please tell me the formula of cot(3x)
Answers
Answered by
0
cos 3x = cos(2x+x)
= cos(2x).cos(x) - sin(2x).sin(x)
= {2cos^2 (x) - 1}.cos(x) - {(2sin(x).cos(x)}.sin(x)
= cos(x) {2cos^2 (x) - 1 - 2sin^2 (x)}
=cos(x) {2cos^2 (x) - 1 - 2(1 - cos^2 (x))}
=cos(x) {2cos^2 (x) - 1 - 2 + 2cos^2 (x)}
=cos(x) {4cos^2 (x) - 3}
=4cos^3 (x) - 3cos(x)
HOPE IT HELPS YOU
MARK ME ON BRAINLIEST
= cos(2x).cos(x) - sin(2x).sin(x)
= {2cos^2 (x) - 1}.cos(x) - {(2sin(x).cos(x)}.sin(x)
= cos(x) {2cos^2 (x) - 1 - 2sin^2 (x)}
=cos(x) {2cos^2 (x) - 1 - 2(1 - cos^2 (x))}
=cos(x) {2cos^2 (x) - 1 - 2 + 2cos^2 (x)}
=cos(x) {4cos^2 (x) - 3}
=4cos^3 (x) - 3cos(x)
HOPE IT HELPS YOU
MARK ME ON BRAINLIEST
sprao534:
cot 3x but not cos3x
Answered by
0
bsidvszsd uh hshdgsudgheid uh gun gsussudsus in ya
Similar questions