# Math Help - [SOLVED] Double Angles-Advanced Functions Trig

1. ## [SOLVED] Double Angles-Advanced Functions Trig

Hi there it would be greatly appreciated if someone could help me out with a few things to do with double angles and proving trig identities:
These are some of my hw questions:

tanx=csc2x-cot2x

1-sin2x=1-tanx
cos2x 1+tanx

sin4x-sin2x = cos3x
sin2x cosx

I seem to start theses problems okay but then i end up getting suck just because im unsure of what i can or cant do
Can someone please help with full solutions just so i can get a better understanding for tomorrow
Thanks

2. Hello, xoAlina!

I assume you know all the half- and double-angle identities . . .

$\tan x \:=\:\csc2x - \cot2x$
The right side is: . $\frac{1}{\sin2x} - \frac{\cos2x}{\sin2x} \;=\;\frac{\overbrace{1-\cos2x}^{\text{This is }2\sin^2\!x}}{\underbrace{\sin2x}_{\text{This is }2\sin x\cos x}} \;=\;\frac{2\sin^2\!x}{2\sin x\cos x} \;=\;\frac{\sin x}{\cos x}\;=\;\tan x$

$\frac{1-\sin2x}{\cos2x} \:=\:\frac{1-\tan x}{1+\tan x}$

The right side is: . $\frac{1-\dfrac{\sin x}{\cos x}}{1 + \dfrac{\sin x}{\cos x}}$

Multiply by $\frac{\cos x}{\cos x}\!:\;\;\frac{\cos x - \sin x}{\cos x + \sin x}$

Multiply by $\frac{\cos x-\sin x}{\cos x-\sin x}\!:\;\;{\color{blue}\frac{\cos x - \sin x}{\cos x - \sin x}} \cdot\frac{\cos x - \sin x}{\cos x + \sin x} \;=\;\frac{(\cos x - \sin x)^2}{(\cos x + \sin x)(\cos x - \sin x)}$

. . $= \;\frac{\overbrace{\cos^2\!x - 2\sin x\cos x + \sin^2\!x}^{\cos^2\!x + \sin^2\!x \:=\:1}}{\underbrace{\cos^2\!x - \sin^2\!x}_{\text{This is }\cos2x}} \;=\;\frac{1-\overbrace{2\sin x\cos x}^{\text{This is }\sin2x}}{\cos2x} \;=\;\frac{1-\sin2x}{\cos2x}$

$\frac{\sin4x - \sin2x}{\sin 2x} \:=\:\frac{\cos3x}{\cos x}$
The best way for this one is a sum-to-product identity:

. . $\sin A - \sin B \:=\:2\cos\left(\frac{A+B}{2}\right)\sin\left(\fra c{A-B}{2}\right)$

Then: . $\frac{\sin4x - \sin2x}{\sin2x} \;=\;\frac{2\!\cdot\!\cos3x\!\cdot\!\sin x}{2\!\cdot\!\sin x\!\cdot\!\cos x} \;=\;\frac{\cos3x}{\cos x}$