Question

Trig!

Answer 1

\[\cos^2(2x) - \sin^2(2x) = 0\]

Answer 2

So... I attempted...
(1 - 2sin^2x) - (1 - 2sin^2x) - (2sinxcosx)(2sinxcosx)
Does that look right so far?

Answer 3

Even if it is right, it looks very messy

Answer 4

Yeah :/

Answer 5

\[cos^2(2x) = cos^2(x) - sin^2(x)\]

I'm sure you can use it backwords to simplify the question and then use another equation for cos^2(2x)

Answer 6

So there might exist a better way to work this

Answer 7

\(\cos^2(2x) - \sin^2(2x) = 0\)

\(\sin^2(2x) = \cos^2(2x) \)

divide cos^2(2x) both sides, what do you get ?

Answer 8

sin^2(2x)/cos^2(2x) = 1

Answer 9

That should work, or @TheSmartOne's method using the identity\[\large \cos^2(2x) = \cos^2(x) - \sin^2(x)\]Replace each x with 2x\[\large \cos^2(2*2x) = \cos^2(2x) - \sin^2(2x)\]
now look at your equation\[\large \cos^2(2x) - \sin^2(2x) = 0\]and use the above to replace\[\large \cos^2(2*2x)= 0\]
\[\large \cos^2(4x)= 0\]square root both sides
\[\large \cos (4x)= 0\]then solve.

Answer 10

Agent to the rescue!
Would the answers be pi/8 + npi/2 and 3pi/8 + npi/2 ?

Answer 11

Looks good.

@ganeshie8's method would've worked fine too, and maybe a little easier than using the identity, but this one isn't really obvious or simple at first glance, no matter which method you take.

Answer 12

I think the identity was what they wanted, I dunno. Thanks for the help Agent :)
Although I will say, I never thought I'd get along with someone who doesn't like cilantro...
;)

Answer 13

Yeah i feel like you've been learning double angle identities etc. Welcome :) I never thought I'd get along with someone who does ;)

Though that'd mean I didn't get along with an awful lot of people :D

Answer 14

Ha xD
Once you try my salsa, you will fall in love with cilantro.

Answer 15

I'm sure I'd like your salsa. A little cilantro is okay; i don't mind it when it doesn't overpower.

Answer 16

What about basil?

Answer 17

That's mine btw :)

Answer 18

Impressive plant!

Basil is fine.