(adsbygoogle = window.adsbygoogle || []).push({}); 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

--------- + ------------

factor quadractic

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?

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

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