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I am attempting to find $\cos(s+t)$ and $\cos(s-t)$

I am given that they are in quadrant II and that $\cos s = -1/5$ and $\sin t = 3/5$. I have no idea what the relation between these numbers $s$ or what I am suppose to do exactly. What is $s$ and what is $t$? I know that $\cos$ is $x/r$ so $x=-1$ and $r=5$, $\sin$ is $y/r$ so $y=3$ but I don't know what to do from here really.

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up vote 3 down vote accepted

$s$ and $t$ are the "names" of two angles. You should use three things:

  1. The addition formulas for cosine: $$\begin{align*} \cos(\alpha+\beta) &= \cos\alpha\cos\beta - \sin\alpha\sin\beta\\ \cos(\alpha-\beta) &= \cos\alpha\cos\beta + \sin\alpha\sin\beta. \end{align*}$$

    But in order to use these formulas, you need to know $\cos s$, $\cos t$, $\sin s$, and $\sin t$. You only know $\cos s$ and $\sin t$, so you need to figure out the other two.

  2. To figure out $\sin s$, you should use the fact that $$\cos^2 s + \sin^2 s = 1.$$ That will tell you that $\sin^2 s = 1-\cos^2 s$; finally, use the fact that $s$ is in quadrant II to decide if you should take the positive or negative square root.

  3. Likewise, to figure out $\cos t$, you should use the fact that $$\cos^2 t + \sin^2 t = 1,$$ and again use the fact that $t$ is in the second quadrant to determine if you use the positive or negative square roots.

The fact that cosine of an angle can be defined to be the length of the adjacent side divided by the length of the hypotenuse (what you describe as $x/r$), and sine as the length of the opposite side divided byt he length of the hypotenuse (what you call $y/r$) does not need to play a role here.

You can use it (it's essentially steps 2 and 3), but you should notice that you would have two different triangles (one with angle $t$ and one with angle $s$, so the $x,y,r$ you use for $\cos s$ would be different from the $x,y,r$ you would use for the $\sin t$).

From $\cos s = -1/5$ you imagine a right triangle with $x=-1$ and $r=5$. Since $x^2+y^2 = r^2$, you get $y^2 = 25-(-1)^2 = 24$; then figure out if you get $y=\sqrt{24}$ or $y=-\sqrt{24}$, using the fact that you are in quadrant II. This gives you $\sin s = y/r$.

You could then use the same idea to find $\cos t$, from $\sin t = 3/5$, $y=3$, $r=5$ (new $y$, because it's a different angle; same $r$ only by chance).

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So the answer I got is4-6(squarerootsix) / 5 which is wrong, I checked my book and it is suppose tobe over 25. I forgot that when multiplying fractions you multiply the top and bottom. – Adam Jun 29 '11 at 20:46
@Adam: Yes, the answer I get for $\cos(s+t)$ is $\displaystyle \frac{4-6\sqrt{6}}{25}$; and the answer I get for $\cos(s-t)$ is $\displaystyle\frac{4+6\sqrt{6}}{25}$. – Arturo Magidin Jun 29 '11 at 20:50
Although Adam's observation that one can think of the angles in terms of reference angles (and corresponding right triangles, using relationships between opposite, adjacent, hypotenuse side isn't necessary for solving this problem, but cos(t), cos(s), sin(t), sin(s) can all be derived this way, and I believe for the sake of reinforcing some of Adam's prior learning and spring-boarding to a new context, like this, can help him understand the validity of some of the identities he's struggling with, and provide him with a means to "check" his computations...just a thought. – amWhy Jun 30 '11 at 0:47
@amWhy: I wasn't criticising you, by the way, I hope it didn't come across that way (I was merely expressing my continuing disappointment with reality). I'll delete my comment, though, but you should keep your first. – Arturo Magidin Jun 30 '11 at 4:04
No, no...your reply to my comment didn't come across as criticism: not at all. My reply to your reply was simply a "nod": I too, (particularly prone to idealism as I am), felt your disappointment with reality...And please know I wasn't criticizing you; I keep tend to try to see/find the morsels of progress in the worst scenarios... – amWhy Jun 30 '11 at 4:15

More than any formula of any kind: Did you draw a picture? Do you have any clue what you are really looking for?

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No I would have no idea how to draw a picture out of this. – Adam Jun 29 '11 at 20:46
@Adam: Here is a picture of the trigonometric circle. Note that $r=1$. I divided the unit radius into 5 parts. – Américo Tavares Jun 29 '11 at 23:59
+1 for mentioning the advantage of a picture. – Américo Tavares Jun 30 '11 at 0:12
Nice, @Americo. How do you insert an image into a comment? I see your link is a web address (your wordpress blog?), but how do you insert it so the URL doesn't appear, but the link is activated by clicking on "Here" (I was going to add an image, but it would have been from my computer, so I would have had to create an answer, or post it in Adam's post). – amWhy Jun 30 '11 at 0:41
@amWhy: the syntax for links in comments (can also be used in posts) is [text of link](address of link). – Arturo Magidin Jun 30 '11 at 4:13

Do you know that $\cos (s+t)=\cos s\cdot \cos t - \sin s \cdot \sin t$? You can just apply that.

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He still needs to use that $\cos^2x+\sin^2 x=1$ and choose the signs for the unknown $\sin,\cos$ looking that in quadrant II $\sin $ is positive and $\cos$ is negative. – Beni Bogosel Jun 29 '11 at 20:38

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