3.12 Equivalent representations of trigonometric functions

The Pythagorean theorem concerning right triangles can be expressed by means of trigonometric ratios.

The following Pythagorean identities can also be used to find trigonometric function values given certain information:\begin{aligned} \left(\sin \theta \right)^2 + \left(\cos \theta \right) ^2 = 1 \quad \text{ or } \quad \sin^2 \theta + \cos^2 \theta = 1 \\\ \left(\tan \theta \right) ^ 2 + 1 = \left(\sec \theta \right) ^ 2 \quad \text{ or } \quad \tan ^2 \theta + 1 = \sec^2 \theta \\\ 1 + \left( \cot \theta \right) ^ 2 = \left( \csc \theta \right) ^2 \quad \text{ or } \quad 1 + \cot^2 \theta = \csc^2 \theta \end{aligned}

Examples

Example 1

Given the following diagram, prove the Pythagorean identity \sin ^2 \theta + \cos ^2 \theta = 1.

Worked Solution

Create a strategy

Use the Pythagorean theorem with the given triangle to begin the proof. Then, make connections to the sine and cosine of the reference angle to get to the Pythagorean identity.

Apply the idea

\displaystyle a^2 + b^2	\displaystyle =	\displaystyle c^2	Pythagorean theorem
\displaystyle \dfrac{a^2}{c^2} + \dfrac{b^2}{c^2}	\displaystyle =	\displaystyle 1	Divide both sides of the equation by c^2
\displaystyle \left( \dfrac{a}{c} \right) ^2 + \left( \dfrac{b}{c} \right) ^2	\displaystyle =	\displaystyle 1	Apply the quotient to a power property of exponents
\displaystyle \cos \theta= \dfrac{a}{c} \text{ and } \sin \theta	\displaystyle =	\displaystyle \dfrac{b}{c}	Evaluate the sine and cosine of the reference angle
\displaystyle \left(\cos \theta \right) ^2 + \left( \sin \theta \right)^2	\displaystyle =	\displaystyle 1	Substitution
\displaystyle \sin ^2 \theta + \cos ^2 \theta	\displaystyle =	\displaystyle 1	Commutative property of addition

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Example 2

Use \sin^2 \theta + \cos^2 \theta = 1 to prove the Pythagorean identity 1 + \cot^2 \theta = \csc^2 \theta.

Worked Solution

Create a strategy

Use reciprocal identities and the cotangent identity to rewrite the expressions in terms of sine and cosine.

Apply the idea

\displaystyle \sin^2 \theta + \cos^2 \theta	\displaystyle =	\displaystyle 1	Pythagorean identity
\displaystyle \dfrac{\sin ^2 \theta}{\sin^2 \theta} + \dfrac{\cos^2 \theta}{\sin^2 \theta}	\displaystyle =	\displaystyle \dfrac{1}{\sin ^2 \theta}	Divide both sides of the equation by \sin ^2 \theta
\displaystyle 1 + \cot ^ 2 \theta	\displaystyle =	\displaystyle \csc^ 2 \theta	Multiplicative inverse property and reciprocal identities

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Example 3

Find the acute value of x, in radians, for the equation \sin^{2} x + \cos^{2} x + \cos x = \dfrac{\sqrt{3}}{2}+1.

Worked Solution

Create a strategy

Use the identity \sin^{2} x + \cos^{2} x = 1 to simplify the equation so that only a single variable term remains, then solve the equation as normal.

