# 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.

Dividing In this problem, it is easier to start from the RHS. We multiply numerator and denominator of the fraction by the conjugate of the denominator.

$$\displaystyle \sin \theta =\frac\,\,\,\,\,\,\,\,\,\,\,\csc \theta =\frac\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \theta =\frac\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec \theta =\frac\,$$ $$\displaystyle \tan \theta =\frac=\frac\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cot \theta =\frac=\frac$$ $$\begin\sin \left( \right)=\sin A\cos B \cos A\sin B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( \right)=\cos A\cos B-\sin A\sin B\\\sin \left( \right)=\sin A\cos B-\cos A\sin B\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cos \left( \right)=\cos A\cos B \sin A\sin B\end$$ $$\displaystyle \theta \theta =1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta 1=\theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta 1=\theta$$ An “identity” is something that is always true, so you are typically either substituting or trying to get two sides of an equation to equal each other.

Think of it as a reflection; like looking in a mirror.

An example of a trig identity is $$\displaystyle \csc (x)=\frac$$; for any value of $$x$$, this equation is true.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.(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.$$\displaystyle \begin\cos \left( \right)\cos \left( \right)&=\frac\\left( \right)\left( \right)&=\frac\\left( \right)\left( \right)&=\frac\\fracx-\fracx&=\frac\\frac\left( \right)-\fracx&=\frac\3-3x-x&=2\-4x&=-1\x&=\frac\\sin x&=\pm \frac\end$$ $$\displaystyle x=\left$$ We use double angle and half angle identities the same way we used sum and difference identities when we need to split up the angle to make it easier to find the values (for example, to find values on the unit circle).We also use the identities in conjunction with other identities to prove and solve trig problems.Note how we work on one side only and pull down the other side when it matches.It doesn’t matter which side we start on, but typically, it’s the most complicated.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!

## Comments Solving Trig Identities Practice Problems

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