution in quadrant III is \(\theta '≈\pi+1.3181≈4.4597\).

The solutions are \(\theta≈1.8235\pm 2\pi k\) and \(\theta≈4.4597\pm 2\pi k\).

Exercise \(\PageIndex{3}\)

Solve \(\cos \theta=−0.2\).

\(\theta≈1.7722\pm 2\pi k\) and \(\theta≈4.5110\pm 2\pi k\)

Solving Trigonometric Equations in Quadratic Form

Solving a quadratic equation may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as \(x\) or \(u\). If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.

Example \(\PageIndex{6A}\): Solving a Trigonometric Equation in Quadratic Form

Solve the equation exactly: \({\cos}^2 \theta+3 \cos \theta−1=0\), \(0≤\theta<2\pi\).

We begin by using substitution and replacing \(\cos \theta\) with \(x\). It is not necessary to use substitution, but it may make the problem easier to solve visually. Let \(\cos \theta=x\). We have

\(x^2+3x−1=0\)

The equation cannot be factored, so we will use the quadratic formula: \(x=\dfrac{−b\pm \sqrt{b^2−4ac}}{2a}\).

\[\begin{align*} x&= \dfrac{ -3\pm \sqrt{ {(-3)}^2-4 (1) (-1) } }{2}\\ &= \dfrac{-3\pm \sqrt{13}}{2}\end{align*}\]

Replace \(x\) with \(\cos \theta \) and solve.

\[\begin{align*} \cos \theta&= \dfrac{-3\pm \sqrt{13}}{2}\\ \theta&= {\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right) \end{align*}\]

Note that only the + sign is used. This is because we get an error when we solve \(\theta={\cos}^{−1}\left(\dfrac{−3−\sqrt{13}}{2}\right)\) on a calculator, since the domain of the inverse cosine function is \([ −1,1 ]\). However, there is a second solution:

\[\begin{align*} \theta&= {\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right)\\ &\approx 1.26 \end{align*}\]

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

\[\begin{align*} \theta&= 2\pi-{\cos}^{-1}\left(\dfrac{-3+\sqrt{13}}{2}\right)\\ &\approx 5.02 \end{align*}\]

Example \(\PageIndex{6B}\): Solving a Trigonometric Equation in Quadratic Form by Factoring

Solve the equation exactly: \(2 {\sin}^2 \theta−5 \sin \theta+3=0\), \(0≤\theta≤2\pi\).

Using grouping, this quadratic can be factored. Either make the real substitution, \(\sin \theta=u\),or imagine it, as we factor:

\[\begin{align*} 2 {\sin}^2 \theta-5 \sin \theta+3&= 0\\ (2 \sin \theta-3)(\sin \theta-1)&= 0 \qquad \text {Now set each factor equal to zero.}\\ 2 \sin \theta-3&= 0\\ 2 \sin \theta&= 3\\ \sin \theta&= \dfrac{3}{2}\\ \sin \theta-1&= 0\\ \sin \theta&= 1 \end{align*}\]

Next solve for \(\theta\): \(\sin \theta≠\dfrac{3}{2}\), as the range of the sine function is \([ −1,1 ]\). However, \(\sin \theta=1\), giving the solution \(\theta=\dfrac{\pi}{2}\).

Make sure to check all solutions on the given domain as some factors have no solution.

Exercise \(\PageIndex{4}\)

Solve \({\sin}^2 \theta=2 \cos \theta+2\), \(0≤\theta≤2\pi\). [Hint: Make a substitution to express the equation only in terms of cosine.]

\(\cos \theta=−1\), \(\theta=\pi\)

Example \(\PageIndex{7A}\): Solving a Trigonometric Equation Using Algebra

Solve exactly: \(2 {\sin}^2 \theta+\sin \theta=0;\space 0≤\theta<2\pi\)

This problem should appear familiar as it is similar to a quadratic. Let \(\sin \theta=x\). The equation becomes \(2x^2+x=0\). We begin by factoring:

\[\begin{align*} 2x^2+x&= 0\\ x(2x+1)&= 0\qquad \text {Set each factor equal to zero.}\\ x&= 0\\ 2x+1&= 0\\ x&= -\dfrac{1}{2} \end{align*}\] Then, substitute back into the equation the original expression \(\sin \theta \) for \(x\). Thus, \[\begin{align*} \sin \theta&= 0\\ \theta&= 0,\pi\\ \sin \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{7\pi}{6},\dfrac{11\pi}{6} \end{align*}\]

The solutions within the domain \(0≤\theta<2\pi\) are \(\theta=0,\pi,\dfrac{7\pi}{6},\dfrac{11\pi}{6}\).

