Question

Use the squared identities to simplify sin^2x cos^2x.
the choices: http://i.imgur.com/B8L5uqp.png

Answer 1

:/

Answer 2

Well the identity for cos(2x) is sin^2 x - cos^2 x so that would the the start i'd say

Answer 3

Which means sin^2 x = cos 2x + cos^2 x hmm would that help

Answer 4

I just get: \[cos^2x-cos^4x\\sin^2x-sin^4x\\\frac{1}{8}(1-cos(4x))\].

Answer 5

2nd: \[sin^2x(1-sin^2x)\] from pythag.

Answer 6

just third*

Answer 7

\[sin^2xcos^2x = \frac{1-cos(4x)}{8}\]

Answer 8

Maybe use
\(cos^2x = \dfrac{cos2x+1}{2}\)

Answer 9

That third option, i would never know why its true in an exam xD

Answer 10

Here's what I think you'd do

\(cos^2x-cos^4x =\left(\dfrac{cos2x+1}{2} - \left(\dfrac{cos2x + 1}{2}\right)^2\right)\)

Answer 11

You only found the answer, but do you know why it is so?

Answer 12

http://www.wolframalpha.com/input/?i=sin%5E2x+cos%5E2x.+

Answer 13

the answer is c

Answer 14

I just need to clarify your doubts, ShantelT1

We know that : \(cos^4x = \dfrac{3}{8} + \dfrac{cos2x}{2} + \dfrac{cos4x}{8}\)

Substitute this trig identity in the first, \(\dfrac{cos2x+1}{2} - cos^4x\)
to get :

\(\dfrac{cos2x+1}{2} - \left(\dfrac{3}{8} + \dfrac{cos2x}{2} + \dfrac{cos4x}{8}\right)\longrightarrow \dfrac{4cos2x+4-3-4cos2x+4x}{8}\)

Therefore,
\(sin^2(x) ~cos^2(x) = \dfrac{1+cos4x}{8}\)

Answer 15

how do you know the first part? :D with cos^4 x ?

Answer 16

Proof's a bit long but it's one of those rare power identities you should know. Check this out: http://en.wikipedia.org/wiki/List_of_trigonometric_identities

Answer 17

You really don't have to cram this up, a concised reference book would do. (: