# Getting exact answer from cos(u+-v) & sin(u+-v). Simple problem.

• Mar 1st 2009, 08:14 PM
vane505
Getting exact answer from cos(u+-v) & sin(u+-v). Simple problem.
Having a simple problem again.

First: I'm really sorry, I've looked and I can't find and post on how to insert the correct "Math-Code" (well, I don't know how to insert degrees, square root and other things).

* = degrees

Example:
Determine the EXACT value of cos(15*) by rewriting cos(15*) = cos(45*-30*):

I can get exact values like:
cos15* = cos(45*-30*) = cos45*cos30* + sin45*sin30* = 0.965925......
But, ofcourse, this won't do. The correct answer is:
(squareroot(6) + squareroot(2))/4

How can I get the correct (well, I guess both are correct if I write the whole answer, but I mean the "better looking" one) later answer instead of the first? Appreciate help!!
• Mar 1st 2009, 08:41 PM
sbcd90
as u could see,that sqrt.(6) is a non-recurring & non-terminating decimal ,so this problem arises.
• Mar 1st 2009, 09:03 PM
vane505
Hi, and thank you for your response.

I'm not sure if I understand you correct. You mean that sqrt.(6) is a never-ending decimal right?

I probably wasn't clear with my problem, or . I can get the correct answer, but I don't know how to calculate/convert the result so that it shows (sqrt(6) + sqrt(2))/4 instead of 0.965925......

I know how to calculate cos(45*-30*) = cos45*cos30*+sin45*sin30* = =0.965925......

This is how the problem is solved in the book:
cos(45* - 30*)= (1/sqrt(2))(sqrt(3)/2) + (1/sqrt(2))(1/2) =
= (sqrt(3) + 1)/(2sqrt(2)) = sqrt(2)(sqrt(3) + 1)/ = (sqrt(6)+sqrt(2))/4

My issue is that I don't know how to rewrite the cos45*cos30*+sin45*sin30* into (1/sqrt(2))(sqrt(3)/2) + (1/sqrt(2))(1/2)
and so on.

I'm sorry if my explanation is bad. I haven't studied much math before.

Again, appreciate help!
• Mar 1st 2009, 09:31 PM
Reckoner
Quote:

Originally Posted by vane505
Example:
Determine the EXACT value of cos(15*) by rewriting cos(15*) = cos(45*-30*):

I can get exact values like:
cos15* = cos(45*-30*) = cos45*cos30* + sin45*sin30* = 0.965925......
But, ofcourse, this won't do. The correct answer is:
(squareroot(6) + squareroot(2))/4

You should know the sine and cosine of $30^\circ$ and $45^\circ.$ If you do not, then you must review the unit circle.

$\cos15^\circ=\cos(45^\circ-30^\circ)=\cos45^\circ\cos30^\circ+\sin45^\circ\si n30^\circ$

$=\left(\frac{\sqrt2}2\right)\left(\frac{\sqrt3}2\r ight)+\left(\frac{\sqrt2}2\right)\left(\frac12\rig ht)$

$=\frac{\sqrt2\sqrt3}4+\frac{\sqrt2}4=\frac{\sqrt6} 4+\frac{\sqrt2}4=\frac{\sqrt6+\sqrt2}4.$

Edit: From your latest post it seems that you do not know your unit circle, so I suggest you review that section in the book. Wikipedia has an overview. These values are found using special properties of $45^\circ\text{-}45^\circ\text{-}90^\circ$ and $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangles, discussions of which can be found in topics 6 and 7 on this page.
• Mar 1st 2009, 10:00 PM
vane505
Thank you for your help! I get it now ;)
• Mar 2nd 2009, 05:42 AM
HallsofIvy
Quote:

Originally Posted by vane505
Hi, and thank you for your response.

I'm not sure if I understand you correct. You mean that sqrt.(6) is a never-ending decimal right?

I probably wasn't clear with my problem, or . I can get the correct answer, but I don't know how to calculate/convert the result so that it shows (sqrt(6) + sqrt(2))/4 instead of 0.965925......

I know how to calculate cos(45*-30*) = cos45*cos30*+sin45*sin30* = =0.965925......

This is how the problem is solved in the book:
cos(45* - 30*)= (1/sqrt(2))(sqrt(3)/2) + (1/sqrt(2))(1/2) =
= (sqrt(3) + 1)/(2sqrt(2)) = sqrt(2)(sqrt(3) + 1)/ = (sqrt(6)+sqrt(2))/4

My issue is that I don't know how to rewrite the cos45*cos30*+sin45*sin30* into (1/sqrt(2))(sqrt(3)/2) + (1/sqrt(2))(1/2)
and so on.

I'm sorry if my explanation is bad. I haven't studied much math before.

Again, appreciate help!

Can we at least assume that you know that cos(45)= $\frac{\sqrt{2}}{2}$, sin(45)= $\frac{\sqrt{2}}{2}$, cos(30)= $\frac{\sqrt{3}}{2}$, and sin(30)= $\frac{1}{2}$? This problem assumes that you already know that!

Then cos(45)cos(30)= $\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}= \frac{\sqrt{6}}{4}$ and sin(45)sin(30)= $\frac{\sqrt{2}}{2}\frac{1}{2}= \frac{\sqrt{2}}{4}$
• Mar 2nd 2009, 06:06 AM
vane505
The strange thing is that the book never mentioned (yet) that it's important to know these specific values, so I thought there was some almost-impossible magic way to get these numbers by doing some kind of calculation.

As said before, I'm a math-beginner :)