Trigonometry! #90

Geometry Level 2

If $a\cos\theta+b\sin\theta=m$ and $a\sin\theta-b\cos\theta=n$ then $a^{2}+b^{2}$ is equal to

This problem is part of the set Trigonometry .

m^{2}+n^{2}

m^{2}n^{2}

m^{2}-n^{2}

n^{2}-m^{2}

1 solution

Tom Engelsman
Mar 29, 2016

Squaring both sides of either equation, one obtains:

a^2 * cos(x)^2 + 2ab*cos(x)sin(x) + b^2 * sin(x)^2 = m^2 (i)

a^2 * sin(x)^2 - 2ab*cos(x)sin(x) + b^2 * cos(x)^2 = n^2 (ii)

Adding (i) to (ii) produces:

a^2 * [cos(x)^2 + sin(x)^2] + b^2 * [cos(x)^2 + sin(x)^2] = m^ + n^2,

or (a^2 + b^2) [cos(x)^2 + sin(x)^2] = (a^2 + b^2) 1 = m^2 + n^2.