Question

cos >3x- cos3x/cosx + sin>3x + sin 3x/ sinx​

Answer 1

Step-by-step explanation:

sin3x=3sinx−4sin

3

x

∴sin

3

x=

4

1

(3sinx−sin3x)

Similarly, cos

3

x=

4

1

(3cosx+cos3x)

L.H.S.=

4

1

[sin3x(3sinx−sin3x)+cos3x(3cosx+cos3x)]

=

4

1

[3cos(3x−x)+(cos

2

3x−sin

2

3x)]

=

4

1

[3cos2x+cos6x]=

4

1

[3cosA+cos3A]

=cos

3

A,by(2)=cos

3

2x ∵A=2x