Introduction

Today, you’ll learn how to rewrite and manipulate trigonometric expressions using equivalent forms. In this unit, you will practice applying fundamental identities. You will also learn how to simplify expressions, rewrite functions in alternative forms, and solve trigonometric equations and inequalities more efficiently.

Trigonometric identities let us rewrite expressions in different but also equal forms.

Example 1: Show that ${\sin }^2\theta +{\cos }^2\theta =1$ for all $\theta$

To solve this problem, first consider the unit circle. On the unit circle, any point has the coordinates $\left(\cos \theta ,\sin \theta \right)$.

The equation of the unit circle is $x^2+y^2=1$. Through this, we can find a way to represent any point on the circle. Substituting $x=\cos \theta$ and $y=\sin \theta$, we achieve the fundamental Pythagorean Identity:

${\sin }^2\theta +{\cos }^2\theta =1$, which is valid for all $\theta$.

This identity allows us to replace one squared trig function with an equivalent form in terms of the other, allowing us to simplify (and thus solve) many problems. Furthermore, by algebraic manipulation, we can also obtain the related forms:

$1+{\tan }^2\theta ={\sec }^2\theta$ (where we divide ${\sin }^2\theta +{\cos }^2\theta =1$ by ${\cos }^2\theta$)

And $1+{\cot }^2\theta ={\csc }^2\theta$ (where we divide by ${\sin }^2\theta$)

These identities hold true whenever the denominators are nonzero (e.g. $\cos \theta \ne 0$ for $1+{\tan }^2\theta ={\sec }^2\theta$)

They allow us to eliminate squares of tan, sec, etc., which allows us to simplify equations into a much more appealing form.

Sum, Difference, Double-Angle, and Half-Angle identities

Using angle-sum identities, any sum or difference of angles can be rewritten in terms of products of sines and cosines. These sine-angle-sum and cosine-angle-sum identities are:

$\sin \left(a\pm b\right)=\sin a\cdot \cos b\pm \cos a\cdot \sin b$

$\cos \left(a\pm b\right)=\cos a\cdot \cos b\mp \sin a\cdot \sin b$

Tangent also has these identities

$\tan \left(a\pm b\right)=\frac{\sin \left(a\pm b\right)}{\cos \left(a\pm b\right)}$

There is also a simplified form in terms of tan only:

$\tan \left(a\pm b\right)=\frac{\tan a\pm \tan b}{1\mp \tan a\tan b}$

Cotangent is just the reciprocal of tangent

$\cot \left(a\pm b\right)=\frac{1\mp \tan a\tan b}{\tan a\pm \tan b}$

Secant and Cosecant follow the same principle for sum identities; they are the reciprocal ($\frac{1}{x}$) of the cosine and sine identities.

Special cases of these give double-angle formulas by setting $b=a$:

$\sin \left(2a\right)=2\sin a\cdot \cos a$

$\cos \left(2a\right)={\cos }^{2}a-{\sin }^{2}a=2{\cos }^{2}a-1=1-2{\sin }^{2}a$

$\tan \left(2\theta \right)=\frac{2\tan \theta }{1-{\tan }^2\theta }$

$\cot \left(2\theta \right)=\frac{{\cot }^2\theta -1}{2\cot \theta }$

Secant and Cosecant are still just the reciprocals.

There are also Half-angle identities (not tested on AP exam).

$\begin{array}{c}\sin \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{1-\cos \left(\theta \right)}{2}}\\ \cos \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{1+\cos \left(\theta \right)}{2}}\\ \tan \left(\frac{\theta }{2}\right)=\frac{\sin \left(\theta \right)}{1+\cos \left(\theta \right)}=\frac{1-\cos \left(\theta \right)}{\sin \left(\theta \right)}\\ \tan \left(\frac{\theta }{2}\right)=\pm \sqrt{\frac{1-\cos \left(\theta \right)}{1+\cos \left(\theta \right)}}\\ \tan \left(\frac{\theta }{2}\right)=\frac{\sin \left(\theta \right)}{1+\cos \left(\theta \right)}=\frac{1-\cos \left(\theta \right)}{\sin \left(\theta \right)}\\ \cot \left(\frac{\theta }{2}\right)=\frac{\sin \left(\theta \right)}{1-\cos \left(\theta \right)}=\frac{1+\cos \left(\theta \right)}{\sin \left(\theta \right)}\end{array}$

