Question

write the expression in terms of sin x and cos x only. sin(2x)+cos(3x)

Answer 1

cos3x = cos (2x+ x)
=cos2x*cosx - sin2x*sinx
=(2*cos^2 (x) - 1) *cosx - 2sinx*cosx*sinx { since cos2x = 2cos^2 (X) -1 , sin2x = 2*sinx*cosx }
=2 cos^3 (x) - cos x - 2 cosx *sin^2 (x)
= 2 cos^3 (x) - cos x (1 + 2 sin^2 (x) )
= 2 cos^3 (x) - cos x (1 + 2 ( 1 - cos^2 (x) ))
= 2 cos^3 (x) - cos x ( 1+ 2 - 2*cos^2(x) )
=2 cos^3 (x) - cos x (3 -2*cos^2(x) )
=2 cos^3 (x) - 3cos x + 2*cos^3(x)
=4 cos^3 (x) - 3cos x

Answer 2

sin(2x+x) = cos(2x)sin(x) + sin(2x)cos(x)
Then substitute the double angle formulae for the 2xs
= (cos^2(x) - sin^2(x))sin(x) + 2cos(x)sin(x)cos(x)
// factor out the sinx
= sin(x) (cos^2(x)-sin^2(x) + 2 cos^2(x))
=sin(x) (3cos^2(x) - sin^2(x))

Answer 3

but for the second one it was sin(2x) not sin(3x) and using the identity the sin(2x) can become 2sin(x)cos(x) when would make the problem turn into 2sin(x)cos(x)+4 cos^3 (x) - 3cos x is that the final answer? or- I'm actually kinda confused on what the question is asking, does the answer need to using just sin or just cos, or can it use both? I copied the question word for word from the book onto here

Answer 4

2sin(x)cos(x)+4 cos^3 (x) - 3cos x is the final answer