sin^4 x = (3-4cos 2x + cos4x)/8
prove on one side please!
4 answers
just simplify both sides! easy as that!!!
how?
RS
= (3 - 4cos(2x) + cos^2(2x) - sin^2(2x))/8
= (3 - 4cos(2x) + cos^2(2x) - (1 - cos^2(2x)))/8
= (2cos^2(2x) - 4cos(2x) + 2)/8
= (2(cos^2(2x) - 2cos(2x) + 1))/8
= (2(cos(2x) - 1)^2)/8
= (2(1 - 2sin^2(x) - 1)^2)/8
= (2(-2sin^2(x))^2)/8
= (8sin^4(x))/8
= sin^4(x)
= LS
hope I made no typos this time
= (3 - 4cos(2x) + cos^2(2x) - sin^2(2x))/8
= (3 - 4cos(2x) + cos^2(2x) - (1 - cos^2(2x)))/8
= (2cos^2(2x) - 4cos(2x) + 2)/8
= (2(cos^2(2x) - 2cos(2x) + 1))/8
= (2(cos(2x) - 1)^2)/8
= (2(1 - 2sin^2(x) - 1)^2)/8
= (2(-2sin^2(x))^2)/8
= (8sin^4(x))/8
= sin^4(x)
= LS
hope I made no typos this time
1-4cos2x+cos4x/8=sin4x