Question

Let alpha and beta be the solutions of the equation f(x)=0,where f(x)=p*cos x+q*sin x,show that f(x) vanishes identically if alpha-beta not equal to n*pi,n belongs to integers.

Answer 1

help pls

Answer 2

yey

Answer 3

First we need a trig identity.

Consider the dot product of two vector (p,q) and (cos x, sin x). Note that (p,q) makes an angle to positive x-axis, phi = arctan(q/p).

Then\[(p,q).(\cos x, \sin x) = p \cos x + q \sin x\]
also
\[(p,q).(\cos x, \sin x) = \sqrt{p^2 + q^2} \cos (x - \phi)\]

So the two expressions are equal.

Answer 4

Suppose now that alpha and beta are zeros of
f(x)=p*cos x+q*sin x

i.e., zeros of
\[f(x) = \sqrt{p^2 + q^2} \cos(x - \phi)\]
The cos part of this function, cos(x - phi) has zeros at x = k.pi - phi for integers k

Answer 5

thus if
\[\alpha - \beta \neq k \pi \text{ , for some integer k}\]
Then at least one of alpha and beta are not zeros of cos(x - phi)

Hence \[\sqrt{p^2 + q^2} = 0 \ \ \ \text{ i.e., } p = q = 0\]

Answer 6

CORRECTION
The function cos(x - phi) has zeros at
\[x = k \pi + \phi \text{, for integers } k\]