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- July 5th 2008, 07:09 AM #1

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- July 5th 2008, 07:47 AM #2

- July 5th 2008, 08:54 AM #3

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- July 5th 2008, 10:08 AM #4
## Here is what you need!

Hi!gracey,hope ur having a great time!

cosx=(cos(x/2))^2-(sin(x/2))^2

since (cos(x/2))^2+(sin(x/2))^2=1

and A=A/1

so cosx=[(cos(x/2))^2-(sin(x/2))^2]/1

=[(cos(x/2))^2-(sin(x/2))^2]/

[(cos(x/2))^2+(sin(x/2))]

dividing numerator and denominator by (cos(x/2))^2 we get

cosx=[1-(tan(x/2))^2]/[1+(tan(x/2))^2]

now cross multiply

[1+(tan(x/2))^2]cosx=1-(tan(x/2))^2

cosx+cosx(tan(x/2))^2=1-(tan(x/2))^2

cosx(tan(x/2))^2+(tan(x/2))^2=1-cosx

[(tan(x/2))^2][cosx+1]=1-cosx

(tan(x/2))^2=(1-cosx)/1+cosx

or

(1-cosx)/1+cosx=(tan(x/2))^2