If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.

\[\begin{align*} {\sin}^2 \theta+\sin \theta&= 0\\ \sin \theta(2\sin \theta+1)&= 0\\ \sin \theta&= 0\\ \theta&= 0,\pi\\ 2 \sin \theta+1&= 0\\ 2\sin \theta&= -1\\ \sin \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{7\pi}{6},\dfrac{11\pi}{6} \end{align*}\]

We can see the solutions on the graph in Figure \(\PageIndex{3}\). On the interval \(0≤\theta<2\pi\),the graph crosses the \(x\) - axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.

Graph of 2*(sin(theta))^2 + sin(theta) from 0 to 2pi. Zeros are at 0, pi, 7pi/6, and 11pi/6.

We can verify the solutions on the unit circle in via the result in the section on Sum and Difference Identities as well.

Example \(\PageIndex{7B}\): Solving a Trigonometric Equation Quadratic in Form

Solve the equation quadratic in form exactly: \(2 {\sin}^2 \theta−3 \sin \theta+1=0\), \(0≤\theta<2\pi\).

We can factor using grouping. Solution values of \(\theta\) can be found on the unit circle.

\[\begin{align*} (2 \sin \theta-1)(\sin \theta-1)&= 0\\ 2 \sin \theta-1&= 0\\ \sin \theta&= \dfrac{1}{2}\\ \theta&= \dfrac{\pi}{6}, \dfrac{5\pi}{6}\\ \sin \theta&= 1\\ \theta&= \dfrac{\pi}{2} \end{align*}\]

Exercise \(\PageIndex{5}\)

Solve the quadratic equation \(2{\cos}^2 \theta+\cos \theta=0\).

\(\dfrac{\pi}{2}, \space \dfrac{2\pi}{3}, \space \dfrac{4\pi}{3}, \space \dfrac{3\pi}{2}\)

Solving Trigonometric Equations Using Fundamental Identities

While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation.

Example \(\PageIndex{8}\): Solving an Equation Using an Identity

Solve the equation exactly using an identity: \(3 \cos \theta+3=2 {\sin}^2 \theta\), \(0≤\theta<2\pi\).

If we rewrite the right side, we can write the equation in terms of cosine:

\[\begin{align*} 3 \cos \theta+3&= 2 {\sin}^2 \theta\\ 3 \cos \theta+3&= 2(1-{\cos}^2 \theta)\\ 3 \cos \theta+3&= 2-2{\cos}^2 \theta\\ 2 {\cos}^2 \theta+3 \cos \theta+1&= 0\\ (2 \cos \theta+1)(\cos \theta+1)&= 0\\ 2 \cos \theta+1&= 0\\ \cos \theta&= -\dfrac{1}{2}\\ \theta&= \dfrac{2\pi}{3},\space \dfrac{4\pi}{3}\\ \cos \theta+1&= 0\\ \cos \theta&= -1\\ \theta&= \pi\\ \end{align*}\]

Our solutions are \(\theta=\dfrac{2\pi}{3},\space \dfrac{4\pi}{3},\space \pi\).

Solving Trigonometric Equations with Multiple Angles

Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as \(\sin(2x)\) or \(\cos(3x)\). When confronted with these equations, recall that \(y=\sin(2x)\) is a horizontal compression by a factor of 2 of the function \(y=\sin x\). On an interval of \(2\pi\),we can graph two periods of \(y=\sin(2x)\),as opposed to one cycle of \(y=\sin x\). This compression of the graph leads us to believe there may be twice as many x -intercepts or solutions to \(\sin(2x)=0\) compared to \(\sin x=0\). This information will help us solve the equation.

Example \(\PageIndex{9}\): Solving a Multiple Angle Trigonometric Equation

Solve exactly: \(\cos(2x)=\dfrac{1}{2}\) on \([ 0,2\pi )\).

We can see that this equation is the standard equation with a multiple of an angle. If \(\cos(\alpha)=\dfrac{1}{2}\),we know \(\alpha\) is in quadrants I and IV. While \(\theta={\cos}^{−1} \dfrac{1}{2}\) will only yield solutions in quadrants I and II, we recognize that the solutions to the equation \(\cos \theta=\dfrac{1}{2}\) will be in quadrants I and IV.

