# Solving Trig Identities Practice Problems

Here are some examples of simple identity proofs with reciprocal and quotient identities.Typically, to do these proofs, you must always start with one side (either side, but usually take the more complicated side) and manipulate the side until you end up with the other side.

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We now proceed to derive two other related formulas that can be used when proving trigonometric identities.

It is suggested that you remember how to find the identities, rather than try to memorise each one.

Sometimes we have to find common denominators, like in the last example.

We didn’t need to turn it into sin and cos, since we only had tan and cot in the identity (although it still would have worked).

(Some teachers will let you go down both sides until the two sides are equal).

The best way to solve these is to turn everything into sin and cos.Here are the double angle and half angle identities, and tricks to help you memorize them: ” out and put it in front of a sin and cos.$$\cos \left( \right)$$ is a little more complicated, especially since it can be written in three ways.Trig identities are sort of like puzzles since you have to “play” with them to get what you want.You will also have to do some memorizing for these, since most of them aren’t really obvious.You may not like Trig Identity problems, since they can resemble the proofs that you had to in Geometry. There are typically two types of problems you’ll have with trig identities: working on one side of an equation to “prove” it equals the other side, and also solving trig problems by substituting identities to make the problem solvable.We’ll start out with the simpler identities that you’ve seen before.And note that there may be more than one way to do these! “Proof“: $$\displaystyle \begin\sin \theta =\frac\,\,\,\,\,\,\,\,\,\,\,\cos \theta =\frac\,\,\,\,\,\,\,\,\,\,\,\, =1\,\,\,\,\left( \right)\\theta \theta = = = =1\end$$ (Note that  is written as $$x$$.) From the first Pythagorean identity, we can derive the other two: $$\begin\theta \theta &=1\\frac \frac&=\frac\\theta 1&=\theta \end$$ $$\begin\theta \theta &=1\\frac \frac&=\frac\1 \theta &=\theta \end$$ The three Pythagorean Identities are: $$\displaystyle \begin\theta \theta =1\\theta 1=\theta \\theta 1=\theta \end$$ $$\require \displaystyle \begin&=\frac\left( \right) \frac\left( \right)\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&=\frac \frac\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&=\frac \frac\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&=\frac\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&=\frac\,\,=\,\frac\\frac\,\,&=\frac\cdot \frac\,\,\,\surd \end$$ $$\displaystyle \begin2 3\cos x-3&=0\2\left( \right) 3\cos x-3&=0\2-2 3\cos x-3&=0\-2 3\cos x-1&=0\2-3\cos x 1&=0\\left( \right)\left( \right)&=0\end$$ $$\displaystyle \cos x=\frac\,\,\,\,\,\,\,\,\,\,\,\,\,\cos x=1\,\,\,\,\,\,\,\,\,\,\,\,\,x=\left$$ $$\displaystyle \begin\frac-\frac&=\frac\\frac-\frac&=0\\frac-\frac&=0\\frac&=0\\sin x-x-\left( \right)&=0\\sin x-1&=0\,\x&=\left\end$$ This is an extraneous solution, since $$\displaystyle \cos \left( \right)=0$$, and we can’t divide by $$\displaystyle \begin\sin x=0\,\,\,\,\,\,\,\,\cos x=-1\\,\,\,\,\,0,\pi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\pi \end$$ The answer, over the reals, is all real numbers except 0 and $$\boldsymbol$$, or $$\displaystyle \left$$, which is the same as $$\displaystyle \left$$. )Note: From these identities, you may be asked to be familiar with the Odd/Even Identities: $$\displaystyle \begin\text\,\,\,\cos \left( \right)=\cos \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text\,\,\,\,\,\,\sin \left( \right)=-\sin \left( x \right)\,\,\,\,\,\,\,\tan \left( \right)=-\tan \left( x \right)\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec \left( \right)=\sec \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\csc \left( \right)=-\csc \left( x \right)\,\,\,\,\,\,\,\cot \left( \right)=-\cot \left( x \right)\end$$ ….the Cofunction Identities in radians (trig functions of an angle is equal to the value of the cofunction of the complement). $$\displaystyle \begin\sin \left( \right)=\cos \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\csc \left( \right)=\sec \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tan \left( \right)=\cot \left( x \right)\\cos \left( \right)=\sin \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec \left( \right)=\csc \left( x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cot \left( \right)=\tan \left( x \right)\end$$ Since $$\sin A\cos B \cos A\sin B=\sin \left( \right)$$, we have: $$\displaystyle \begin\sin \left( \right)\cos \left( \right) \cos \left( \right)\sin \left( \right)\=\sin \left( \right)=\sin \left( \right)\end$$ Divide up $$\displaystyle \frac$$ into two fractions that can be found on the Unit Circle: $$\displaystyle \begin\cos \left( \right)&=\cos \left( \right)=\cos \left( \right)\&=\cos \left( \right)\cos \left( \right) \sin \left( \right)\sin \left( \right)\&=\left( \right)\left( \right) \left( \right)\left( \right)\&=\frac\end$$ Divide up $$\displaystyle -\frac$$ into two fractions that can be found on the Unit Circle; use cos first, and then take reciprocal: $$\displaystyle \begin\cos \left( \right)&=\cos \left( \right)=\cos \left( \right)\&=\cos \left( \right)\cos \left( \right) \sin \left( \right)\sin \left( \right)\&=\left( \right)\left( \right) \left( \right)\left( \right)\&=\frac\end$$ $$\displaystyle \begin\sec \left( \right)&=\frac=\frac\left( \right)\&=\sqrt-\sqrt\end$$ Divide up $$\displaystyle -\frac$$ into two fractions that can be found on the Unit Circle; use tan first, and then take the reciprocal: $$\displaystyle \begin\tan \left( \right)&=\tan \left( \right)=\tan \left( \right)\&=\frac=\frac\&=\frac=\frac\end$$ $$\displaystyle \cot \left( \right)=\frac=\frac$$ $$\displaystyle \begin\cos x\cos \pi -\sin x\sin \pi &=\\left( \right)\left( \right)-\left( \right)\left( 0 \right)&=\\left( \right)\left( \right)-0&=-\cos x\,\,\,\,\,\,\surd \end$$ Note: Evaluate any expressions that turn into constants (like $$\sin \pi$$).To help memorize this, I remember that since cos is even, we have the cos’s together and the sin’s together on the right side. $$\displaystyle \begin\color\\\left( \right)\left( \right)=\-=\\left( \right)\left( \right)-\left( \right)\left( \right)=\\left( \right)\left( \right)-\left( \right)\left( \right)=\\alpha -\cancel-\\beta \cancel=\alpha -\beta \,\,\,\,\,\surd \end$$ $$\displaystyle \begin\frac&=\\frac&=\\frac\cdot \frac&=\\frac&=\\frac&=\\frac=\frac\end$$ Note: We thought we’d try to multiply the top and bottom by $$\sec x\sec y$$ to get the right-hand side numerator; it worked!I memorize the $$A-A$$ part and then remember that I can use the Pythagorean Identity $$A A=1$$ to substitute and do the algebra to arrive at the other expressions.For $$\tan \left( \right)$$, the identity only has tan’s in it, with the “These are a little more complicated.

## Comments Solving Trig Identities Practice Problems

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