Question

Rewrite with only sin x and cos x.

cos 3x

cos x - 4 cos x sin^2 x
-sin^3 x + 2 sin x cos x
-sin^2 x + 2 sin x cos x
2 sin^2 x cos x - 2 sin x cos x

Answer 1

correct Answer is A,
`cos3x=cosx -4sinx^(2)* cosx`

Explanation:

Theory:

`cos(A+B)=cosA* cosB-sinA *`sinB.

`cos(A-B)=cosA* cosB+sinA *`sinB.

`sin(A+B)=sinA* cosB+cosA *`sinB

`sin(A-B)=sinA* cosB-cosA *`sinB

By using above formula, we write as,

`cos(2x)=cos(x+x)= cosx* cosx-sinx* sinx.`

`cos2x=cosx^(2) - sinx^(2)`

Also,
`sin2x=sin(x+x)=sinx* cosx+cosx* sinx`

`sin2x=2sinx* cosx`

Now,

`cos(3x)=cos(2x+x)=cos2x* cosx-sin2x*sinx`

`cos3x={cosx^(2) - sinx^(2)* cosx}-{2sinx* cosx* sinx}`

`cos3x= {cosx^(3) - sinx^(2)* cosx} - {2sinx^(2)* cosx}`

`cos3x= cosx^(3) - 3sinx^(2)* cosx`

`cos3x= cosx^(2)* cosx -3sinx^(2)* cosx`

`cos3x= (1-sinx^(2))* cosx -3sinx^(2)* cosx`

`cos3x= (cosx -sinx^(2)* cosx) -3sinx^(2)* cosx`

`cos3x=cosx -4sinx^(2)* cosx`

thus, correct Answer is A,
`cos3x=cosx -4sinx^(2)* cosx`