Cofunction shifts

A particularly useful family of sum/difference identities arises when you add or subtract a quarter turn, $\frac{\pi }{2}$ ($90^{\circ }$). This is called a phase shift of trig functions. These follow directly from the angle-sum formulas and the special values of $\cos \left(\frac{\pi }{2}\right)=0,\sin \left(\frac{\pi }{2}\right)=1$.

Example 2: Show that $\sin \left(x+\frac{\pi }{2}\right)=\cos x$

To solve this problem, apply the sine-angle-sum formula

$\sin \left(x+\frac{\pi }{2}\right)$

$=\sin x\cdot \cos \left(\frac{\pi }{2}\right)+\cos x\cdot \sin \left(\frac{\pi }{2}\right)$

$=\sin x\cdot \left(0\right)+\cos x\cdot \left(1\right)=\cos x$

Example 3: Show that $\sin \left(x-\frac{\pi }{2}\right)=-\cos x$

Similar to the last question, apply the sine-angle-formula

$\sin \left(x-\frac{\pi }{2}\right)$

$=\sin x\cdot \cos \left(\frac{\pi }{2}\right)-\cos x\cdot \sin \left(\frac{\pi }{2}\right)=$

$\sin x\cdot \left(0\right)-\cos x\cdot \left(1\right)=-\cos x$

This applies to all trig functions. The table below provides a list:

List of $\pm \frac{\pi }{2}$ shift identities:

$\begin{array}{cc}\sin \left(x+\frac{\pi }{2}\right)=\cos \left(x\right)& \sin \left(x-\frac{\pi }{2}\right)=-\cos \left(x\right)\\ \cos \left(x+\frac{\pi }{2}\right)=-\sin \left(x\right)& \cos \left(x-\frac{\pi }{2}\right)=\sin \left(x\right)\\ \tan \left(x+\frac{\pi }{2}\right)=-\cot \left(x\right)=-\frac{1}{\tan \left(x\right)}& \tan \left(x-\frac{\pi }{2}\right)=-\cot \left(x\right)=-\frac{1}{\tan \left(x\right)}\\ \sec \left(x+\frac{\pi }{2}\right)=-\csc \left(x\right)& \sec \left(x-\frac{\pi }{2}\right)=\csc \left(x\right)\\ \csc \left(x+\frac{\pi }{2}\right)=\sec \left(x\right)& \csc \left(x-\frac{\pi }{2}\right)=-\sec \left(x\right)\\ \cot \left(x+\frac{\pi }{2}\right)=-\tan \left(x\right)& \cot \left(x-\frac{\pi }{2}\right)=-\tan \left(x\right)\end{array}$

Inverse-function relationships (not tested on AP exam):

Example 4: Show that $\arccos \left(\sqrt{1-x^2}\right)=\left|\arcsin x\right|$

Start from the Pythagorean identity, we can derive a formula for ${\cos }^2\theta =1-{\sin }^2\theta =1-x^2$, so $\cos \theta =\pm \sqrt{1-x^2}$.

Because $\arcsin \left(x\right)$ lies in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ where cosine is nonnegative, we must take the positive root and obtain the identity

$\cos \left(\arcsin \left(x\right)\right)=\sqrt{1-x^2}$

Applying arccos to both sides gives

$\arccos \left(\sqrt{1-x^2}\right)=\arccos \left(\cos \left(\arcsin \left(x\right)\right)\right)$

Since $\arccos \left(\cos u\right)$ returns the principal value in $\left[0,\pi \right]$, we have

$\arccos \left(\cos \left(\arcsin \left(x\right)\right)\right)=\left|\arcsin \left(x\right)\right|$

Thus, when putting it all together, we have:

$\arccos \left(\sqrt{1-x^2}\right)=\left|\arcsin \left(x\right)\right|$ for $-1\le x\le 1$.