Therefore, the possible angles are \(\theta=\dfrac{\pi}{3}\) and \(\theta=\dfrac{5\pi}{3}\). So, \(2x=\dfrac{\pi}{3}\) or \(2x=\dfrac{5\pi}{3}\), which means that \(x=\dfrac{\pi}{6}\) or \(x=\dfrac{5\pi}{6}\). Does this make sense? Yes, because \(\cos\left(2\left(\dfrac{\pi}{6}\right)\right)=\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}\).

Are there any other possible answers? Let us return to our first step.

In quadrant I, \(2x=\dfrac{\pi}{3}\), so \(x=\dfrac{\pi}{6}\) as noted. Let us revolve around the circle again:

\[\begin{align*} 2x&= \dfrac{\pi}{3}+2\pi\\ &= \dfrac{\pi}{3}+\dfrac{6\pi}{3}\\ &= \dfrac{7\pi}{3}\\ x&= \dfrac{7\pi}{6}\\ \text {One more rotation yields}\\ 2x&= \dfrac{\pi}{3}+4\pi\\ &= \dfrac{\pi}{3}+\dfrac{12\pi}{3}\\ &= \dfrac{13\pi}{3}\\ \end{align*}\]

\(x=\dfrac{13\pi}{6}>2\pi\), so this value for \(x\) is larger than \(2\pi\), so it is not a solution on \([ 0,2\pi )\).

In quadrant IV, \(2x=\dfrac{5\pi}{3}\), so \(x=\dfrac{5\pi}{6}\) as noted. Let us revolve around the circle again:

\[\begin{align*} 2x&= \dfrac{5\pi}{3}+2\pi\\ &= \dfrac{5\pi}{3}+\dfrac{6\pi}{3}\\ &= \dfrac{11\pi}{3} \end{align*}\]

so \(x=\dfrac{11\pi}{6}\).

One more rotation yields

\[\begin{align*} 2x&= \dfrac{5\pi}{3}+4\pi\\ &= \dfrac{5\pi}{3}+\dfrac{12\pi}{3}\\ &= \dfrac{17\pi}{3} \end{align*}\]

\(x=\dfrac{17\pi}{6}>2\pi\),so this value for \(x\) is larger than \(2\pi\),so it is not a solution on \([ 0,2\pi )\)  .

Our solutions are \(x=\dfrac{\pi}{6}, \space \dfrac{5\pi}{6}, \space \dfrac{7\pi}{6}\), and \(\dfrac{11\pi}{6}\). Note that whenever we solve a problem in the form of \(sin(nx)=c\), we must go around the unit circle \(n\) times.

Key Concepts

Contributors and Attributions

Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a  Creative Commons Attribution License 4.0  license. Download for free at  https://openstax.org/details/books/precalculus .

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Trigonometry Review and Practice Questions

Trigonometry is the branch of mathematics dealing with the relations of the sides and angles of triangles and with the relevant functions of any angles.

Quick Review with Examples

If we are observing a right triangle, where  a  and  b  are its legs and  c  is its hypotenuse, we can use trigonometric functions to make a relationship between angles and sides of the right triangle.

trigonometry quadrant questions pdf

If the right angle of the right triangle ABC is at the point C, then the sine ( sin ) and the cosine ( cos ) of the angles α (at the point A) and β (at the point B) can be found like this:

             sinα = a/c         sinβ = b/c

                        cosα = b/c         cosβ = a/c

Notice that  sinα  and  cosβ  are the equal, and same goes for  sinβ  and  cosα.  So, to find sine of the angle, we divide the side that is opposite of that angle and the hypotenuse. To find cosine of the angle, we divide the side that makes that angle (adjacent side) by the hypotenuse.

There are 2 more important trigonometric functions, tangent and cotangent:

tgα = sinα/cosα = a/b

                        ctgα = cosα/sinα = b/a

For the functions sine and cosine, there is a table with values for some of the angles, which is to be memorized as it is very useful for solving various trigonometric problems. Here is that table:

Let’s see one example:

If  a  is 9 cm and  c  is 18 cm, find  α.

We can use the sine for this problem:

sinα = a/c = 9/18 = 1/2

We can see from the table that if  sinα  is 1/2, then angle  α  is 30⁰.

