3
$\begingroup$

I don't know how to show that $a+b\sin(x)+c\cos(x)=0$ has no solutions...

How should I go about doing this? Is the only way using the $\sin^2+\cos^2=1$ identity and substituting that?

  • 0
    Thank you @Théophile for these clarifications. I admit I was a bit rushed when I was typing this. Although you might already noticed this, but originally I was asking this for the intent of showing the linear independence. Hence, I meant to say "has no non-trivial solutions."2012-05-28

1 Answers 1

6

Substitute values of $x$ in the supposed identity $a+b\sin x +c\cos x=0$. Use for example $x=0$, $x=\pi/2$, $x=-\pi/2$. Then conclude that $a=b=c=0$.

The above is probably the simplest approach, and works nicely in other situations. But other ideas can be brought to bear on the problem. For example, if $a+b\sin x+c\cos x$ is identically $0$, then, by differentiating, we can conclude that $b\cos x-c\sin x$ is identically $0$. Put $x=0$. We conclude that $b=0$. Differentiate again, set $x=0$. We conclude that $c=0$. Then looking back at $a+b\sin x +c\cos x$, we conclude that $a=0$.

  • 0
    Ok. This implies that $a,b,c= 0$. the only case where we have solution.2012-05-27