Solving for we get where we look at the quadrant of to decide if it's positive or negative. Likewise, we can use the fact that to find a half angle identity for sine. Then, to find a half angle identity for tangent, we just use the fact that and plug in the half angle identities for sine and cosine. Grade 11 trigonometry problems and questions with answers and solutions are presented. Problems and Questions A ferris wheel with a radius of 25 meters makes one rotation every 36 seconds. When working with trigonometric identities, it may be useful to keep the following tips in mind: Draw a picture illustrating the problem if it involves only the basic trigonometric functions. If the problem expresses an identity between trigonometric functions, try working on one side of the identity to write the trigonometric functions from one side in terms of trigonometric functions.

1. Trigonometric Identititiesmath Problem Solving Problems
2. Solving Trigonometric Equations With Identities

Learning Objectives

• Verify the fundamental trigonometric identities.
• Simplify trigonometric expressions using algebra and the identities.

In espionage movies, we see international spies with multiple passports, each claiming a different identity. However, we know that each of those passports represents the same person. The trigonometric identities act in a similar manner to multiple passports—there are many ways to represent the same trigonometric expression. Just as a spy will choose an Italian passport when traveling to Italy, we choose the identity that applies to the given scenario when solving a trigonometric equation.

In this section, we will begin an examination of the fundamental trigonometric identities, including how we can verify them and how we can use them to simplify trigonometric expressions.

Verifying the Fundamental Trigonometric Identities

Identities enable us to simplify complicated expressions. They are the basic tools of trigonometry used in solving trigonometric equations, just as factoring, finding common denominators, and using special formulas are the basic tools of solving algebraic equations. In fact, we use algebraic techniques constantly to simplify trigonometric expressions. Basic properties and formulas of algebra, such as the difference of squares formula and the perfect squares formula, will simplify the work involved with trigonometric expressions and equations. We already know that all of the trigonometric functions are related because they all are defined in terms of the unit circle. Consequently, any trigonometric identity can be written in many ways.

To verify the trigonometric identities, we usually start with the more complicated side of the equation and essentially rewrite the expression until it has been transformed into the same expression as the other side of the equation. Sometimes we have to factor expressions, expand expressions, find common denominators, or use other algebraic strategies to obtain the desired result. In this first section, we will work with the fundamental identities: the Pythagorean identities, the even-odd identities, the reciprocal identities, and the quotient identities.

We will begin with the Pythagorean identities (Table (PageIndex{1})), which are equations involving trigonometric functions based on the properties of a right triangle. We have already seen and used the first of these identifies, but now we will also use additional identities.

The second and third identities can be obtained by manipulating the first. The identity (1+{cot}^2 theta={csc}^2 theta) is found by rewriting the left side of the equation in terms of sine and cosine.

Prove: (1+{cot}^2 theta={csc}^2 theta)

[begin{align*} 1+{cot}^2 theta&= (1+dfrac{{cos}^2}{{sin}^2})qquad text{Rewrite the left side} &= left(dfrac{{sin}^2}{{sin}^2}right)+left (dfrac{{cos}^2}{{sin}^2}right)qquad text{Write both terms with the common denominator} &= dfrac{{sin}^2+{cos}^2}{{sin}^2} &= dfrac{1}{{sin}^2} &= {csc}^2 end{align*}]

Similarly,(1+{tan}^2 theta={sec}^2 theta)can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives

[begin{align*} 1+{tan}^2 theta&= 1+{left(dfrac{sin theta}{cos theta}right )}^2qquad text{Rewrite left side} &= {left (dfrac{cos theta}{cos theta}right )}^2+{left (dfrac{sin theta}{cos theta}right)}^2qquad text{Write both terms with the common denominator} &= dfrac{{cos}^2 theta+{sin}^2 theta}{{cos}^2 theta} &= dfrac{1}{{cos}^2 theta} &= {sec}^2 theta end{align*}]

Recall that we determined which trigonometric functions are odd and which are even. The next set of fundamental identities is the set of even-odd identities. The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle (Table (PageIndex{2})).

Recall that an odd function is one in which (f(−x)= −f(x)) for all (x) in the domain off. f. The sine function is an odd function because (sin(−theta)=−sin theta). The graph of an odd function is symmetric about the origin. For example, consider corresponding inputs of (dfrac{pi}{2}) and (−dfrac{pi}{2}). The output of (sinleft (dfrac{pi}{2}right )) is opposite the output of (sin left (−dfrac{pi}{2}right )). Thus,

[begin{align*} sinleft (dfrac{pi}{2}right)&=1 [4pt] sinleft (-dfrac{pi}{2}right) &=-sinleft (dfrac{pi}{2}right) [4pt] &=-1 end{align*}]

This is shown in Figure (PageIndex{2}).

