Question #9d630

Question

857 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-21T03:52:56

See below

$cos^3x*sin^2x$

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