Compound Angle Identities

**Morphayne** · December 11th 2008, 01:34 PM

Problem:

Determine the exact value of each trigonometric ratio.

$tan\frac{23\pi}{12}<br />$

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

I'm stuck after that. Please help me.

**nzmathman** · December 11th 2008, 01:51 PM

First find $\tan{\left(\frac{\pi}{12}\right)}$

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

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

Now find $\tan{\left(\frac{23\pi}{12}\right)}$ by expanding
$\tan{\left(2\pi - \frac{\pi}{12}\right)}$

**skeeter** · December 11th 2008, 01:53 PM

hint ...

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

last two values are on the unit circle ...

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

$\tan\left(\frac{5\pi}{4}\right) = 1$

**Soroban** · December 11th 2008, 02:10 PM

Hello, Morphayne!

Determine the exact value of: . $\tan\frac{23\pi}{12}$

Your game plan is excellent.

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

Make a list . . .

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

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

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

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

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

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

**chabmgph** · December 11th 2008, 02:11 PM

Originally Posted by Morphayne

Problem:

Determine the exact value of each trigonometric ratio.

$tan\frac{23\pi}{12}<br />$

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

I'm stuck after that. Please help me.

Hi Morphayne,

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

$\tan \frac{23\pi}{12}=\tan (-\frac{\pi}{12})=-\tan \frac{\pi}{12}<br />$

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

See if you can do it from here.

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