We can get a more general form, though, that expresses a signed relationship (rather than a magnitude)

Example 5: Show that $\arcsin x+\arccos x=\frac{\pi }{2}$

We can rearrange the original equation into

$\sin \left(\frac{\pi }{2}-\arccos \left(x\right)\right)$

Applying the phase shift shown earlier, it can be shown that

$\sin \left(\frac{\pi }{2}-\arccos \left(x\right)\right)=\cos \left(\arccos \left(x\right)\right)=x$

where $-1\le x\le 1$

Additionally, we know that $\frac{\pi }{2}-\arccos \left(x\right)$ lies in $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ because $\frac{\pi }{2}$ subtracted by the max and min value of $\arccos \left(x\right)$ yields $-\frac{\pi }{2}$ and $\frac{\pi }{2}$. This is the exact range of the $\arcsin$ function; $\arcsin$ only returns angles in this interval.

Therefore, applying $\arcsin$ on both sides of the equation, we get $\frac{\pi }{2}-\arccos \left(x\right)=\arcsin x$, or

$\arcsin \left(x\right)+\arccos \left(x\right)=\frac{\pi }{2}$ for $-1\le x\le 1$.

Such co-function relations can be useful with inverse trig functions. Other relationships are included in the table below:

$\begin{array}{c}\arcsin \left(x\right)+\arccos \left(x\right)=\frac{\pi }{2}\\ \arctan \left(x\right)+\mathrm{arccot}\left(x\right)=\frac{\pi }{2}\\ \cos \left(\arcsin \left(x\right)\right)=\sqrt{1-x^2}\\ \arccos \left(\sqrt{1-x^2}\right)=\left|\arcsin \left(x\right)\right|\\ \mathrm{arcsec}\left(x\right)=\arccos \left(\frac{1}{x}\right)\\ \mathrm{arccsc}\left(x\right)=\arcsin \left(\frac{1}{x}\right)\\ \arcsin \left(-x\right)=-\arcsin \left(x\right)\\ \arctan \left(-x\right)=-\arctan \left(x\right)\\ \arccos \left(-x\right)=\pi -\arccos \left(x\right)\\ \mathrm{arccot}\left(-x\right)=\pi -\mathrm{arccot}\left(x\right)\\ \mathrm{arcsec}\left(-x\right)=\pi -\mathrm{arcsec}\left(x\right)\\ \mathrm{arccsc}\left(-x\right)=-\mathrm{arccsc}\left(x\right)\end{array}$

Verifying Trigonometric Identities

To verify a trig identity/equation, pick one side (usually the side where you see more trig functions, a.k.a., the more “complex” side) and algebraically transform it to look like the other side. This is not a suggestion; you are not allowed to change both sides. For these problems, you can only modify one side throughout the entire solving process.

A common strategy is to convert all the functions to sines and cosines and apply known identities. Then, go ahead and simplify.

Let’s take an example.

$\frac{1-{\cos }^2\left(x\right)}{\sin \left(x\right)}=\sin \left(x\right)$

And there we have it! We just turned $\frac{1-{\cos }^2\left(x\right)}{\sin \left(x\right)}=\sin \left(x\right)$ into something as simple as $\sin \left(x\right)$! Yipee!

Common Mistakes/Warnings

Solving Trig Inequalities

Solving trig inequalities is not as bad as it seems. You just do what you’d normally do: First, gather all terms on one side, leaving one side as zero. Then, find all solutions.

After you have found all your solutions (on the specified domain), simply create a sign chart. If your domain is all real numbers, simply find one domain and apply the periodicity of the function.

Overall, to be successful with problems that involve content from this unit, there’s only one way to be prepared: you must practice.

Practice Problems

Simplify the expressions as much as possible:

Verify each identity (remember, you may only modify one side):

Solve the Trigonometric Equations:

Solve the Trigonometric Inequality (assume all to be of within the domain $0<x<2\pi$ unless told otherwise)