# Thread: Trigonometry help

1. ## Trigonometry help

I have done some of these and attempted others that I am stuck on....I have to simplify so that both sides are equal (RS and LS)...is there anyone who can take a look at my work and give me a little guidance?

2. Hello, aikenfan!

#3 and #6 are fine . . . good work!

$\displaystyle 9)\;\;\sin^2\!a-\sin^4\!a \:=\:\cos^2\!a - \cos^4\!a$
Factor the left side: .$\displaystyle \sin^2\!a\underbrace{(1-\sin^2\!a)}$

. . . . . . . . . . . . $\displaystyle = \quad\;\;\sin^2\!a\cdot\cos^2\!a$

. . . . . . . . . . . . $\displaystyle =\;\overbrace{(1-\cos^2\!a)}\cos^2\!a$

. . . . . . . . . . . . $\displaystyle = \;\cos^2\!a - \cos^4\!a$

What is the right side of #12?

$\displaystyle 15)\;\;\sin^{\frac{1}{2}}\!x\cos x = \sin^{\frac{5}{2}}\!x\cos x \:=\:\cos^3\!x\sqrt{\sin x}$
Factor the left side: .$\displaystyle \sin^{\frac{1}{2}}\!x\cos x\left(1-\sin^2\!x\right) \;=\; \sin^{\frac{1}{2}}\!x\cos x\cdot\cos^2x$

. . $\displaystyle =\;\cos^3\!x\sin^{\frac{1}{2}}\!x \;=\;\cos^3\!x\sqrt{\sin x}$

$\displaystyle 18)\;\;\frac{\sec x - 1}{1-\cos x} \:=\:\sec x$

The left side is: .$\displaystyle \frac{\dfrac{1}{\cos x} - 1}{1-\cos x}$

Multiply by $\displaystyle \frac{\cos x}{\cos x}\!:\;\;\frac{\cos x}{\cos x}\cdot\frac{\frac{1}{\cos x} - 1}{1 - \cos x} \;=\;\frac{1-\cos x}{\cos x(1-\cos x)} \;=\;\frac{1}{\cos x} \;=\;\sec x$

$\displaystyle 21)\;\;\csc x\cos x \:=\:\sin x\tan x$
This one isn't true . . . Is there a typo?

3. Wow! Thank you very much for looking at my work...on number 21, it is minus on the left side not times.