Recall that an even function is one in which

(f(−x)=f(x)) for all (x) in the domain of (f)

The graph of an even function is symmetric about the y-axis. The cosine function is an even function because (cos(−theta)=cos theta). For example, consider corresponding inputs (dfrac{pi}{4}) and (−dfrac{pi}{4}). The output of (cosleft (dfrac{pi}{4}right)) is the same as the output of (cosleft (−dfrac{pi}{4}right)). Thus,

[begin{align*} cosleft (−dfrac{pi}{4}right ) &=cosleft (dfrac{pi}{4}right) [4pt] &≈0.707 end{align*}]

See Figure (PageIndex{3}).

For all (theta) in the domain of the sine and cosine functions, respectively, we can state the following:

• Since (sin(−theta)=−sin theta),sine is an odd function.
• Since (cos(−theta)=cos theta),cosine is an even function.

The other even-odd identities follow from the even and odd nature of the sine and cosine functions. For example, consider the tangent identity,(tan(−theta)=−tan theta). We can interpret the tangent of a negative angle as

[tan (−theta)=dfrac{sin (−theta)}{cos (−theta)}=dfrac{−sin theta}{cos theta}=−tan theta. nonumber]

Tangent is therefore an odd function, which means that (tan(−theta)=−tan(theta)) for all (theta) in the domain of the tangent function.

The cotangent identity, (cot(−theta)=−cot theta),also follows from the sine and cosine identities. We can interpret the cotangent of a negative angle as

[cot(−theta)=dfrac{cos(−theta)}{sin(−theta)}=dfrac{cos theta}{−sin theta}=−cot theta.nonumber]

Cotangent is therefore an odd function, which means that (cot(−theta)=−cot(theta)) for all (theta) in the domain of the cotangent function.

The cosecant function is the reciprocal of the sine function, which means that the cosecant of a negative angle will be interpreted as

[csc(−theta)=dfrac{1}{sin(−theta)}=dfrac{1}{−sin theta}=−csc theta. nonumber]

The cosecant function is therefore odd.

Finally, the secant function is the reciprocal of the cosine function, and the secant of a negative angle is interpreted as

[sec(−theta)=dfrac{1}{cos(−theta)}=dfrac{1}{cos theta}=sec theta. nonumber]

The secant function is therefore even.

To sum up, only two of the trigonometric functions, cosine and secant, are even. The other four functions are odd, verifying the even-odd identities.

The next set of fundamental identities is the set of reciprocal identities, which, as their name implies, relate trigonometric functions that are reciprocals of each other. (Table (PageIndex{3})). Recall that we first encountered these identities when defining trigonometric functions from right angles in Right Angle Trigonometry.

The final set of identities is the set of quotient identities, which define relationships among certain trigonometric functions and can be very helpful in verifying other identities (Table (PageIndex{4})).

The reciprocal and quotient identities are derived from the definitions of the basic trigonometric functions.

SUMMARIZING TRIGONOMETRIC IDENTITIES

The Pythagorean identities are based on the properties of a right triangle.

[{cos}^2 theta+{sin}^2 theta=1]

[1+{cot}^2 theta={csc}^2 theta]

[1+{tan}^2 theta={sec}^2 theta]

The even-odd identities relate the value of a trigonometric function at a given angle to the value of the function at the opposite angle.

[tan(−theta)=−tan theta]

Trigonometric Identititiesmath Problem Solving Problems

[cot(−theta)=−cot theta]

[sin(−theta)=−sin theta]

[csc(−theta)=−csc theta]

[cos(−theta)=cos theta]

[sec(−theta)=sec theta]

The reciprocal identities define reciprocals of the trigonometric functions.

[sin theta=dfrac{1}{csc theta}]

[cos theta=dfrac{1}{sec theta}]

[tan theta=dfrac{1}{cot theta}]

[csc theta=dfrac{1}{sin theta}]

[sec theta=dfrac{1}{cos theta}]

[cot theta=dfrac{1}{tan theta}]

The quotient identities define the relationship among the trigonometric functions.

[tan theta=dfrac{sin theta}{cos theta}]

[cot theta=dfrac{cos theta}{sin theta}]

Example (PageIndex{1}): Graphing the Equations of an Identity

Graph both sides of the identity (cot theta=dfrac{1}{tan theta}). In other words, on the graphing calculator, graph (y=cot theta) and (y=dfrac{1}{tan theta}).

Solution

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Solving Trigonometric Equations With Identities

See Figure (PageIndex{4}).