In addition to degrees we can write angles using π, where π represents 180⁰. For example, angle π/2 means a right angle of 90⁰.

trigonometry quadrant questions pdf

Practice Questions

1. If sides a and b of a right triangle are 8 and 6, respectively, find cosine of α.

a. 1/5 b. 5/3 c. 3/5 d. 2/5

2. Find tangent of a right triangle, if a is 3 and c is 5.

a. 1/4 b. 5/3 c. 4/3 d. 3/4

3. If α=30 0 , find sin30 0  + cos60 0 .

a. 1/2 b. 2/3 c. 1 d. 3/2

4. Calculate  (sin 2 30 0  – sin0 0 )/(cos90 0  – cos60 0 ).

A. -1/2 B. 2/3 C. 0 D. 1/2

5. Find cotangent of a right angle.

A. -1 B. 0 C. 1/2 D. -1/2

Answer Key 

a = 8 b = 6 a 2 + b 2 = c 2 8 2 + 6 2 = c 2 64 + 36 = c 2 c 2 = 100 c = 10 cosα = b/c = 6/10 = 3/5

2. D 3/4 a = 3 c = 5 a 2 + b 2 = c 2 3 2 + b 2  = 5 2 b 2  = 25 – 9 b 2  = 16 b = 4 tgα = a/b = 3/4

3. C 1 α = 30 0 sin30 0  + cos60 0  = 1/2 + 1/2 = 1

(sin 2 30 0 – sin0 0 ) / (cos90 0  – cos60 0 ) = ((1/2) 2  – 0) / (0 – 1/2) = (1/4) / (-1/2) = -1/2

α = 90 0 ctg90 0  = cos90 0 / sin90 0  = 0/1 = 0

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Trigonometry Questions

Trigonometry questions given here involve finding the missing sides of a triangle with the help of trigonometric ratios and proving trigonometry identities. We know that trigonometry is one of the most important chapters of Class 10 Maths. Hence, solving these questions will help you to improve your problem-solving skills.

What is Trigonometry?

The word ‘trigonometry’ is derived from the Greek words ‘tri’ (meaning three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). Trigonometry is the study of relationships between the sides and angles of a triangle.

The basic trigonometric ratios are defined as follows.

sine of ∠A = sin A = Side opposite to ∠A/ Hypotenuse

cosine of ∠A = cos A = Side adjacent to ∠A/ Hypotenuse

tangent of ∠A = tan A = (Side opposite to ∠A)/ (Side adjacent to ∠A)

cosecant of ∠A = cosec A = 1/sin A = Hypotenuse/ Side opposite to ∠A

secant of ∠A = sec A = 1/cos A = Hypotenuse/ Side adjacent to ∠A

cotangent of ∠A = cot A = 1/tan A = (Side adjacent to ∠A)/ (Side opposite to ∠A)

Also, tan A = sin A/cos A

cot A = cos A/sin A

Also, read: Trigonometry

Trigonometry Questions and Answers

1. From the given figure, find tan P – cot R.

Trigonometry Questions Q1

From the given,

In the right triangle PQR, Q is right angle.

By Pythagoras theorem,

PR 2 = PQ 2 + QR 2

QR 2 = (13) 2 – (12) 2

= 169 – 144

tan P = QR/PQ = 5/12

cot R = QR/PQ = 5/12

So, tan P – cot R = (5/12) – (5/12) = 0

2. Prove that (sin 4 θ – cos 4 θ +1) cosec 2 θ = 2

L.H.S. = (sin 4 θ – cos 4 θ +1) cosec 2 θ

= [(sin 2 θ – cos 2 θ) (sin 2 θ + cos 2 θ) + 1] cosec 2 θ

Using the identity sin 2 A + cos 2 A = 1,

= (sin 2 θ – cos 2 θ + 1) cosec 2 θ

= [sin 2 θ – (1 – sin 2 θ) + 1] cosec 2 θ

= 2 sin 2 θ cosec 2 θ

= 2 sin 2 θ (1/sin 2 θ)

3. Prove that (√3 + 1) (3 – cot 30°) = tan 3 60° – 2 sin 60°.

LHS = (√3 + 1)(3 – cot 30°)

= (√3 + 1)(3 – √3)

= 3√3 – √3.√3 + 3 – √3

= 2√3 – 3 + 3

RHS = tan 3 60° – 2 sin 60°

= (√3) 3 – 2(√3/2)

= 3√3 – √3

Therefore, (√3 + 1) (3 – cot 30°) = tan 3 60° – 2 sin 60°.

Hence proved.

4. If tan(A + B) = √3 and tan(A – B) = 1/√3 ; 0° < A + B ≤ 90°; A > B, find A and B.

tan(A + B) = √3

tan(A + B) = tan 60°

A + B = 60°….(i)

tan(A – B) = 1/√3

tan(A – B) = tan 30°

A – B = 30°….(ii)

Adding (i) and (ii),

A + B + A – B = 60° + 30°

Substituting A = 45° in (i),

45° + B = 60°

B = 60° – 45° = 15°

Therefore, A = 45° and B = 15°.

