# Thread: Proof there is no real roots....

1. ## Proof there is no real roots....

Hi

Given that $a^2 + b^2 < c^2$, 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.

2. Hi

$a \cos x + b \sin x = c$

Divide by $\sqrt{a^2+b^2}$

$\frac{a}{\sqrt{a^2+b^2}}\: \cos x + \frac{b}{\sqrt{a^2+b^2}}\: \sin x = \frac{c}{\sqrt{a^2+b^2}}$

$\frac{a}{\sqrt{a^2+b^2}}$ and $\frac{b}{\sqrt{a^2+b^2}}$ are two reals such that the sum of their square is equal to 1

$\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2 + \left(\frac{b}{\sqrt{a^2+b^2}}\right)^2 = \frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2} = 1$

Therefore there exists $y$ such that

$\cos y = \frac{a}{\sqrt{a^2+b^2}}$ and $\sin y = \frac{b}{\sqrt{a^2+b^2}}$

The equation becomes

$\cos x \:\cos y + \sin x \:\sin y = \frac{c}{\sqrt{a^2+b^2}}$

$\cos (x-y) = \frac{c}{\sqrt{a^2+b^2}}$

And since $\cos (x-y) \leq 1$ and $\frac{c}{\sqrt{a^2+b^2}} > 1$ there are no solution