Analysis

We see only one graph because both expressions generate the same image. One is on top of the other. This is a good way to prove any identity. If both expressions give the same graph, then they must be identities.

How to: Given a trigonometric identity, verify that it is true.

1. Work on one side of the equation. It is usually better to start with the more complex side, as it is easier to simplify than to build.
2. Look for opportunities to factor expressions, square a binomial, or add fractions.
3. Noting which functions are in the final expression, look for opportunities to use the identities and make the proper substitutions.
4. If these steps do not yield the desired result, try converting all terms to sines and cosines.

Example (PageIndex{2}): Verifying a Trigonometric Identity

Verify (tan theta cos theta=sin theta).

Solution

We will start on the left side, as it is the more complicated side:

[ begin{align*} tan theta cos theta &=left(dfrac{sin theta}{cos theta}right)cos theta [4pt] &=sin theta end{align*}]

Analysis

This identity was fairly simple to verify, as it only required writing (tan theta) in terms of (sin theta) and (cos theta).

Exercise (PageIndex{1})

Verify the identity (csc theta cos theta tan theta=1).

[ begin{align*} csc theta cos theta tan theta=left(dfrac{1}{sin theta}right)cos thetaleft(dfrac{sin theta}{cos theta}right) [4pt] & =dfrac{cos theta}{sin theta}(dfrac{sin theta}{cos theta}) [4pt] & =dfrac{sin theta cos theta}{sin theta cos theta} [4pt] &=1 end{align*}]

Example (PageIndex{3A}): Verifying the Equivalency Using the Even-Odd Identities

Verify the following equivalency using the even-odd identities:

((1+sin x)[1+sin(−x)]={cos}^2 x)

Solution

Working on the left side of the equation, we have

( (1+sin x)[1+sin(−x)]=(1+sin x)(1-sin x))

Since

[begin{align*} sin(-x)&= -sin x [5pt] &=1-{sin}^2 xqquad text{Difference of squares} [5pt] &={cos}^2 x {cos}^2 x&= 1-{sin}^2 x end{align*}]

Example (PageIndex{3B}): Verifying a Trigonometric Identity Involving ({sec}^2 theta)

Verify the identity (dfrac{{sec}^2 theta−1}{{sec}^2 theta}={sin}^2 theta)

Solution

As the left side is more complicated, let’s begin there.

[begin{align*}
dfrac{{sec}^2 theta-1}{{sec}^2 theta}&= dfrac{({tan}^2 theta +1)-1}{{sec}^2 theta}
{sec}^2 theta&= {tan}^2 theta +1
&= dfrac{{tan}^2 theta}{{sec}^2 theta}
&= {tan}^2 thetaleft (dfrac{1}{{sec}^2 theta}right )
&= {tan}^2 theta left ({cos}^2 thetaright )
{cos}^2 theta&= dfrac{1}{{sec}^2 theta}
&= left (dfrac{{sin}^2 theta}{{cos}^2 theta}right )
{tan}^2 theta&= dfrac{{sin}^2 theta}{{cos}^2 theta}
&= {sin}^2 theta
end{align*}]

There is more than one way to verify an identity. Here is another possibility. Again, we can start with the left side.

[begin{align*} dfrac{{sec}^2 theta-1}{{sec}^2 theta}&= dfrac{{sec}^2 theta}{{sec}^2 theta}-dfrac{1}{{sec}^2 theta} &= 1-{cos}^2 theta &= {sin}^2 theta end{align*}]

Analysis

In the first method, we used the identity ({sec}^2 theta={tan}^2 theta+1) and continued to simplify. In the second method, we split the fraction, putting both terms in the numerator over the common denominator. This problem illustrates that there are multiple ways we can verify an identity. Employing some creativity can sometimes simplify a procedure. As long as the substitutions are correct, the answer will be the same.

Exercise (PageIndex{2})

Show that (dfrac{cot theta}{csc theta}=cos theta).

[begin{align*} dfrac{cot theta}{csc theta}&= dfrac{tfrac{cos theta}{sin theta}}{dfrac{1}{sin theta}} &= dfrac{cos theta}{sin theta}cdot dfrac{sin theta}{1} &= cos theta end{align*}]

Example (PageIndex{4}): Creating and Verifying an Identity

Create an identity for the expression (2 tan theta sec theta) by rewriting strictly in terms of sine.

