Question

Rewrite with only sin x and cos x.

cos 3x

Answer 1

hint : 3x =2x+x , then 2x = x+x
and use the formula \(\cos(A+B)=\cos A \cos B - \sin A\sin B\)
twice, once when A=2x, B=x....then when A=x, B=x

Answer 2

.
.
.
=cosx( 2(cosx)^2 -1 -2(sinx)^2 ) ??

Answer 3

then what would i get?

Answer 4

cos(2x+x)=cos2xcosx-sin2xsinx;
cos(x+x)=cosxcosx-sinxsinx=(cosx)^2 - (sinx)^2;

sin(2x+x)=sin2xcosx+sinxcos2x;
sin(x+x)=sinxcox+sinxcosx=2sinxcosx;

cos3x=cos2xcosx-sin2xsinx;
=((cosx)^2 - (sinx)^2)cosx - (2sinxcosx)sinx
=( 2(cosx)^2 -1 )cosx -2cosx(sinx)^2
=2(cosx)^3-cosx-2cosx(sinx)^2
=(cosx) [ 2(cosx)^2 - 1 -2(sinx)^2) ]
=(cosx) [ 4(cosx)^2 -3 ]

Answer 5

It's either
A. cosx -4 cosx sin^2x
B. -sin^3x+2 sinx cosx
C. -sin^2x+ 2sinx cosx
D. 2sin^2xcosx-2sinxcosx

Answer 6

A

Answer 7

from my answer
=(cosx) [ 4(cosx)^2 -3 ]
=4(cosx)^3-3cosx

substitute (cosx)^2 = 1 - (sinx)^2 into the question

=4cosx(cosx)^2-3cosx
=4cosx [ 1 - (sinx)^2 ]-3cosx
=4cosx - 4cosx(sinx)^2-3cosx
=cosx- 4cosx(sinx)^2..
A

Answer 8

thank you :)

Answer 9

your wc :D