0
$\begingroup$

This question has been troubling me for days, I really haven't got a clue how to handle it:

$f(x) = -3+2\cos(x)$

$g(x) = \cos(x-\dfrac{1}{4}\pi)-2 $

Get the sum ($s(x)=f(x)+g(x)$) and difference ($d(x)=f(x)-g(x)$) of these functions.

Can you guys please explain how to tackle these problems IN GENERAL, because I don't know the action scheme for solving a question like this one.

I have a TI-84+ with graphing abilities (calc intersect, min/max, dy/dx etc.) which I'm allowed to use. Please help me, I need urgent help with this question!

  • 0
    The constant terms just add directly, of course. For the sum of two cosines, does the intuitive explanation of a similar question [here](http://math.stackexchange.com/questions/181676/how-to-see-sin-x-cos-x/181841#181841) help?2012-09-16

1 Answers 1

1

$g(x)=\cos(x-\frac{\pi}{4})-2=\cos x\cos \frac{\pi}{4} + \sin x\sin \frac{\pi}{4} -2$ $=\frac{1}{\sqrt2}\cos x+\frac{1}{\sqrt2}\sin x -2$

So, $s(x)=f(x)+g(x)=3+2\cos x + \frac{1}{\sqrt2}\cos x+\frac{1}{\sqrt2}\sin x -2$ $=-5+\frac{2\sqrt2+1}{\sqrt2}\cos x+\frac{1}{\sqrt2}\sin x$

Let $r\cos y=\frac{2\sqrt2+1}{\sqrt2}$ and $r\sin y = \frac{1}{\sqrt2} $

$\frac{2\sqrt2+1}{\sqrt2}\cos x+\frac{1}{\sqrt2}\sin x=r(\cos y\cos x+\sin y\sin x)$ $=r\cos(x-y)$

Squaring and adding we get, $r^2=\frac{(2\sqrt2+1)^2+1}{2}=5+2\sqrt2$

On division, $\tan y=\frac{1}{2\sqrt2+1}\implies y=\tan^{-1}(\frac{1}{2\sqrt2+1})$

So, $s(x)=f(x)+g(x)=-5+\sqrt{5+2\sqrt2}\cos(x-\tan^{-1}(\frac{1}{2\sqrt2+1}))$

  • 0
    @ZafarS: if you need to understand that method specifically, you should probably add a comment along those lines to the question.2012-09-17