Apply the idea

\displaystyle \sin^{2} x + \cos^{2} x + \cos x	\displaystyle =	\displaystyle \dfrac{\sqrt{3}}{2}+1	Write the equation
\displaystyle (\sin^{2} x + \cos^{2} x) + \cos x	\displaystyle =	\displaystyle \dfrac{\sqrt{3}}{2}+1	Group the terms that form one side of the identity
\displaystyle 1 + \cos x	\displaystyle =	\displaystyle \dfrac{\sqrt{3}}{2}+1	Replace the grouped terms with 1
\displaystyle \cos x	\displaystyle =	\displaystyle \dfrac{\sqrt{3}}{2}	Subtract 1 from both sides
\displaystyle x	\displaystyle =	\displaystyle \cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)	Take the inverse cosine of both sides
\displaystyle x	\displaystyle =	\displaystyle \dfrac{\pi}{6}	Use the unit circle to find the exact value

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The sum and difference identities in trigonometry are formulas that give relationships between the sine, cosine, and tangent of sums and differences of angles. These identities are useful when simplifying complex trigonometric expressions, solving trigonometric equations, and proving other trigonometric identities.

Sum and difference identities for sine:

1. \sin (\alpha + \beta)= \sin \alpha \cos \beta + \cos \alpha \sin \beta
2. \sin (\alpha - \beta)= \sin \alpha \cos \beta - \cos \alpha \sin \beta

Sum and difference identities for cosine:

1. \cos (\alpha + \beta)= \cos \alpha \cos \beta - \sin \alpha \sin \beta
2. \cos (\alpha - \beta)= \cos \alpha \cos \beta + \sin \alpha \sin \beta

Sum and difference identities for tangent:

1. \tan (\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}
2. \tan (\alpha - \beta) = \dfrac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}

Double-angle identities are specific trigonometric identities that help us express trigonometric functions of double angles in terms of single angles. They are derived from the sum identities of trigonometric functions.

By letting \beta = 2\alpha we obtain three corresponding half-angle identities. These identities are sometimes referred to in alternate forms which have been manipulated by applying the Pythagorean identities. The relationships remain the same

Examples

Recall that we can rewrite trig expressions by factoring them, or using substitution and different polynomial identities. We can apply these techniques and the zero product property to solve trig equations that mimic polynomial equations.

The following polynomial identities also apply to trigonometric expressions of the same form:

Recall the basic trigonometric identities:

\begin{aligned} \sin \theta = \dfrac{1}{\csc \theta} \qquad \cos \theta = \dfrac{1}{\sec \theta} \qquad \tan \theta = \dfrac{1}{\cot \theta} \qquad \tan \theta = \dfrac{ \sin \theta}{\cos \theta} \\\\ \csc \theta= \dfrac{1}{\sin \theta} \qquad \sec \theta = \dfrac{1}{\cos \theta} \qquad \cot \theta = \dfrac{1}{\tan \theta} \qquad \cot \theta = \dfrac{\cos \theta}{\sin \theta} \end{aligned}

Cofunction identities are relationships between functions of complementary angles. These relationships allow us to express one trigonometric function in terms of another function of the complement of the angle. Here are the basic cofunction identities for sine, cosine, tangent, cotangent, secant, and cosecant:

\begin{aligned} \sin \alpha = \cos(90\degree - \alpha) \qquad \tan \alpha = \cot(90\degree - \alpha) \qquad \sec \alpha = \csc(90\degree - \alpha) \\ \qquad \cos \alpha = \sin(90\degree - \alpha) \qquad \cot \alpha = \tan (90\degree - \alpha) \qquad \csc \alpha = \sec(90\degree - \alpha) \end{aligned}

The key when solving problems involving these identities is to figure out which to use and when.

Examples

Example 6

Solve \cot x\cos ^2x=2\cot x.

Worked Solution

Create a strategy

We can substitute the \cot x=A and \cos x=B. This gives us the following equation: AB^2=2A.

Apply the idea

\displaystyle \cot x\cos ^2x	\displaystyle =	\displaystyle 2\cot x	Write the equation
\displaystyle AB^2	\displaystyle =	\displaystyle 2A	Let \cot x=A and \cos x=B
\displaystyle AB^2-2A	\displaystyle =	\displaystyle 0	Subtract 2A from both sides
\displaystyle A\left(B^2-2\right)	\displaystyle =	\displaystyle 0	Factor out the greatest common factor of A
\displaystyle \cot x(\cos ^2x-2)	\displaystyle =	\displaystyle 0	Substitute the trigonometric functions

Set each factor equal to zero using the zero product property.

