Question

A sin^3theta+b cos^3theta=sintheta costheta and a sintheta-b costheta=0 then prove a^2+b^2=1​

Answer 1

Answer:

Proof in explanation

Step-by-step explanation:

Trigonometric Identities

The basic trigonometric identity is:

$\sin^2\theta+\cos^2\theta=1$

We'll use it and some basic algebra to prove that, given:

$a sin^3\theta+b cos^3\theta=sin\theta cos\theta$

And

$a\sin\theta-b\cos\theta=0$

Then

$a^2+b^2=1$

From the equation:

$a\sin\theta-b\cos\theta=0$

We have:

$a\sin\theta=b\cos\theta\qquad [1]$

The equation

$a sin^3\theta+b cos^3\theta=sin\theta cos\theta$

Can be rewritten as

$a\sin\theta \sin^2\theta+b \cos^3\theta=\sin\theta \cos\theta$

Replacing [1]:

$b\cos\theta \sin^2\theta+b \cos^3\theta=\sin\theta \cos\theta$

Taking the common factor:

$b\cos\theta (\sin^2\theta+ \cos^2\theta)=\sin\theta \cos\theta$

The expression in parentheses is 1, thus:

$b\cos\theta =\sin\theta \cos\theta$

Dividing by $\cos\theta$

$b=\sin\theta$

Replacing in

$a\sin\theta=b\cos\theta$

We have

$a\sin\theta=\sin\theta\cos\theta$

Dividing by $\sin\theta$

$a=\cos\theta$

Now:

$a^2+b^2=(\cos\theta)^2+(\sin\theta)^2$

This expression is 1, thus it's proven:

$\boxed{a^2+b^2=1}$