Question

Really hard trig question... Anyone wanna try and help?

Answer 1

\[\sin^2 x \sin^2 x\]

Answer 2

First option:
\[\frac{ 3+4\cos(2x)+\cos(4x) }{ 8 }\]
Second option:
\[\frac{ 3-4\cos(2x)+\cos(4x) }{ 8 }\]
Third option:
\[\frac{ 3+4\cos(2x)-\cos(4x) }{ 8 }\]
Fourth option:
\[\frac{ 3-4\cos(2x)-\cos(4x) }{ 8 }\]

Answer 3

@TheSmartOne @wio do you know if you can help me?

Answer 4

I think you need to use half angle formula.

Answer 5

I'm kinda clueless when it comes to trig. i have heard of the half angle formula i just dont know how to apply that to this problem. could you help me step by step?

Answer 6

is it the half angle formula for sine?

Answer 7

@wio am i like on the right track?

Answer 8

\[
\sin^2(x) = \frac{1-\cos(2x)}{2}
\]

Answer 9

thats the half angle formula right?

Answer 10

You have to square this value, and then you'll have to use the half angle for cosine.\[
\cos^2(x) = \frac{1+\cos(2x)}{2}
\]

Answer 11

They are called half angle formula, because the initial angle \(x\) is half of the resulting angle \(2x\).

Answer 12

oh ok so

\[\sqrt{\cos^2x}= \sqrt{\frac{ 1+\cos(2x) }{ 2 }}\]

Answer 13

Yes, that is true.

Answer 14

so would it be sqrt cos^2x = cos x

Answer 15

No, \[
\sqrt{\cos^2(x)} = |\cos (x)|
\]But there are no square roots in this equations anyway.

Answer 16

oh ok so then what would it be? im kinda confused.... then how could you have no square roots?

Answer 17

\[
(\sin^2 x) (\sin^2 x)
=
\left(\frac{1-\cos(2x)}{2}\right)
\left(\frac{1-\cos(2x)}{2}\right)
\]First foil this out.

Answer 18

What do you get when you multiply them?

Answer 19

ok so sin^4x?

Answer 20

like sin^2x*sin^2x=sin^4x?

Answer 21

@wio is that right?

Answer 22

Yes

Answer 23

ok so how would that relate to any of the options that was provided to me? thats the part im confused about...

Answer 24

That is why I have been giving you half angle formula. Why did you completely disregard it to talk about \(\sin^4(x)\)?

Answer 25

='( i'm really sorry but im really confused :_(

Answer 26

\[
(\sin^2 x) (\sin^2 x)
=
\left(\frac{1-\cos(2x)}{2}\right)
\left(\frac{1-\cos(2x)}{2}\right)
\]Multiply this

Answer 27

im not quite sure how to do that correctly...

Answer 28

It's a fraction.