# Thread: [Urgent - Trig] How would I prove the followin equation with double angle formulas?

1. ## [Urgent - Trig] How would I prove the followin equation with double angle formulas?

cotx = csc2x + cot2x

My Attempt:

RS:
= 1/sin2x + sin2x/cos2x
= (cos2x + sin2x) / (sin2x)(cos2x)
= (2cos^2x - 1 + 2sinxcosx) / (2sinxcosx + 1 - 2sin^2x)
= 2cos^2x - 1 / 1 - 2sin^2x

2. Originally Posted by Macleef
cotx = csc2x + cot2x

My Attempt:

RS:
= 1/sin2x + sin2x/cos2x
= (cos2x + sin2x) / (sin2x)(cos2x)
= (2cos^2x - 1 + 2sinxcosx) / (2sinxcosx + 1 - 2sin^2x)
= 2cos^2x - 1 / 1 - 2sin^2x
ALWAYS write everything in terms of sin and cos

$\displaystyle \frac{cos x}{sin x} = \frac{1}{sin 2x} + \frac{cos 2x}{sin 2x}$

RHS:

$\displaystyle = \frac{1 + cos 2x}{sin 2x}$

$\displaystyle = \frac{1 + ( 2cos^2 x - 1)}{2 \ Sin x \ Cos x}$

$\displaystyle = \frac{2cos^2 x}{2 \ sin x \ Cos x}$

$\displaystyle = \frac{cos x}{sin x}$

$\displaystyle = LHS$

$\displaystyle \rightarrow \mathfrak{Quad \ Erat \ Demonstrandum}$

3. Hello, Macleef!

By now, you've seen your mistake . . .

Prove: .$\displaystyle \cot x \:= \:\csc2x + \cot2x$

My Attempt:
$\displaystyle RS \:=\:\frac{1}{\sin2x} + {\color{red}\frac{\sin2x}{\cos2x}}$ . ← Here!

We have: .$\displaystyle \frac{1}{\sin2x} + \frac{\cos2x}{\sin2x} \;= \;\frac{1 + \cos2x}{\sin2x} \;=\;\frac{2\cos^2x}{2\sin x\cos x} \;=\;\frac{\cos x}{\sin x} \;=\;\cot x$