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

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

$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!

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1 Answer 1

up vote 1 down vote accepted

We can simply add or subtract, to get a correct answer. So for the sum, we have $$s(x)=(-5+2\cos (x))+(\cos(x-\pi/4)-2)=2\cos(x)+\cos(x-\pi/4)-7,$$ and for the difference we have $$d(x)=(-5+2\cos (x))-(\cos(x-\pi/4)-2)=2\cos(x)-\cos(x-\pi/4)-3.$$

But I imagine you are expected to do more. You may be expected to proceed as follows:

$1$. Use the addition/subtraction law for cosines to express $\cos(x-\pi/4)$ as $(\cos x)(\cos(\pi/4))+(\sin x)(\sin(\pi/4))$. So we get $$\cos(x-\pi/4)=\frac{1}{\sqrt{2}}\cos x+\frac{1}{\sqrt{2}}\sin x.$$

$2.$ Now add $f(x)$ and the modified version of $g(x)$. (Or, for the other part of the question, subtract.) So for the sum we would have $$f(x)+g(x)=\left(2+\frac{1}{\sqrt{2}}\right)\cos x +\frac{1}{\sqrt{2}}\sin x -7.$$

$3.$ In certain Physics-oriented courses, you may be expected to do further processing. After doing $(2.)$, we have something of the shape $k+a\sin x+b\cos x$. You may be expected to express $a\sin x+b\cos x$ as a single trigonometric function. The idea is that $$a\sin x+b\cos x=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\sin x+ \frac{b}{\sqrt{a^2+b^2}}\cos x\right).$$

Now let $\theta$ be any angle whose cosine is $\frac{a}{\sqrt{a^2+b^2}}$ and whose sine is $\frac{b}{\sqrt{a^2+b^2}}$. Then $$a\sin x+b\cos x=\sqrt{a^2+b^2}\sin(x+\theta).$$

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The question, as quoted, seems to be answered satisfactorily in your first sentence. –  Lubin Sep 16 '12 at 4:59
    
@Lubin: The question has been asked at least twice recently, with some urgency. I do not know what kind of answer the OP is looking for, so I thought I would at least supply some relevant information. –  André Nicolas Sep 16 '12 at 5:05
    
Yes, it is very urgent. I had pneumonia for a month so I had to revise 1 entire chapter in 1 day and it has worked out, I'm just having trouble with these types of questions. Your answer seems to be of a higher level of mathematics than we are at now. So Lubin was right. However, can you work out how to add/subtract 2 functions, I have no idea how to.. –  ZafarS Sep 16 '12 at 9:42
    
@ZafarS: I have modified the answer, putting in detail at the beginning. It sounds as if part $(3)$ is irrelevant to you at this time, and even maybe part $(2)$. –  André Nicolas Sep 16 '12 at 12:23

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