To prove that a trigonometric equation is an identity, one typically starts by trying to show that either one side of the proposed equality can be transformed into the other, or that both sides can be transformed into the same expression.

In other words, suppose $A$ and $B$ are some trigonometric expressions and we are trying to determine if $A=B$.

We hope that both expressions will simplify to some common form $C$, as if we can show the following:

$$\begin{array}{rclcrcl} A &=& A_1 \quad && \quad B &=& B_1\\ &=& A_2 && &=& B_2\\ &=& \cdots & \textrm{and} & &=& B_3\\ &=& A_n && &=& B_4\\ &=& C && &=& \cdots\\ & & && &=& B_m\\ & & && &=& C \end{array}$$then we will know

$$A = A_1 = A_2 = \cdots = A_n = C = B_m = \cdots = B_2 = B_1 = B$$and thus,

$$A = B$$That is our general "plan of attack" -- although, we might get lucky and the sequence of $A_1, A_2, \cdots$ will terminate in $B$, or the sequence $B_1, B_2, \cdots$ will terminate in $A$, which then shortens our argument a bit.

There are some basic strategies to help us get to that common form $C$ as efficiently as possible:

Only manipulate one side of the proposed identity at a time. Start by attempting to simplify the more complicated side first, as which steps one should take will likely be more obvious for this side.

First "trigonometrically simplify" the side in question.

One should try rewriting all of trigonometric functions involved in terms of sines and cosines, unless there is a compelling reason not to do this.

$$\begin{array}{rcl} \tan \theta + 2\csc \theta &=& \left( \frac{\sin \theta}{\cos \theta} \right) + 2\left( \frac{1}{\sin \theta} \right)\\ &=& \cdots \end{array}$$**Example**Likewise, if you see trigonometric functions involving more than one angle measure, try to use known identities to rewrite things so that only a single angle measure is involved.

$$\begin{array}{rcl} \frac{\cos 2\theta + \sin \theta}{\sin 2\theta + \sin(-\theta)} &=& \frac{ (\cos^2 \theta - \sin^2 \theta) + \sin \theta}{2\sin \theta \cos \theta - \sin \theta}\\ &=& \cdots \end{array}$$**Example**If one side can be written in terms involving (perhaps multiple occurances of) a single trigonometric function of a single angle measure, doing so may help. In particular, this often helps in cases where one side of the proposed identity is already in this form, and the other side consists of a mixture of trigonometric functions.

$$\begin{array}{rcll} \frac{\cos 2\theta}{\cos(\pi/2-\theta)} &=& \frac{1 - 2\sin^2 \theta}{\sin \theta} & \overset{\normalsize{\textrm{assuming we were trying to get}}}{\scriptsize{\textrm{things in terms of only sine functions}}}\\ &=& \cdots \end{array}$$**Example**

As the previous examples attest, trigonometrically simplifying an expression sometimes makes it algebraically more cumbersome. This can actually be a good thing -- as it gives us a direction to proceed. Specifically -- after trigonometrically simplifying a side of the proposed identity -- one can next focus on "algebraically simplifying" it.

Complex fractions (i.e., fractions with fractions in either the numerator or denominator) should be collapsed

$$\begin{array}{rcll} \frac{\displaystyle{\frac{\sin \theta}{\cos^2 \theta}}}{\displaystyle{1 + \frac{1}{\cos \theta}}} &=& \frac{\sin \theta}{\cos \theta + 1} & \scriptsize{\textrm{ after multiplying by } \displaystyle{\frac{\cos^2 \theta}{\cos^2 \theta}}}\\ &=& \cdots \end{array}$$**Example**When fractional expressions appear in a sum or difference, these terms should be combined into a single fraction, finding common denominators as necessary. The resulting fractional expression may, upon factoring, admit a common factor that can be cancelled -- or may be simplified in some other fashion.

$$\begin{array}{rcll} \frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta} + \frac{\sin \theta \cos^2 \theta - 1}{\sin^2 \theta \cos^2 \theta} &=& \frac{\cos^2 \theta + \sin^2 \theta - 1 + \sin \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}\\\\ &=& \frac{1 -1 + \sin \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}\\\\ &=& \frac{\sin \theta \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}\\\\ &=& \frac{1}{\sin \theta}\\\\ &=& \cdots \end{array}$$**Example**$\require{cancel}$

Try to manipulate the side in question into the same "form" as the other side. For example, suppose one is attempting to simplify the left side of a proposed identity and the right side is a product. Then one should attempt to factor the left side, so that it is also expressed as a product.

Also, when the left and right sides get to the point where there is a partial "match", one should leave the matched parts alone from that point forward, and only manipulate the parts that still don't look like one another.

**Example**Show $\sin \theta \cos^2 \theta - \sin \theta = -\sin^3 \theta$ is an identity.

Starting with the left side, we have...

$$\begin{array}{rcll} \sin \theta \cos^2 \theta - \sin \theta &=& \sin \theta (\cos^2 \theta - 1) & \overset{\normalsize{\textrm{notice the right side, } -\sin^3 \theta, \textrm{ is a product, so}}}{\scriptsize{\textrm{we factor the left side to have the same form}}}\\\\ &=& \cdots\\\\ &=& \sin \theta (-\sin^2 \theta) & \overset{\normalsize{\textrm{working backwards from the sought expression below}}}{\scriptsize{\textrm{we expose a "match" of } \sin \theta \textrm{ with the above}}}\\\\ &=& -\sin^3 \theta\\\\ \end{array}$$To fill in the missing steps above, we just need to "massage" $(\cos^2 \theta - 1)$ into $(-\sin^2 \theta)$. This of course is immediate, given the Pythagorean identity $\cos^2 \theta + \sin^2 \theta = 1$.

It may be the case that in the course of trying to prove a given equation is an identity, one begins to suspect that it is not.

In such situations, one should test whether the equation's left and right sides are actually equal by plugging in some values for the variables it contains. Remember, one only needs a single counter-example to prove an equation is not an identity.

If however, one tests a particular value (or set of values) and the left and right sides of the given equation agree in value, that particular test is inconclusive -- and a decision must be made whether to continue the search for a counter-example and test additional values, or to return to trying to prove the given equation is an identity.

To have the best chance of selecting values that will show a given equation is not an identity, one should keep the following in mind:

While picking angle measures that are integer multiples of $\pi/2$ will lend itself to easy evaluation of the expressions involved in the equation, such values will often fail to reveal an equation is not an identity due to the fact that either the sine or cosine for these angle measures will be zero.

**Example**The following is clearly not an identity: $\sin \theta = \cos \theta + \sin \theta$. However, if one tests this with either $\theta = \pi/2$ or $\theta = 3\pi/2$, the results will be inconclusive, as the left and right sides will have the same value (i.e., $1$ in the first case, $-1$ in the second).

A similar problem presents itself when testing a proposed identity with angle measures that are odd integer multiples of $\pi/4$, but for a different reason. For these angle measures, recall the sine and cosine values are either identical or differ only in sign. This too can create inconclusive results for an equation that is not actually an identity.

**Example**The following is clearly not an identity: $2\sin \theta= \cos \theta + \sin \theta$. However, if one tests this with either $\theta = \pi/4$ or $\theta = 5\pi/4$, the results will be inconclusive, as both sides evaluate to $\sqrt{2}$.

Testing a proposed identity with angle measures that are integer multiples of $\pi/6$ or $\pi/3$ (when reduced) can be good first choices, as the exact values of the trigonometric functions are easy to find, and the problems seen above don't occur.