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# p2 trig watch

1. using the half angle formulae, how would you get from 1 + cosx to 1 + ( 2 cos^2 x/2 - 1 )

And similarly how would you get from 1 - cosx to 1 - ( 1- 2sin^2 x/2 )?

Anyone?
2. (Original post by Jackal123)
using the half angle formulae, how would you get from 1 + cosx to 1 + ( 2 cos^2 x/2 - 1 )
And similarly how would you get from 1 - cosx to 1 - ( 1- 2sin^2 x/2 )?
Anyone?
You know that cos2x = 2cos^2 x - 1 = 1-2sin^2 x
Replace x with (x/2) [and hence 2x with x.]

cosx=2cos^2(x/2) - 1 = 1-2sin^2(x/2)
3. Well as you know 1+cos(x+y)=1+cosxcosy-sinxsiny

=> 1+cos(2x)=1+cos(x+x)=1+cosxcosx-sinxsinx

Therefore:

1+cos(2x)=1+((cosx)^2)-((sinx)^2)

cos(x) can be obtained by multiplying 2x by (1/2). So you multiply every "x" by (1/2) (half angle):

1+cos(x)=1+((cos(x/2))^2)-((sin(x/2)^2) (i)

Substituting into (i) ((sin(x/2)^2)=1-((cos(x/2))^2)

Gives 1+cos(x)=1+(2((cos(x/2))^2)-1) Q. E. D.

Substituting into (i) ((cos(x/2))^2)=1-((sin(x/2)^2)

Gives: 1-cos(x)=1-(1-2((sin(x/2)^2)) Q. E. D.

Newton.

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