Question

Rewrite with only sin x and cos x.

Answer 1

Answer:

correct Answer is A, $cos3x=cosx -4sinx^{2}\times cosx$

Step-by-step explanation:

Theory:

$cos(A+B)=cosA\times cosB-sinA \times$sinB.

$cos(A-B)=cosA\times cosB+sinA \times$sinB.

$sin(A+B)=sinA\times cosB+cosA \times$sinB

$sin(A-B)=sinA\times cosB-cosA \times$sinB

By using above formula, we write as,

$cos(2x)=cos(x+x)= cosx\times cosx-sinx\times sinx.$

$cos2x=cosx^{2} - sinx^{2}$

Also,$sin2x=sin(x+x)=sinx\times cosx+cosx\times sinx$

$sin2x=2sinx\times cosx$

Now,

$cos(3x)=cos(2x+x)=cos2x\times cosx-sin2x\timessinx$

$cos3x={cosx^{2} - sinx^{2}\times cosx}-{2sinx\times cosx\times sinx}$

$cos3x= {cosx^{3} - sinx^{2}\times cosx} - {2sinx^{2}\times cosx}$

$cos3x= cosx^{3} - 3sinx^{2}\times cosx$

$cos3x= cosx^{2}\times cosx -3sinx^{2}\times cosx$

$cos3x= (1-sinx^{2})\times cosx -3sinx^{2}\times cosx$

$cos3x= (cosx -sinx^{2}\times cosx) -3sinx^{2}\times cosx$

$cos3x=cosx -4sinx^{2}\times cosx$

thus, correct Answer is A, $cos3x=cosx -4sinx^{2}\times cosx$