Thread: Using the identity cos^2x + sin^2x = 1

1. Using the identity cos^2x + sin^2x = 1

Using the identity cos^2x + sin^2x = 1, or otherwise, find an equivalent expression to 1+cosx/2sinx + sinx/2+2cosx

If anyone is able to help. Please do.

2. 1+cosx/2sinx + sinx/2+2cosx

If that is
(1 +cosX)/(2sinX) +(sinX)/(2 +2cosX),
then,

Combine the two fractions into one fraction only.
The common denominator is 2(sinX)(1+cosX),

= [(1+cosX)^2 +(sinX)^2] / [2(sinX)(1+cosX)]
= [1 +2cosX +cos^2(X) +sin^2(X)] / [2(sinX)(1+cosX)]
= [1 +2cosX +1] / [2(sinX)(1+cosX)]
= [2 +2cosX] / [(sinX)(2 +2cosX)]
= 1 / sinX

3. Hello, haku!

Another approach . . .

Using the identity $\displaystyle \cos^2\!x + \sin^2\!x \:= \:1$, or otherwise,

find an equivalent expression to: .$\displaystyle \frac{1+\cos x}{2\sin x} + \frac{\sin x}{2+2\cos x}$
We have: .$\displaystyle \frac{1+\cos x}{2\sin x} + \frac{\sin x}{2(1+\cos x)}$

Multiply the second fraction by: $\displaystyle \frac{1-\cos x}{1-\cos x}$

. . $\displaystyle \frac{\sin x}{2(1+ \cos x)}\cdot\frac{1-\cos x}{1-\cos x} \;=\;\frac{\sin x(1-\cos x)}{2(1-\cos^2\!x)} \;=\;\frac{\sin x(1-\cos x)}{2\sin^2\!x} \;=\;\frac{1-\cos x}{2\sin x}$

The problem becomes: .$\displaystyle \frac{1 + \cos x}{2\sin x} + \frac{1-\cos x}{2\sin x} \;=\;\frac{2}{2\sin x} \;=\;\frac{1}{\sin x} \;=\;\boxed{\csc x}$

4. Thanks for the help!