How to prove that sin square x cos square x is identical

1. Oct 20, 2011

jigoku_snow

1. The problem statement, all variables and given/known data
1) how to prove that sin square x cos square x is identical to 1/8 (1-
cos 4x)?

2) f(x) = x square +7x -6
----------------, show that when x is sufficiently small for
(x-1)(x-2)(x+1)
x^4 and higher powers to be neglected.

2. Relevant equations
1) cos (s + t) = cos s cos t – sin s sin t
cos 2t = cos2 t – sin2 t = 2 cos2 t – 1 = 1 – 2 sin2 t
sin 2t = 2 sin t cos t

2) A Bx + C
--------- + ------------

3. The attempt at a solution
1) from RHS:
1- cos (2x +2x) 1-[ cos 2x cos 2x - sin 2x sin 2x]
-------------- = ----------------------------------
8 8
1- cos^4 x + sin^4 x + 6 sin^2 x cos^2 x
----------------------------------------
8
* how to do the next step? is my solutions are correct so far?

2) (x^2+7x-6)(x-1)^-1 ( x-2)^-1 (x+1)^-1
expand (x-1)^-1 = 1+x+x^2+x^3
(x-2)^-1 = 1/2 + x/4 + x^2/8 + x^3/16
(1+x)^-1 = 1- x + x^2 - x^3
multiply all : (x^2+7x-6)(1+x+x^2+x^3)( 1/2 + x/4 + x^2/8 + x^3/16)
(1- x + x^2 - x^3)
= 3x^2/2 - x^3/4 +2x -3
* however , the answer I obtain is wrong. which part of my solution is
wrong?

2. Oct 20, 2011

obafgkmrns

Re: polynomials

For the trig problem, try using the 2nd line of your relevant equations to get an expression for (cos x)^2 in terms of cos 2x. Then plug that expression into the Pythagorean identity to get an expression for (sin x)^2.

Now, multiply those to get an expression for ((cos x)^2)*((sin x)^2) in terms of cos 2x. The next step should easy.

I'm not sure what you're after in the rational function problem. Looks like if x is sufficiently small, you can ignore powers of x greater than 1 -- in the limit, the expression goes to -3.

3. Oct 20, 2011

Staff: Mentor

Re: polynomials

You have not made it clear what you are trying to do, but I think you are trying to write a Maclaurin series representation for your rational function.

I see a mistake in this line:
It should be
(x - 2)-1 = -1/(1 - x/2) = -(1 + x/2 + x2/4 + x3/8 + ...)