\displaystyle \cot x	\displaystyle =	\displaystyle 0
\displaystyle \dfrac{1}{\tan {x}}	\displaystyle =	\displaystyle 0
\displaystyle \tan x	\displaystyle =	\displaystyle \dfrac{1}{0}
\displaystyle x	\displaystyle =	\displaystyle \tan^{-1}\dfrac{1}{0}

\displaystyle \cos ^2x-2	\displaystyle =	\displaystyle 0
\displaystyle \cos ^2x	\displaystyle =	\displaystyle 2
\displaystyle \cos x	\displaystyle =	\displaystyle \pm \sqrt{2}
\displaystyle x	\displaystyle =	\displaystyle \cos ^{-1}\pm \sqrt{2}

Using the unit circle, we know the angle \dfrac{\pi }{2} radians will produce \tan \dfrac{\pi }{2}=\dfrac{1}{0}. There is no angle on the unit circle that will produce \cos x=\pm \sqrt{2}. Using the graphing calculator, we discover that there is no solution for\cos x=\pm \sqrt{2} because \pm \sqrt{2} is outside the range of the \cos function.

So, the answer is x=\dfrac{\pi }{2} radians.

Reflect and check

Using the graphing calculator to find the period, we determine that the period of \cot x\cos ^2x=2\cot x is \pi. Remember, a period of a function is the interval of x-values where the graph repeated in both directions along the x-axis. The general form of the solution is found by adding multiples of \pi to x=\dfrac{\pi }{2}.

x=\dfrac{\pi }{2}+k\pi where k is an integer.

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Example 7

Consider \sin \theta = -\dfrac{2}{3}, where 0 < \theta < 2 \pi.

a

State any quadrant angle \theta lies in.

Worked Solution

Create a strategy

Since \sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}, and the hypotenuse always represents positive value of length, -2 represents the side length opposite to \theta or the y-coordinate and 3 represents the hypotenuse of the triangle on the coordinate plane. Since -2 is negative, the angle must be in any quadrant where the y-coordinate is negative.

Apply the idea

Quadrants 3 or 4.

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b

Find \cos \theta when \pi < \theta < \dfrac{3 \pi}{2}.

Worked Solution

Create a strategy

Note that \theta is restricted to the third quadrant. Use the Pythagorean identity to solve for \cos \theta.

Apply the idea

\displaystyle \sin^{2} \theta + \cos^{2} \theta	\displaystyle =	\displaystyle 1	Pythagorean identity
\displaystyle \left(-\dfrac{2}{3}\right)^{2} + \cos^{2} \theta	\displaystyle =	\displaystyle 1	Substitute \sin \theta = -\dfrac{2}{3}
\displaystyle \dfrac{4}{9} + \cos^{2} \theta	\displaystyle =	\displaystyle 1	Evaluate the square
\displaystyle \cos ^2 \theta	\displaystyle =	\displaystyle 1 - \dfrac{4}{9}	Subtract \dfrac{4}{9} from both sides
\displaystyle \cos^{2} \theta	\displaystyle =	\displaystyle \dfrac{5}{9}	Evaluate the subtraction
\displaystyle \cos \theta	\displaystyle =	\displaystyle \pm \dfrac{\sqrt{5}}{3}	Take the square root of both sides

Since cosine is negative in the third quadrant, we have \cos \theta= - \dfrac{\sqrt{5}}{3}.

Reflect and check

We could use the Pythagorean theorem to draw a right triangle and find its missing side, then use the domain of the cosine function to determine the sign of the ratio.

