Question

Question #9d630

Answer 1

See below

Explanation:

`cos^3x*sin^2x`

We know `sin^2x = 1 - cos^2x`
Put this value in the equation

= `cos^3x(1-cos^2x)`
= `cos^3x - cos^5x`

Also you can put any value of x to verify
I am putting x = 60°
`sin60° = sqrt3/2`
`cos60° = 1/2`

L.H.S
`cos^3x*sin^2x`

= `cos^3 60°*sin^2 60°`

= `(1/2)^3*(sqrt3/2)^2`

= `1/8*3/4`

= `3/32`

R.H.S
`cos^3x - cos^5x`

= `cos^3 60° - cos^5 60°`

= `(1/2)^3 - (1/2)^5`

= `1/8 - 1/32`

= `(4 - 1)/32`

= `3/32`

L.H.S = R.H.S

Hence verified