Hi

Given that $\displaystyle a^2 + b^2 < c^2$, show that the equation acosx + bsin x = c has no real roots.

SOLUTION:

acosx + bsiny = c

as cos(x-y) = c/sqrt(a^2 + b^2)

where cos y = a/sqrt(a^2+b^2) and siny = b/sqrt(a^2 + b^2)

Since a^2 + b^2 < c^2

cos(x-y)>1

Hence, no real solution

Could someone please explain to me how the solution works? I'm confused.