Solution

There are a number of ways to begin, but here we will use the quotient and reciprocal identities to rewrite the expression:

[begin{align*} 2 tan theta sec theta&= 2left (dfrac{sin theta}{cos theta}right )left(dfrac{1}{cos theta}right ) &= dfrac{2sin theta}{{cos}^2 theta} &= dfrac{2sin theta}{1-{sin}^2 theta}qquad text{Substitute } 1-{sin}^2 theta text{ for } {cos}^2 theta end{align*}]

Thus,

(2 tan theta sec theta=dfrac{2 sin theta}{1−{sin}^2 theta})

Example (PageIndex{5}): Verifying an Identity Using Algebra and Even/Odd Identities

Verify the identity:

(dfrac{{sin}^2(−theta)−{cos}^2(−theta)}{sin(−theta)−cos(−theta)}=cos theta−sin theta)

Solution

Let’s start with the left side and simplify:

[begin{align*} dfrac{{sin}^2(-theta)-{cos}^2(-theta)}{sin(-theta)-cos(-theta)}&= dfrac{{[sin(-theta)]}^2-{[cos(-theta)]}^2}{sin(-theta)-cos(-theta)} &= dfrac{{(-sin theta)}^2-{(cos theta)}^2}{-sin theta -cos theta} ;; ; , sin(-x) = -sinspace xtext { and } cos(-x)=cos space x &= dfrac{{(sin theta)}^2-{(cos theta)}^2}{-sin theta -cos theta}qquad text{Difference of squares} &= dfrac{(sin theta-cos theta)(sin theta+cos theta)}{-(sin theta+cos theta)} &= cos theta-sin theta end{align*}]

Exercise (PageIndex{3})

Verify the identity (dfrac{{sin}^2 theta−1}{tan theta sin theta−tan theta}=dfrac{sin theta+1}{tan theta}).

[begin{align*} dfrac{{sin}^2 theta-1}{tan theta sin theta-tan theta}&= dfrac{(sin theta +1)(sin theta -1)}{tan theta(sin theta -1)} &= dfrac{sin theta+1}{tan theta} end{align*}]

Example (PageIndex{6}): Verifying an Identity Involving Cosines and Cotangents

Verify the identity: ((1−{cos}^2 x)(1+{cot}^2 x)=1).

Solution

[begin{align*} (1-{cos}^2 x)(1+{cot}^2 x)&= (1-{cos}^2 x)left(1+dfrac{{cos}^2 x}{{sin}^2 x}right) &= (1-{cos}^2 x)left(dfrac{{sin}^2 x}{{sin}^2 x}+dfrac{{cos}^2 x}{{sin}^2 x}right )qquad text{Find the common denominator} &= (1-{cos}^2 x)left(dfrac{{sin}^2 x +{cos}^2 x}{{sin}^2 x}right) &= ({sin}^2 x)left (dfrac{1}{{sin}^2 x}right ) &= 1 end{align*}]

Using Algebra to Simplify Trigonometric Expressions

We have seen that algebra is very important in verifying trigonometric identities, but it is just as critical in simplifying trigonometric expressions before solving. Being familiar with the basic properties and formulas of algebra, such as the difference of squares formula, the perfect square formula, or substitution, will simplify the work involved with trigonometric expressions and equations.

For example, the equation ((sin x+1)(sin x−1)=0) resembles the equation ((x+1)(x−1)=0), which uses the factored form of the difference of squares. Using algebra makes finding a solution straightforward and familiar. We can set each factor equal to zero and solve. This is one example of recognizing algebraic patterns in trigonometric expressions or equations.

Another example is the difference of squares formula, (a^2−b^2=(a−b)(a+b)), which is widely used in many areas other than mathematics, such as engineering, architecture, and physics. We can also create our own identities by continually expanding an expression and making the appropriate substitutions. Using algebraic properties and formulas makes many trigonometric equations easier to understand and solve.

Example (PageIndex{7A}): Writing the Trigonometric Expression as an Algebraic Expression

Write the following trigonometric expression as an algebraic expression: (2{cos}^2 theta+cos theta−1).

Solution

Notice that the pattern displayed has the same form as a standard quadratic expression,(ax^2+bx+c). Letting (cos theta=x),we can rewrite the expression as follows:

(2x^2+x−1)

This expression can be factored as ((2x+1)(x−1)). If it were set equal to zero and we wanted to solve the equation, we would use the zero factor property and solve each factor for (x). At this point, we would replace (x) with (cos theta) and solve for (theta).

Example (PageIndex{7B}): Rewriting a Trigonometric Expression Using the Difference of Squares

Rewrite the trigonometric expression using the difference of squares: (4{cos}^2 theta−1).

Solution

Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the difference of squares.

[begin{align*} 4{cos}^2 theta-1&= {(2cos theta)}^2-1 &= (2cos theta-1)(2cos theta+1) end{align*}]

Analysis

If this expression were written in the form of an equation set equal to zero, we could solve each factor using the zero factor property. We could also use substitution like we did in the previous problem and let (cos theta=x), rewrite the expression as (4x^2−1), and factor ((2x−1)(2x+1)). Then replace (x) with (cos theta) and solve for the angle.

