Question

Explain this trig identity?

Answer 1

I'm confused on #6. They didn't really show their work... how did they use the half angle formula? Isn't the half angle formula:
cos^2x = 1/2(1 - sin(2x))

Answer 2

They don't mention a half-angle...

Answer 3

https://latex.artofproblemsolving.com/d/e/3/de328344a61ed865d729d9458e6af88a1ffaddb5.png

second line

Answer 4

Oh, that makes more sense... I don't know how I read that as half angle.
How would I use the double angle identity in this case? cos2x = 2cos^2x - 1 is the one to use?

Answer 5

Hey

Answer 6

If you're concerned about the working, will you understand if I provide it?

Answer 7

Yes, I'm just unsure how they did it :/
thanks!

Answer 8

\[\large \cos^2 (\text {stuff} ) -\sin^2 (\text {stuff}) = \cos( 2* \text {stuff})\]

Answer 9

@abbles when i said 2nd line, i meant the 2nd line, in the diagram.

Answer 10

In this case, your 'stuff' is 2x.

Answer 11

cos2x = cos^2x - sin^2x?

Answer 12

cos2x = cos^2x - sin^2x? yes.

Answer 13

\[\large\rm cos²y=cos²y-sin²y\\]\[\large\rm y=2x\\]\[\large\rm cos²2x-sin²2x=√3/2\\]\[\large\rm cos²y-sin²y=√3/2\\]\[\large\rm cos2y =√3/2\\]
Recall that\[\large\rm y=2x\\]\[\large\rm cos(2(2x)) =√3/2\\]\[\large\rm cos4x= √3/2\\]

Answer 14

Figure this one out Pebbles?

Answer 15

Pebbles :3
No, still kinda confused... :/

Answer 16

\[\large \cos^2 (2x) - \sin^2(2x) = \frac{ \sqrt 3 }{ 2}\]You just have to apply the earlier identity\[\large \cos 2 \alpha = \cos^2 \alpha - \sin ^2 \alpha \]notice you have \[\large \cos^2 (2x) - \sin^2(2x) \]so in this case, \(\large \alpha = 2x \) right?

Answer 17

So... \(\large \cos^2 (2x) - \sin^2(2x) = \cos 2(2x)\)

Answer 18

\[\large \cos^2 (2x) - \sin^2(2x) = \frac{ \sqrt 3 }{ 2} \]or\[\large \cos 2(2x) = \frac{ \sqrt 3 }{ 2}\]\[\large \cos (4x) = \frac{ \sqrt 3 }{ 2}\]

Answer 19

Ohh! I see. I was working the other way around.

Thanks agent :)

Answer 20

:)