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Trigonometric Identities

Trigonometric identities are expressions and equations that allow us to express trigonometric functions in many different ways. By using these identities, we can simplify complex trigonometric expressions, solve equations, and prove relationships between functions.

Pythagorean Identities​

The pythagorean identities are equations involving trigonometric functions based on the properties of a right triangle.

We can derive the first identity from the unit circle which can be defined as x2+y2=1x^2 + y^2 = 1 because it is a circle with a radius of 11. The coordinates of a point on this unit circle can also be expressed as x=cos⁡θx = \cos \theta and y=sin⁡θy = \sin \theta. So, by substituting these values into the equation of the unit circle, we get the first pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1.

The second identity 1+cot⁡2θ=csc⁡2θ1 + \cot^2 \theta = \csc^2 \theta can be derived by manipulating the first identity and we can verify the identity by rewriting 1+cot⁡2θ1 + \cot^2 \theta as csc⁡2θ\csc^2 \theta...

1+cot⁡2θ=1+(cos⁡θsin⁡θ)2=(sin⁡θsin⁡θ)2+(cos⁡θsin⁡θ)2=sin⁡2θsin⁡2θ+cos⁡2θsin⁡2θ=cos⁡2θ+sin⁡2θsin⁡2θ=1sin⁡2θ=csc⁡2θ\begin{array}{ccccc} 1 + \cot^2 \theta &=& 1 + (\dfrac{\cos \theta}{\sin \theta})^2 \\[1em] &=& (\dfrac{\sin \theta}{\sin \theta})^2 + (\dfrac{\cos \theta}{\sin \theta})^2 \\[1em] &=& \dfrac{\sin^2 \theta}{\sin^2 \theta} + \dfrac{\cos^2 \theta}{\sin^2 \theta} \\[1em] &=& \dfrac{\cos^2 \theta + \sin^2 \theta}{\sin^2 \theta} \\[1em] &=& \dfrac{1}{\sin^2 \theta} \\[1em] &=& \csc^2 \theta \\[1em] \end{array}

The third and final pythagorean identity is 1+tan⁡2θ=sec⁡2θ1 + \tan^2 \theta = \sec^2 \theta which can also be derived by manipulating the first identity and we can verify the identity by rewriting 1+tan⁡2θ1 + \tan^2 \theta as sec⁡2θ\sec^2 \theta...

1+tan⁡2θ=1+(sin⁡θcos⁡θ)2=(cos⁡θcos⁡θ)2+(sin⁡θcos⁡θ)2=cos⁡2θcos⁡2θ+sin⁡2θcos⁡2θ=sin⁡2θ+cos⁡2θcos⁡2θ=1cos⁡2θ=sec⁡2θ\begin{array}{ccccc} 1 + \tan^2 \theta &=& 1 + (\dfrac{\sin \theta}{\cos \theta})^2 \\[1em] &=& (\dfrac{\cos \theta}{\cos \theta})^2 + (\dfrac{\sin \theta}{\cos \theta})^2 \\[1em] &=& \dfrac{\cos^2 \theta}{\cos^2 \theta} + \dfrac{\sin^2 \theta}{\cos^2 \theta} \\[1em] &=& \dfrac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta} \\[1em] &=& \dfrac{1}{\cos^2 \theta} \\[1em] &=& \sec^2 \theta \\[1em] \end{array}

This gives us all three forms of the pythagorean identity...

sin⁡2θ+cos⁡2θ=11+cot⁡2θ=csc⁡2θ1+tan⁡2θ=sec⁡2θ\begin{array}{cccccc} \sin^2 \theta + \cos^2 \theta &=& 1 \\[1em] 1 + \cot^2 \theta &=& \csc^2 \theta \\[1em] 1 + \tan^2 \theta &=& \sec^2 \theta \\[1em] \end{array}

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. Except for y=0y = 0, every other function in mathematics can be categorized as odd, even, or neither.