Exercise (PageIndex{4})

Rewrite the trigonometric expression using the difference of squares: (25−9{sin}^2 theta).

This is a difference of squares formula: (25−9{sin}^2 theta=(5−3sin theta)(5+3sin theta)).

Example (PageIndex{8}): Simplify by Rewriting and Using Substitution

Simplify the expression by rewriting and using identities:

({csc}^2 theta−{cot}^2 theta)

Solution

We can start with the Pythagorean identity.

[begin{align*} 1+{cot}^2 theta&= {csc}^2 theta text{Now we can simplify by substituting } 1+{cot}^2 theta text{ for } {csc}^2 theta {csc}^2 theta-{cot}^2 theta&= 1+{cot}^2 theta-{cot}^2 theta &= 1 end{align*}]

Exercise (PageIndex{5})

Use algebraic techniques to verify the identity: (dfrac{cos theta}{1+sin theta}=dfrac{1−sin theta}{cos theta}).

(Hint: Multiply the numerator and denominator on the left side by (1−sin theta).)

[begin{align*} dfrac{cos theta}{1+sin theta}left(dfrac{1-sin theta}{1-sin theta}right)&= dfrac{cos theta (1-sin theta)}{1-{sin}^2 theta} &= dfrac{cos theta (1-sin theta)}{{cos}^2 theta} &= dfrac{1-sin theta}{cos theta} end{align*}]

Media

Access these online resources for additional instruction and practice with the fundamental trigonometric identities.

• Fundamental Trigonometric Identities
• Verifying Trigonometric Identities

Key Equations

Key Concepts

• There are multiple ways to represent a trigonometric expression. Verifying the identities illustrates how expressions can be rewritten to simplify a problem.
• Graphing both sides of an identity will verify it. See Example (PageIndex{1}).
• Simplifying one side of the equation to equal the other side is another method for verifying an identity. See Example (PageIndex{2}) and Example (PageIndex{3}).
• The approach to verifying an identity depends on the nature of the identity. It is often useful to begin on the more complex side of the equation. See Example (PageIndex{4}).
• We can create an identity and then verify it. See Example (PageIndex{5}).
• Verifying an identity may involve algebra with the fundamental identities. See Example (PageIndex{6}) and Example (PageIndex{7}).
• Algebraic techniques can be used to simplify trigonometric expressions. We use algebraic techniques throughout this text, as they consist of the fundamental rules of mathematics. See Example (PageIndex{8}), Example (PageIndex{9}), and Example (PageIndex{10}).

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.

Section 1-4 : Solving Trig Equations

Without using a calculator find the solution(s) to the following equations. If an interval is given find only those solutions that are in the interval. If no interval is given find all solutions to the equation.

1. (4sin left( {3t} right) = 2) Solution
2. (4sin left( {3t} right) = 2) in (displaystyle left[ {0,frac{{4pi }}{3}} right]) Solution
3. (displaystyle 2cos left( {frac{x}{3}} right) + sqrt 2 = 0) Solution
4. (displaystyle 2cos left( {frac{x}{3}} right) + sqrt 2 = 0) in (left[ { - 7pi ,7pi } right]) Solution
5. (4cos left( {6z} right) = sqrt {12} ) in (displaystyle left[ {0,frac{pi }{2}} right]) Solution
6. (displaystyle 2sin left( {frac{{3y}}{2}} right) + sqrt 3 = 0) in (displaystyle left[ { - frac{{7pi }}{3},0} right]) Solution
7. (8tan left( {2x} right) - 5 = 3) in (displaystyle left[ { - frac{pi }{2},frac{{3pi }}{2}} right]) Solution
8. (16 = - 9sin left( {7x} right) - 4) in (displaystyle left[ { - 2pi ,frac{{9pi }}{4}} right]) Solution
9. (displaystyle sqrt 3 tan left( {frac{t}{4}} right) + 5 = 4) in (left[ {0,4pi } right]) Solution
10. (sqrt 3 csc left( {9z} right) - 7 = - 5) in (displaystyle left[ { - frac{pi }{3},frac{{4pi }}{9}} right]) Solution
11. (displaystyle 1 - 14cos left( {frac{{2x}}{5}} right) = - 6) in (displaystyle left[ {5pi ,frac{{40pi }}{3}} right]) Solution
12. (displaystyle 15 = 17 + 4cos left( {frac{y}{7}} right)) in (left[ {10pi ,15pi } right]) Solution