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Doglover2011 · Oct 11, 2009, 02:35 PM

I'm trying to study for a test, but I get stuck a lot.
I'm doing a problem that says to find the exact value of Cos 15 using sum and difference formulas. Well, I'm almost done but got a little stuck at the last part. Here is what I have using Cos(45-30):
(1/sqre. Root of 2)(sqr. Root of 3/2) + (sqre. Root of 2/2) (1/2)
Which is (Do I have this right? I think so... ) sqre. Root of 3/2 times sqre. Root of 2 + sqre. Root of 2/4.

Which equals 6/??

Perito · Oct 11, 2009, 03:04 PM

Originally Posted by Doglover2011

I'm trying to study for a test, but I get stuck alot.
I'm doing a problem that says to find the exact value of Cos 15 using sum and difference formulas. Well, I'm almost done but got a little stuck at the last part. Here is what I have using Cos(45-30):
(1/sqre. root of 2)(sqr. root of 3/2) + (sqre. root of 2/2) (1/2)
Which is (Do I have this right? I think so...) sqre. root of 3/2 times sqre. root of 2 + sqre. root of 2/4.

Which equals 6/????

$sin(a-b) = (sin\, a) (cos\, b) - (cos\, a) (sin\, b)$

$cos(45-30) = sin(45)cos(30) - cos(45)sin(30)$

$cos(45-30) = \frac {\sqrt{2}}{2}\,\frac {2\sqrt {3}}{3} - \frac {\sqrt{2}}{2}\,\frac 12$

$cos(45-30) = \frac {\sqrt {6}}{3} - \frac {\sqrt 2}{4}$

You can simplify it a bit because $\sqrt {6} = \sqrt {2 \,\times\, 3}$ and you can combine them under a common denominator, but you will always have an irrational value.

galactus · Oct 11, 2009, 03:05 PM

$\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\cdot \frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$

Doglover2011 · Oct 11, 2009, 03:14 PM

Galactus- Shouldn't 1/sqre root of to be sqre root of 2/2?

galactus · Oct 11, 2009, 03:58 PM

That's the same thing.

Unknown008 · Oct 14, 2009, 10:33 AM

Hey hey hey!! :mad:

$cos(A-B) \cancel{=} sinAcosB - cosAsinB$

$cos(A-B) =cosAcosB + sinAsinB$

Then;

$cos(45-30) =cos45cos30 + sin45sin30 = \frac{1}{\sqrt2} \frac{\sqrt3}{2} + \frac{1}{\sqrt2}\frac12 = \frac{\sqrt3}{2\sqrt2} + \frac{1}{2\sqrt2} = \frac{1+\sqrt3}{2\sqrt2}$

Then, you can go on rationalising if you want:

$\frac{1+\sqrt3}{2\sqrt2} = \frac{\sqrt2+\sqrt{3\times2}}{2\sqrt{2\times2}} = \frac{\sqrt2+\sqrt6}{4}$