5. If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A.

sin 3A = cos(A – 26°); 3A is an acute angle

cos(90° – 3A) = cos(A – 26°) {since cos(90° – A) = sin A}

⇒ 90° – 3A = A – 26

⇒ 3A + A = 90° + 26°

⇒ 4A = 116°

⇒ A = 116°/4

6. If A, B and C are interior angles of a triangle ABC, show that sin (B + C/2) = cos A/2.

We know that, for a given triangle, the sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° ….(1)

B + C = 180° – A

Dividing both sides of this equation by 2, we get;

⇒ (B + C)/2 = (180° – A)/2

⇒ (B + C)/2 = 90° – A/2

Take sin on both sides,

sin (B + C)/2 = sin (90° – A/2)

⇒ sin (B + C)/2 = cos A/2 {since sin(90° – x) = cos x}

7. If tan θ + sec θ = l, prove that sec θ = (l 2 + 1)/2l.

tan θ + sec θ = l….(i)

We know that,

sec 2 θ – tan 2 θ = 1

(sec θ – tan θ)(sec θ + tan θ) = 1

(sec θ – tan θ) l = 1 {from (i)}

sec θ – tan θ = 1/l….(ii)

tan θ + sec θ + sec θ – tan θ = l + (1/l)

2 sec θ = (l 2 + 1)l

sec θ = (l 2 + 1)/2l

8. Prove that (cos A – sin A + 1)/ (cos A + sin A – 1) = cosec A + cot A, using the identity cosec 2 A = 1 + cot 2 A.

LHS = (cos A – sin A + 1)/ (cos A + sin A – 1)

Dividing the numerator and denominator by sin A, we get;

= (cot A – 1 + cosec A)/(cot A + 1 – cosec A)

Using the identity cosec 2 A = 1 + cot 2 A ⇒ cosec 2 A – cot 2 A = 1,

= [cot A – (cosec 2 A – cot 2 A) + cosec A]/ (cot A + 1 – cosec A)

= [(cosec A + cot A) – (cosec A – cot A)(cosec A + cot A)] / (cot A + 1 – cosec A)

= cosec A + cot A

9. Prove that: (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)

[Hint: Simplify LHS and RHS separately]

LHS = (cosec A – sin A)(sec A – cos A)

= (cos 2 A/sin A) (sin 2 A/cos A)

= cos A sin A….(i)

RHS = 1/(tan A + cot A)

= (sin A cos A)/ (sin 2 A + cos 2 A)

= (sin A cos A)/1

= sin A cos A….(ii)

From (i) and (ii),

i.e. (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)

10. If a sin θ + b cos θ = c, prove that a cosθ – b sinθ = √(a 2 + b 2 – c 2 ).

a sin θ + b cos θ = c

Squaring on both sides,

(a sin θ + b cos θ) 2 = c 2

a 2 sin 2 θ + b 2 cos 2 θ + 2ab sin θ cos θ = c 2

a 2 (1 – cos 2 θ) + b 2 (1 – sin 2 θ) + 2ab sin θ cos θ = c 2

a 2 – a 2 cos 2 θ + b 2 – b 2 sin 2 θ + 2ab sin θ cos θ = c 2

a 2 + b 2 – c 2 = a 2 cos 2 θ + b 2 sin 2 θ – 2ab sin θ cos θ

a 2 + b 2 – c 2 = (a cos θ – b sin θ ) 2

⇒ a cos θ – b sin θ = √(a 2 + b 2 – c 2 )

Video Lesson on Trigonometry

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Practice Questions on Trigonometry

Solve the following trigonometry problems.

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    Explore the surplus collection of trigonometry worksheets that cover key skills in quadrants and angles, measuring angles in degrees and radians, conversion between degrees, minutes and radians, understanding the six trigonometric ratios, unit circles, frequently used trigonometric identities, evaluating, proving and verifying trigonometric …

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    0/1900 Mastery points Unit circle introduction Radians The Pythagorean identity Special trigonometric values in the first quadrant Trigonometric values on the unit circle Graphs of sin (x), cos (x), and tan (x) Amplitude, midline, and period Transforming sinusoidal graphs Graphing sinusoidal functions Sinusoidal models Long live Tau

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    The Pythagorean identity. Trigonometric values of special angles. Quiz 1: 5 questions Practice what you've learned, and level up on the above skills. Graphs of sin (x), cos (x), and tan (x) Amplitude, midline and period. Transforming sinusoidal graphs. Quiz 2: 6 questions Practice what you've learned, and level up on the above skills.

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