Prove that (sin4t)/4 -= cos^3tsint-sin^3tcost ?
1 Answer
Feb 14, 2018
We seek to show that:
(sin4t)/4 -= cos^3tsint-sin^3tcost
Consider the LHS:
LHS = (sin4t)/4
Using the sine and cosine double angle formulas:
sin2A -= 2sinAcosA
cos2A -= cos^2A-sin^2A
we have:
LHS = (sin(2(2t)))/4
\ \ \ \ \ \ \ \ = (2sin2tcos2t)/4
\ \ \ \ \ \ \ \ = (2(2sintcost)(cos^2t-sin^2t))/4
\ \ \ \ \ \ \ \ = sintcost(cos^2t-sin^2t)
\ \ \ \ \ \ \ \ = sintcostcos^2t-sintcostsin^2t
\ \ \ \ \ \ \ \ = sintcos^3t-costsin^3t
\ \ \ \ \ \ \ \ = RHS \ \ \ \ QED
