Trigonometry Proof

Question

Please help me answer the following question! I'm stuck!

 

Prove that cos⁶ß + sin⁶ß = 1/4 + 3/4cos²ß

 

I've tried simplifying out the LHS into (cos²ß)³ + (sin²ß)³, but I don't know what to do from there.

 

Thank you so much!

Answer 1

use the identity x^3+y^3=(x+y)^3-3xy(x+y)

use n for ß

prove cos^6n+sin^6n=1/4+(3/4)cos^2(2n) (do you mean 2n instead of n ???; 2ß instead of ß ???)

(cos^2n)^3+(sin^2n)^3=1/4+(3/4)cos^2(2n)

using the identity from above...

(cos^2n+sin^2n)^3-3cos^2nsin^2n(cos^2n+sin^2n)=

1-3cos^2nsin^2n(1)=       (sin^2+cos^2=1)

1-(3/4)(4cos^2nsin^2n)=

1-(3/4)(2cosnsinn)^2=

2cosnsinn=sin2n

1-(3/4)(sin2n)^2=

1-(3/4)(sin^22n)=

sin^22n+cos^22n=1

sin^22n=1-cos^22n

1-(3/4)(1-cos^22n)=

1-3/4+(3/4)(cos^22n)=

(1/4)+(3/4)(cos^22n)

 

Answer 2

Factor using general form for sum of cubes

Leading factor = 1

In trinomial use power reduction identities