Compound Angle Identity.sync-conflict-20240906-210443-7M7WZA3

Mathematically, the following expressions are true

$sin(x+y)=sin(x)cos(y)+cos(x)sin(y)$ $sin(x-y)=sin(x)cos(y)-cos(x)sin(y)$ $cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$ $cos(x-y)=cos(x)cos(y)+sin(x)sin(y)$ $tan(x+y)$ = $\frac{tan(x)+tan(y)}{1-\tan (x)\tan (y)}$ $\tan (x-y)=\frac{\tan (x)\tan (y)}{1+\tan (x)\tan (y)}$t

Proof

If the subtending length of B + subtending length of A is the subtending length of angle A+B Such is true for the adjacent side too.

Compound Formula for Unverifiable Angles

In the old days, angles that were not special angles were not verifiable. But, we can break those angles down using compond identity to find the exact ratio.

Example

determine exact value for cos $75^o$

Solution

cos(A+B) = cos(A)cos(B) - sin(A)sin(B) cos(45 + 30) = cos(45)cos(30) - sin(45)sin(30) $=\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\frac{1}{2}$ $-\frac{\sqrt{6}\sqrt{2}}{4}$

Example 2

determine exact value for tan $\frac{7\pi }{12}$

Solution

our compound angles are: $\frac{\pi }{3}+\frac{\pi }{4}$ $\tan (\frac{\pi }{3}+\frac{\pi }{4})$ $\tan (\frac{\pi }{3}+\frac{\pi }{4})$ $1-\tan (\frac{\pi }{3})\tan (\frac{\pi }{4})$ $\frac{\sqrt{3}+1}{1-\sqrt{3}}\ast \frac{1+\sqrt{3}}{1+\sqrt{3}}$ $-2-\sqrt{3}$