prove that 1-sinx/secx=cos3x/1+sinx

Question

I need the answer asap

Answer 1

Hi Kevin,

 

Looks to me like the problem was not transcribed correctly, but the gist of the problem will be what to do with cos 3x.

 

cos3x= cos(2x+x)= cos2xcosx - sin2xsinx (trig identity)

and there are other trig identities for cos2x and sin2x. Use those identities to further transform the above line.

 

Hopefully you can take it from there.

 

Here you go.

 

My guess is that this is really cos³x (rather than cos3x).

 

So:

 

(1-sinx)/secx =? cos³x/(1+sinx)

 

Transform the right side to prove it is identical to the left side.

 

cosx (cos²x)/( 1+sinx) = cosx (1-sin²x) / (1+sinx) = cosx (1-sinx)(1+sinx) / (1+sinx) (difference of squares)

 

Now the factors (1+sinx) cancel and we have:

 

cosx (1-sinx) = (1-sinx)/secx (Q.E.D.)