Question

Verify the given identity using double-angle formulas.

Did I do this right?

Answer 1

You got this one @AlexandervonHumboldt2 ? :)

Answer 2

you got wut to do JOLG?

Answer 3

Nah bruh ^_^ I'll go help someone else. ^_^ good luck

Answer 4

can't see the question.

Answer 5

question: \[\frac{ cosx+\cos3x }{ 2\cos2x }=cosx\]


work: \[\frac{ cosx+\cos(2x+x)}{ 2\cos2x }\]
= \[\frac{ cosx+cos2xcosx-\sin^2x)-\sin^2xsinx}{ 2cos2x }\]
= \[\frac{ cosx+(1-2\sin^2x)cosx-(2sinxcosx)sinx }{ 2\cos2x }\]
=\[\frac{ cosx+cosx(1-2\sin) }{ 2\cos2x }\]
=\[\frac{ 2cosx-4\sin^2xcosx }{ 2(1-2\sin^2x)}\]
= cosx

Answer 6

The page refreshed on me ;-;

Answer 7

also I think I put in some typos...