how do you verify this identity

Sin^3βcosβ+cos^3βsinβ=sinβcosβ

LS = Sin^3βcosβ+cos^3βsinβ

= sinβcosβ(sin^2 β + cos^2 β)
= sinβcosβ (1)
= RS

Now wasn't that easy ?

thank you !!!!

welcome!

To verify the given identity:

Sin^3βcosβ + cos^3βsinβ = sinβcosβ

we will use basic trigonometric identities to manipulate the expressions on both sides of the equation until they become equal.

Let's start by rewriting sin^3β and cos^3β using the power-reducing identities:

sin^3β = (sinβ)^2*sinβ
cos^3β = (cosβ)^2*cosβ

Now we can substitute these expressions back into the original equation:

(sinβ)^2*sinβ*cosβ + (cosβ)^2*cosβ*sinβ = sinβ*cosβ

Next, we rearrange the terms by factoring out sinβ*cosβ:

(sinβ*cosβ) * [(sinβ)^2 + (cosβ)^2] = sinβ*cosβ

Remember that sin^2θ + cos^2θ = 1 for any angle θ. Applying this identity to our equation, we have:

(sinβ*cosβ) * [1] = sinβ*cosβ

Simplifying further:

sinβ*cosβ = sinβ*cosβ

Therefore, we have verified the identity.