Question

If $x=asin$ $\theta$ $+bcos$ $\theta$, $y=acos$ $\theta$ $-bsin$ $\theta$ ,

then show that $(ax+ay{)}^2$ + $(bx-ay{)}^2$ = $(a^2+b^2{)}^2$.

Answer 1

Find the values of $cos\theta$ and $sin\theta$.

$x=a\sin \theta +b\cos \theta$ ...(i)

$y=a\cos \theta -b\sin \theta$ ...(ii)

Multiplying eq.(i) by b and eq. (ii) by a

$bx=ab\sin \theta +b^2\cos \theta$

$ay=-ab\sin \theta +a^2\cos \theta$

Adding resulting equations,

$\frac{bx+ay}{b^2+a^2}=\cos \theta$

similarly, $\sin \theta =\frac{ax-by}{b^2+a^2}$

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

$(ax-by{)}^2+(bx+ay{)}^2=(a^2+b^2{)}^2$

Solving above equation will give:

$\Rightarrow$$cos\theta =\frac{bx+ay}{b^2+a^2}$ and

$\Rightarrow$$sin\theta =\frac{ax-by}{a^2+b^2}$

Squaring and adding both will give -

$\Rightarrow$$(ax-by{)}^2+(bx+ay{)}^2=(a^2+b^2{)}^2$