Question

cos theta = [(3a)/(a^2+b^2)] - 2 ; sin theta = (-3b)/(a^2+b^2). Find (a^2+b^2) in terms of a and b only.

Answer 1

4a-3

Answer 2

Given the expressions for $\cos \theta$ and $\sin \theta$:

$\cos \theta = \frac{3a}{a^2+b^2} - 2$

$\sin \theta = \frac{-3b}{a^2+b^2}$

We use the fundamental trigonometric identity: $\cos^2 \theta + \sin^2 \theta = 1$

Substitute the given expressions into the identity: $\left(\frac{3a}{a^2+b^2} - 2\right)^2 + \left(\frac{-3b}{a^2+b^2}\right)^2 = 1$

Expand the first term using $(x-y)^2 = x^2 - 2xy + y^2$: $\left(\frac{3a}{a^2+b^2}\right)^2 - 2 \times \frac{3a}{a^2+b^2} \times 2 + (-2)^2 + \frac{(-3b)^2}{(a^2+b^2)^2} = 1$

$\frac{9a^2}{(a^2+b^2)^2} - \frac{12a}{a^2+b^2} + 4 + \frac{9b^2}{(a^2+b^2)^2} = 1$

Combine the terms with the common denominator $(a^2+b^2)^2$: $\frac{9a^2 + 9b^2}{(a^2+b^2)^2} - \frac{12a}{a^2+b^2} + 4 = 1$

$\frac{9(a^2 + b^2)}{(a^2+b^2)^2} - \frac{12a}{a^2+b^2} + 4 = 1$

Simplify the first term by cancelling one factor of $(a^2+b^2)$ from the numerator and denominator (assuming $a^2+b^2 \neq 0$, which must be true for the original expressions to be defined): $\frac{9}{a^2+b^2} - \frac{12a}{a^2+b^2} + 4 = 1$

Combine the terms with the common denominator $(a^2+b^2)$: $\frac{9 - 12a}{a^2+b^2} + 4 = 1$

Subtract 4 from both sides: $\frac{9 - 12a}{a^2+b^2} = 1 - 4$

$\frac{9 - 12a}{a^2+b^2} = -3$

Multiply both sides by $(a^2+b^2)$: $9 - 12a = -3(a^2+b^2)$

Divide both sides by -3: $\frac{9 - 12a}{-3} = a^2+b^2$

$-3 + 4a = a^2+b^2$

Rearrange the equation to express $(a^2+b^2)$: $a^2+b^2 = 4a - 3$