An odd function is one in which f(−x)=−f(x)f(-x) = -f(x) for all xx in the domain of ff causing the graph of the function to be symmetric about the origin. An example of an odd function is sin⁡θ\sin \theta because sin⁡(−θ)=−sin⁡θ\sin (-\theta) = -\sin \theta.

On the other hand, an even function is one in which f(−x)=f(x)f(-x) = f(x) for all xx in the domain of ff causing the graph of the function to be symmetric about the yy-axis. An example of an even function is cos⁡θ\cos \theta because cos⁡(−θ)=cos⁡θ\cos (-\theta) = \cos \theta.

Given the fact that sin⁡θ\sin \theta is an odd function and cos⁡θ\cos \theta is an even function, we can derive the following even-odd identities...

tan⁡(−θ)=sin⁡(−θ)cos⁡(−θ)=−sin⁡θcos⁡θ=−tan⁡θ(ODD)cot⁡(−θ)=cos⁡(−θ)sin⁡(−θ)=cos⁡θ−sin⁡θ=−cot⁡θ(ODD)csc⁡(−θ)=1sin⁡(−θ)=1−sin⁡θ=−csc⁡θ(ODD)sec⁡(−θ)=1cos⁡(−θ)=1cos⁡θ=sec⁡θ(EVEN)\begin{array}{ccccccccc} \tan(-\theta) &=& \dfrac{\sin (-\theta)}{\cos (-\theta)} &=& \dfrac{-\sin \theta}{\cos \theta} &=& -\tan \theta & \text{(ODD)} \\[1em] \cot(-\theta) &=& \dfrac{\cos (-\theta)}{\sin (-\theta)} &=& \dfrac{\cos \theta}{-\sin \theta} &=& -\cot \theta & \text{(ODD)} \\[1em] \csc(-\theta) &=& \dfrac{1}{\sin (-\theta)} &=& \dfrac{1}{-\sin \theta} &=& -\csc \theta & \text{(ODD)} \\[1em] \sec(-\theta) &=& \dfrac{1}{\cos (-\theta)} &=& \dfrac{1}{\cos \theta} &=& \sec \theta & \text{(EVEN)} \\[1em] \end{array}

Putting everything together gives us the following even-odd identities...

sin⁡(−θ)=−sin⁡θcos⁡(−θ)=cos⁡θcsc⁡(−θ)=−csc⁡θsec⁡(−θ)=sec⁡θcot⁡(−θ)=−cot⁡θtan⁡(−θ)=−tan⁡θ\begin{array}{ccccccccc} \sin(-\theta) &=& -\sin \theta &&& \cos(-\theta) &=& \cos \theta \\[1em] \csc(-\theta) &=& -\csc \theta &&& \sec(-\theta) &=& \sec \theta \\[1em] \cot(-\theta) &=& -\cot \theta &&& \tan(-\theta) &=& -\tan \theta \\[1em] \end{array}

...where sin⁡θ\sin \theta, tan⁡θ\tan \theta, csc⁡θ\csc \theta, and cot⁡θ\cot \theta are odd functions while cos⁡θ\cos \theta and sec⁡θ\sec \theta are even functions.

Reciprocal Identities​

Reciprocal identities are identities that relate the trigonometric functions that are reciprocals of each other. We have already encounted these identities when defining trigonometric functions from right angles and they are as follows...

sin⁡θ=1csc⁡θcsc⁡θ=1sin⁡θcos⁡θ=1sec⁡θsec⁡θ=1cos⁡θtan⁡θ=1cot⁡θcot⁡θ=1tan⁡θ\begin{array}{ccccccccc} \sin \theta &=& \dfrac{1}{\csc \theta} &&& \csc \theta &=& \dfrac{1}{\sin \theta} \\[1em] \cos \theta &=& \dfrac{1}{\sec \theta} &&& \sec \theta &=& \dfrac{1}{\cos \theta} \\[1em] \tan \theta &=& \dfrac{1}{\cot \theta} &&& \cot \theta &=& \dfrac{1}{\tan \theta} \\[1em] \end{array}

