Question

prove the identity
sin^2 acos^2=1/8(1-cos4a)

Answer 1

Uhh... this one?
\[\Large \sin^2(a) \cos^2(a) = \frac1{8}[1-\cos^4(a)]\]

Answer 2

@terenzreignz Thank you for asking that. That's what I was wondering too :D @eyna_nuzrey

Answer 3

@terenzreignz @genius12 nope it is cos4(a)

Answer 4

Whoops\[\Large \sin^2(a) \cos^2(a) = \frac1{8}[1-\cos(4a)]\]

Answer 5

@terenzreignz yes that's it!

Answer 6

Well, there's not much we can do about the right-side, however, we can do something to the left...
I suggest expanding cos(4a) using the double angle identities and the fact that
4a = 2(2a)

Answer 7

whoops, got it backwards... I meant there's not much we can do about the *left* side, we can do something with the *right*
sorry, my bad...

Answer 8

ouh ok then?

Answer 9

Yes, so... do that, expand cos(4a) and tell me what you get.

Answer 10

To refresh you on the double angle identity for cosine, we have
\[\large \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)\]
With that in mind, what is
\[\Large \cos(4a) = \cos[2(2a)]=\color{red}?\]

Answer 11

erm just substitute the formula?

Answer 12

Yes, substitute (2a) for \(\theta\) and you get...?

Answer 13

By the way, our main objective is to somehow turn the right-side of the equation into the left side.

Answer 14

\[\cos[2(\cos ^{2}a-\sin ^{2}a]\]

Answer 15

No. First, expand this
\[\Large \sin^2(a) \cos^2(a) = \frac1{8}[1-\color{red}{\cos(4a)}]\]
Using this idea
\[\large \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)\]

With
\[\Large \theta = 2a\]

Answer 16

then?

Answer 17

Then? You did incorrectly :)
I said expand
\[\Large \cos(4a)= \cos[2(2a)]=\color{red}?\]
Using this principle \(\large \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)\)

Answer 18

i don't understand which one i should expand?

Answer 19

ouh i know!

Answer 20

Cosine of 'twice something' regardless of what that something is, say, represent it with this triangle...
\[\Large \cos(2\Delta)\]
is just
\[\Large \cos^2(\Delta) -\sin^2(\Delta)\]
It just so happened that in this case, our 'triangle' is 2a.
\[\Large \cos(4a) = \cos[2(2a)]\]

Answer 21

ok i'm starting become blurr..

Answer 22

What is
cos(2a) ?

Answer 23

Expand it, using the double-angle identity...

Answer 24

Step by step solution

http://www.mathskey.com/question2answer/4229/prove-the-identity