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.