Trigonometric Identities

On the unit circle, the point at angle $\theta$ is $(\cos \theta ,\sin \theta )$, and every point on that circle satisfies $x^2+y^2=1$ (radius $1$, centred at the origin) by the Pythagorean theorem. Substituting gives the most important trig identity in the course. You can also see it in any right triangle: divide ${\text{opp}}^2+{\text{adj}}^2={\text{hyp}}^2$ through by ${\text{hyp}}^2$ and you get ${\left(\frac{\text{opp}}{\text{hyp}}\right)}^2+{\left(\frac{\text{adj}}{\text{hyp}}\right)}^2=1$, which is ${\sin }^2\theta +{\cos }^2\theta =1$. The unit-circle version is stronger, because it works for angles beyond $90°$ as well, including those where sine or cosine is negative (squaring removes the sign). Rearranging gives two more forms you will use constantly: ${\sin }^2\theta =1-{\cos }^2\theta$ and ${\cos }^2\theta =1-{\sin }^2\theta$. A note on notation: ${\sin }^2\theta$ means ${(\sin \theta )}^2$, the sine value squared. It does NOT mean $\sin ({\theta }^2)$.

Since $\sin \theta =\frac{y}{1}=y$ and $\cos \theta =\frac{x}{1}=x$ on the unit circle, and $\tan \theta =\frac{y}{x}$ by definition, dividing sine by cosine automatically reproduces tangent. In a right triangle the same thing happens: $\frac{\sin \theta }{\cos \theta }=\frac{\text{opp}/\text{hyp}}{\text{adj}/\text{hyp}}=\frac{\text{opp}}{\text{adj}}\text{,}$ and the hypotenuse cancels. The identity also explains where tangent breaks down: when $\cos \theta =0$ (at $\theta =90°$ and $270°$) you would be dividing by zero, so $\tan \theta$ is undefined there. Any identity involving tangent is only claimed for angles where every expression in it is defined.

To prove that two trig expressions are equal for every angle, work with the Left Side (LS) and Right Side (RS) SEPARATELY — never move something across the equals sign, since that would assume the very thing you're trying to prove. Here is why that rule exists. Start from the false statement $-1=1$ and square both sides: you get $1=1$, which is true. Reasoning backwards from a true line does not make the starting line true. Operations that act on both sides at once can hide a false claim, so a proof must show that one side can be TRANSFORMED into the other. Simplify each side independently (usually by rewriting everything in terms of $\sin$ and $\cos$) until they visibly match, and write the two sides in separate columns so a reader can follow each chain of equal expressions.

1. Pick the more complicated side to simplify first (more terms, fractions, or a $\tan \theta$).
2. Rewrite any $\tan \theta$ as $\frac{\sin \theta }{\cos \theta }$ so everything is in sine and cosine.
3. Look for a ${\sin }^2\theta +{\cos }^2\theta$ pattern to replace with $1$ (or the reverse: replace a lone $1$ with ${\sin }^2\theta +{\cos }^2\theta$).
4. Use ordinary algebra: common denominators for sums of fractions, expanding brackets, common factoring, and difference of squares such as ${\cos }^4\theta -{\sin }^4\theta =({\cos }^2\theta -{\sin }^2\theta )({\cos }^2\theta +{\sin }^2\theta )\text{.}$
5. Cancel only common FACTORS, never terms that are being added.
6. Stop when LS and RS are the same expression, and write a concluding line: LS = RS, so the identity is true.

How to Prove Trigonometric Identities (and how not to)