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# Maths compound angles. Please help!?

Hi, I am trying to solve the following sums, and cannot. I am keen not to fall behind, as previously I was find this work relatively easy. Anyway here are the sums:

Prove each identity, ignoring any restrictions: (they are NOT equations)

3. tanx – 1/(tanx) = -2/(tan2x)

5. cos^4x – sin^4x = cos2x

11. 8sinA.cosA.cos2A.cos4A = sin8A

Thanks a lot

### 3 Answers

- 1 decade agoFavorite Answer
3. Tan x = sinx/cosx and 1/tanx = cos x/sin x. Use these facts to rewrite the left side as

sin x/cos x-cos x/sin x = -2/(tan 2x) get a common denominator

(sin^2 x - cos^2 x)/(sin x cos x) = -2/(tan2x) Factor out a negative one from the numerator

-(cos^2 x - sin^2 x)/(sin x cos x) = -2/(tan2x) Recognize the double angle identity for cosine

-cos 2x/(sin x cos x) = -2/(tan2x) Notice the denominator is close to being the double angle identity for sine, but is missing a 2. Multiply top and bottom by 2 and change to sin 2x.

-2cos 2x/sin 2x = -2/(tan 2x) Cos over sin is reciprocal of tan. So tan 2x goes in denominator

-2/tan 2x = -2/tan 2x

5. Start by factoring the left side like you would a difference of two squares.

(cos^2 x -sin^2 x)(cos^2 x + sin^2 x) = cos2x

You should know that the second set of parentheses equals 1. So:

cos^2 x -sin^2 x = cos 2x

The left side is now one of the double angle identities.

cos2x = cos 2x

7. This one is easier to simplify the right side. You use the double angle identity for sine over and over again.

8 sin A cosAcos2A cos 4A = sin 8A

8 sin A cosAcos2A cos 4A =2sin4A cos4A

8 sin A cosAcos2A cos 4A = 2(2sin2A cos2A) cos4A

8 sin A cosAcos2A cos 4A = 2[2(2sinA cosA) cos2A] cos4A

8 sin A cosAcos2A cos 4A = 8 sinAcosA cos2A cos4A

- CDALv 61 decade ago
You need to be familiar with the 'half-angles ' formulas.

3)

tan2x = 2tanx / (1-tan^2x) [presumably you know this formula]

So - 2/tan2x = -2 (1 - tan^2x) / 2tanx

= ( tan^2x - 1) / tanx

= tanx - 1/tanx

5)

cos^4x - sin^4x = (cos^2x - sin^2x)(cos^2x + sin^2x)

The last bracketed expression = 1

So cos^4x - sin^4x = cos^2x - sin^2x

= cos^2x - (1 - cos^2x)

= 2cos^2x - 1

= cos(2x)

7)

sin8A = 2sin4Acos4A [half angle formula]

= 4sin2Acos2A. cos4A

= 8sinAcosA. cos2A.cos4A

Done!

Source(s): standard trig half-angle formulas - JOS JLv 71 decade ago
tanx – 1/(tanx) = -2/(tan2x)

( tanx – 1/(tanx) )+(2/(tan2x) = 0

------------------------------------------------------------------

cos^4x – sin^4x = cos2x

(cos^4x – sin^4x) -( cos2x) = 0

-----------------------------------------------

8sinA.cosA.cos2A.cos4A = sin8A

(8sinA.cosA.cos2A.cos4A -(sin8A) = 0