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March 6, 2015

March 6, 2015

Posted by **Alex** on Tuesday, November 27, 2007 at 11:17pm.

g) 8sin^2x-4

I just don't get this one. I know it's got something to do with the 1-2sin^2x double angle formula. It's the opposite though? :S

h) 1-2sin^2 (π/4-x/2)

= 1-sin^2(π/4-x/2)-sin^2(π/4-x/2)

= cos^2(π/4-x/2)-sin^2(π/4-x/2)

I got all the way up to cos (π/4 - x/2 + π/4 - x/2)

The answer is supposed to be sin x. I have no clue how they got that.

- Math 2nd question -
**Reiny**, Tuesday, November 27, 2007 at 11:38pmg) 8sin^2x-4

= 4(2sin^2 x - 1)

= 4(-cos 2x)

= -4cos 2x

h) 1-2sin^2 (π/4-x/2)

= cos 2(π/4-x/2)

= cos (π/2-x)

= cos(π/2)cosx + sin(π/2)sinx

= 0(sinx) + 1(sinx)

= sinx

- Math 2nd question -
**Alex**, Tuesday, November 27, 2007 at 11:52pmThanks a lot!

But can you please explain how you got 4(2sin^2 x - 1) for g and cos 2(π/4-x/2) for h?

I think I might understand h because 1-sin^2x = cosx but wouldn't it just be cos(π/4-x/2) cos(π/4-x/2)?

- Math 2nd question -
- Math 2nd question -
**Reiny**, Wednesday, November 28, 2007 at 1:26am"But can you please explain how you got 4(2sin^2 x - 1)"

I took out a common factor of 4

"..and cos 2(π/4-x/2) for h"

Ok, let's work it in reverse.

You know that cos 2A = cos^2 A - sin^2 A, or**cos 2A = 1 - 2sin^2 A**,right?

so I simply let A = (π/4-x/2)

then 2A = 2(π/4-x/2)

= π/2 - x

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