since sin^2θ = (1-cos2θ)/2, we have
(1-cos20)/2 + (1-cos100)/2 + (1-cos140)/2 = 3/2
cos20+cos100+cos140 = 0
Using the sum-to-product rules,
cos20 + cos100+cos140
= cos20 + 2cos120cos20
= cos20 - cos20
= 0
How do we show (sin 10)^2 + (sin 50)^2 + (sin 70)^2 = 3/2 ???
I can't even think of a way to start this..
2 answers
Thank you very much Steve