Answer:
Proof in explanation
Step-by-step explanation:
Trigonometric Identities
The basic trigonometric identity is:
[tex]\sin^2\theta+\cos^2\theta=1[/tex]
We'll use it and some basic algebra to prove that, given:
[tex]a sin^3\theta+b cos^3\theta=sin\theta cos\theta[/tex]
And
[tex]a\sin\theta-b\cos\theta=0[/tex]
Then
[tex]a^2+b^2=1[/tex]
From the equation:
[tex]a\sin\theta-b\cos\theta=0[/tex]
We have:
[tex]a\sin\theta=b\cos\theta\qquad [1][/tex]
The equation
[tex]a sin^3\theta+b cos^3\theta=sin\theta cos\theta[/tex]
Can be rewritten as
[tex]a\sin\theta \sin^2\theta+b \cos^3\theta=\sin\theta \cos\theta[/tex]
Replacing [1]:
[tex]b\cos\theta \sin^2\theta+b \cos^3\theta=\sin\theta \cos\theta[/tex]
Taking the common factor:
[tex]b\cos\theta (\sin^2\theta+ \cos^2\theta)=\sin\theta \cos\theta[/tex]
The expression in parentheses is 1, thus:
[tex]b\cos\theta =\sin\theta \cos\theta[/tex]
Dividing by [tex]\cos\theta[/tex]
[tex]b=\sin\theta[/tex]
Replacing in
[tex]a\sin\theta=b\cos\theta[/tex]
We have
[tex]a\sin\theta=\sin\theta\cos\theta[/tex]
Dividing by [tex]\sin\theta[/tex]
[tex]a=\cos\theta[/tex]
Now:
[tex]a^2+b^2=(\cos\theta)^2+(\sin\theta)^2[/tex]
This expression is 1, thus it's proven:
[tex]\boxed{a^2+b^2=1}[/tex]