Quotient Identities​

Quotient identities are identities that a set of quotient identities which relate the trigonometric functions that are quotients of each other which can and have been useful in verifying other identities. The quotient identities are as follows...

tan⁡θ=sin⁡θcos⁡θcot⁡θ=cos⁡θsin⁡θ\begin{array}{ccccccccc} \tan \theta &=& \dfrac{\sin \theta}{\cos \theta} &&& \cot \theta &=& \dfrac{\cos \theta}{\sin \theta} \\[1em] \end{array}

Simplifying Equations​

Trigonometric identities are critical in simplifying trigonometric equations but just as critical is the ability to recognize and use the basic properties and formulas of algebra, such as the difference of squares formula, the perfect square formula, or substitution to simplify trigonometric equations. For example, the equation (sin⁡x+1)(sin⁡x−1)=0(\sin x + 1)(\sin x - 1) = 0 resembles the equation (x+1)(x−1)=0(x + 1)(x - 1) = 0 which is in the form needed to apply the difference of squares formula. Using algebraic properties and formulas like this makes many trigonometric equations easier to understand and solve which is valuable in all fields of mathematics, science, and engineering.

Example‾\underline{Example}

Use algebraic techniques to verify the identity: cos⁡θ1+sin⁡θ=1−sin⁡θcos⁡θ\dfrac{\cos \theta}{1 + \sin \theta} = \dfrac{1 - \sin \theta}{\cos \theta}.

We can verify this identity by manipulating the left side of the equation, cos⁡θ1+sin⁡θ\dfrac{\cos \theta}{1 + \sin \theta}, to rewrite it as the right side of the equation, 1−sin⁡θcos⁡θ\dfrac{1 - \sin \theta}{\cos \theta}.

Let's start by multiplying the numerator and denominator of cos⁡θ1+sin⁡θ\dfrac{\cos \theta}{1 + \sin \theta} by 1−sin⁡θ1 - \sin \theta. This gives us cos⁡θ(1−sin⁡θ)(1+sin⁡θ)(1−sin⁡θ)\dfrac{\cos \theta (1 - \sin \theta)}{(1 + \sin \theta)(1 - \sin \theta)}.

The denominator, (1+sin⁡θ)(1−sin⁡θ)(1 + \sin \theta)(1 - \sin \theta) resembles the form (a+b)(a−b)(a + b)(a - b) which is the difference of squares. The formula states that (a+b)(a−b)=a2−b2(a + b)(a - b) = a^2 - b^2. So, we can rewrite the denominator as 1−sin⁡2θ1 - \sin^2 \theta.

The current form of the equation is cos⁡θ(1−sin⁡θ)1−sin⁡2θ\dfrac{\cos \theta (1 - \sin \theta)}{1 - \sin^2 \theta}. We can now use the pythagorean identity 1−sin⁡2θ=cos⁡2θ1 - \sin^2 \theta = \cos^2 \theta which we get by subtracting sin⁡2θ\sin^2 \theta from both sides of the first pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1. This gives us cos⁡θ(1−sin⁡θ)cos⁡2θ\dfrac{\cos \theta (1 - \sin \theta)}{\cos^2 \theta}.

Finally, we can simplify the equation by dividing cos⁡θ\cos \theta in the numerator by cos⁡2θ\cos^2 \theta in the denominator. This gives us 1−sin⁡θcos⁡θ\dfrac{1 - \sin \theta}{\cos \theta} which is the right side of the equation.

Therefore, we have verified the identity cos⁡θ1+sin⁡θ=1−sin⁡θcos⁡θ\dfrac{\cos \theta}{1 + \sin \theta} = \dfrac{1 - \sin \theta}{\cos \theta}.