# Compound Angle Identities

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• Dec 11th 2008, 01:34 PM
Morphayne
Compound Angle Identities
Problem:

Determine the exact value of each trigonometric ratio.

$\displaystyle tan\frac{23\pi}{12}$

I know that i have to use either: $\displaystyle tan(A+B)=\frac{tan A + tan B}{1-tan Atan B}$ OR $\displaystyle tan(A-B)=\frac{tan A-tan B}{1+tan A tan B}$

I'm stuck after that. Please help me.(Crying)
• Dec 11th 2008, 01:51 PM
nzmathman
First find $\displaystyle \tan{\left(\frac{\pi}{12}\right)}$

using the expansion for $\displaystyle \tan{\left(\frac{\pi}{3} - \frac{\pi}{4}\right)}$

because 1/3 - 1/4 = 1/12.

Now find $\displaystyle \tan{\left(\frac{23\pi}{12}\right)}$ by expanding
$\displaystyle \tan{\left(2\pi - \frac{\pi}{12}\right)}$
• Dec 11th 2008, 01:53 PM
skeeter
hint ...

$\displaystyle \frac{23\pi}{12} = \frac{8\pi}{12} + \frac{15\pi}{12} = \frac{2\pi}{3} + \frac{5\pi}{4}$

last two values are on the unit circle ...

$\displaystyle \tan\left(\frac{2\pi}{3}\right) = -\sqrt{3}$

$\displaystyle \tan\left(\frac{5\pi}{4}\right) = 1$
• Dec 11th 2008, 02:10 PM
Soroban
Hello, Morphayne!

Quote:

Determine the exact value of: .$\displaystyle \tan\frac{23\pi}{12}$
Your game plan is excellent.

Now we must express $\displaystyle \frac{23\pi}{12}$ as the sum of two "familiar" angles.

Make a list . . .

. . $\displaystyle \frac{23\pi}{12} \;=\;\begin{array}{ccccccc}\frac{\pi}{12} + \frac{22\pi}{12} &=& \frac{\pi}{12} + \frac{11\pi}{6} & & \text{We don't know }\frac{\pi}{12}\\ \\[-3mm]\frac{2\pi}{12} + \frac{21\pi}{12} &=&\frac{\pi}{6} + \frac{7\pi}{4} && \text{But we know these two!}\\ \\[-3mm] \frac{3\pi}{12} + \frac{20\pi}{12} \\ \vdots \end{array}$

So we have: .$\displaystyle \tan\frac{23\pi}{12} \;=\;\tan\left(\frac{\pi}{6} + \frac{7\pi}{4}\right) \;=\;\frac{\tan\frac{\pi}{6} + \tan\frac{7\pi}{4}}{1 - \tan\frac{\pi}{6}\tan\frac{7\pi}{4}}$

. . . . . . . . . . . . . . . $\displaystyle = \;\frac{\frac{1}{\sqrt{3}} + (-1)}{1 - \left(\frac{1}{\sqrt{3}}\right)(-1)} \;=\;\frac{\frac{1}{\sqrt{3}} - 1}{1 + \frac{1}{\sqrt{3}}}$

Multiply by $\displaystyle \frac{\sqrt{3}}{\sqrt{3}}\!:\;\;\frac{\sqrt{3}\lef t(\frac{1}{\sqrt{3}} - 1\right)} {\sqrt{3}\left(1 + \frac{1}{\sqrt{3}}\right)} \;=\;\frac{1-\sqrt{3}}{\sqrt{3}+1}$

Multiply by $\displaystyle \frac{1-\sqrt{3}}{1-\sqrt{3}}\!:\quad\frac{1-\sqrt{3}}{1 + \sqrt{3}}\cdot\frac{1-\sqrt{3}}{1-\sqrt{3}} \;=\;\frac{1 - 2\sqrt{3} + 3}{1 - 3} \;=\;\frac{4-2\sqrt{3}}{-2}$

. . $\displaystyle =\;\frac{-2(\sqrt{3}-2)}{-2} \;=\;\sqrt{3}-2$

• Dec 11th 2008, 02:11 PM
chabmgph
Quote:

Originally Posted by Morphayne
Problem:

Determine the exact value of each trigonometric ratio.

$\displaystyle tan\frac{23\pi}{12}$

I know that i have to use either: $\displaystyle tan(A+B)=\frac{tan A + tan B}{1-tan Atan B}$ OR $\displaystyle tan(A-B)=\frac{tan A-tan B}{1+tan A tan B}$

I'm stuck after that. Please help me.(Crying)

Hi Morphayne,

Since $\displaystyle \frac{23\pi}{12}=\frac{24\pi-\pi}{12}=2\pi-\frac{\pi}{12}$,

$\displaystyle \tan \frac{23\pi}{12}=\tan (-\frac{\pi}{12})=-\tan \frac{\pi}{12}$

And you will probably need to use the half-angle formula for tangent here.
$\displaystyle \tan \frac{\theta}{2}=\frac{\sin \theta}{1+cos\theta}$

See if you can do it from here.