Question

Prove the following identity:

cos2θ = cos2θ − sin2θ = 2cos2θ − 1 = 1 − 2sin2θ

Answer 1

??
you sure you've got that right??

Answer 2

Way too many equal signs .

Answer 3

i think some of those should contain an exponent - like sin^2θ

Answer 4

yes, and the others are pluses

Answer 5

I would think it is

\(\normalsize\color{black}{ \cos2θ + \cos2θ − \sin2θ + 2\cos2θ − 1 = 1 − 2\sin2θ}\)

Answer 6

Depends on which `=` is a `+`. (Where the user forgot to use `SHIFT` )

Answer 7

@Katieholmes please re-write the question.

Answer 8

Ambiguity in question...

Answer 9

@Katieholmes , are you there? Please re-write the question.

Answer 10

No I rechecked it and that is the question. Ill add a screenshot

Answer 11

I can think of a way to do this if we assume the identity

cos ( A + B) = cos A cos B - sinA sinB

but i'm not sure if thats acceptable

Answer 12

you'll also need to apply sin^2A + cos^2 A = 1

Answer 13

cos2x = cos(x +x)= cosx.cosx - sinx.sinx = (cosx)^2 - (sinx)^2
therefor cos2x = (cosx)^2 - (sinx)^2 {x = theta}

and cos2x = {1 - (sinx)^2) - (sinx)^2 = 1 - 2(sinx)^2
and cos2x = (cos2x)^2 - {1 - (cosx)^2} = 2(cosx)^2 - 1

Answer 14

- yes thats basically the way I suggested

Answer 15

yea i know what you guys are talking about. It confused my because of all the different = signs and I though it just didnt seem right

Answer 16

yes also it was confusing because you didnt right the 'squared' bit as '^2'

Answer 17

Katie, talk to your teacher get the correct question, and then ask it in a different question.