\displaystyle a^2 + b^2	\displaystyle =	\displaystyle c^2	Pythagorean theorem
\displaystyle a^2+2^2	\displaystyle =	\displaystyle 3^2	Substitution
\displaystyle a^2+4	\displaystyle =	\displaystyle 9	Evaluate the exponents
\displaystyle a^2	\displaystyle =	\displaystyle 5	Subtract 4 from both sides of the equation
\displaystyle a	\displaystyle =	\displaystyle \sqrt{5}	Evaluate the square root of both sides of the equation

\cos \theta = \dfrac{\sqrt{5}}{3}, and since cosine is negative on the coordinate plane when 0 < \theta < \dfrac{3 \pi}{2}, \cos \theta= - \dfrac{\sqrt{5}}{3}.

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c

Find the value of \tan \theta.

Worked Solution

Create a strategy

Assuming that \theta is in the third quadrant, we can use the tangent identity to determine \tan \theta.

Apply the idea

\tan \theta = \dfrac{\sin \theta}{\cos \theta} = \dfrac{ \dfrac{-2}{3}}{ -\dfrac{\sqrt{5}}{3}} = \dfrac{-2}{3} \cdot - \dfrac{3}{\sqrt{5}}= \dfrac{2}{\sqrt{5}}

Reflect and check

We could also use the side lengths of the right triangle drawn in part (b) to determine \tan \theta in the third quadrant.

\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{2}{\sqrt{5}}. In the third quadrant, \tan \theta is positive, since both sine and cosine are negative and \tan \theta = \dfrac{\sin \theta}{\cos \theta}.

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Example 8

Consider the equation \left(\sin \theta + \dfrac{\sqrt{3}}{2}\right) \left(\cos \theta + \dfrac{\sqrt{3}}{2}\right) = 0 for 0 \degree \lt \theta \lt 90 \degree.

How many solutions for \theta does the equation have?

A

Zero

B

One

C

Two

D

Infinitely many

Worked Solution

Create a strategy

Remember that if the product of two expressions is zero, then at least one of the two expressions is zero.

Apply the idea

For values of \theta in the region 0 \degree \lt \theta \lt 90 \degree, the values of \sin \theta and \cos \theta are both positive.

If we rearrange the equations, we have that:

• \sin \theta = -\dfrac{\sqrt{3}}{2}
• \cos \theta = -\dfrac{\sqrt{3}}{2}

There is no solution.

So, the correct answer is option A.

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Example 9

Find one possible solution, in degrees, for the equation \sec \alpha = \csc\left(\dfrac{\alpha}{2} + 20 \degree\right).

Worked Solution

Create a strategy

We can use the identity \sec \alpha = \csc(90\degree - \alpha) to rewrite the equation in terms of only one trigonometric function.

Apply the idea

\displaystyle \sec \alpha	\displaystyle =	\displaystyle \csc\left(\dfrac{\alpha}{2} + 20 \degree\right)	Write the equation
\displaystyle \csc(90\degree - \alpha)	\displaystyle =	\displaystyle \csc\left(\dfrac{\alpha}{2} + 20 \degree\right)	Use the identity \sec \alpha = \csc(90\degree - \alpha)
\displaystyle (90\degree - \alpha)	\displaystyle =	\displaystyle \left(\dfrac{\alpha}{2} + 20 \degree\right)	Set the angles equal
\displaystyle -\dfrac{3\alpha}{2}	\displaystyle =	\displaystyle -70\degree	Subtract 90\degree and \dfrac{\alpha}{2} from both sides
\displaystyle \alpha	\displaystyle =	\displaystyle \left(\dfrac{140}{3}\right)\degree	Multiply both sides by -\dfrac{2}{3}

Reflect and check

Notice that after the third step (setting the angles to be equal) the result is a linear equation in \alpha which has only one solution.

Since the previous step has two cosecant functions equated, there are a lot more possible solutions. We could find these other solutions by considering the other quadrants and the periodic nature of the cosecant function.

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