f) $$ \tan \frac{23 \pi}{12} $$

To find $$ \tan \frac{23 \pi}{12} $$, we can use the angle sum identity for tangent. First, we can express $$ \frac{23 \pi}{12} $$ in terms of angles whose tangents we know. We can rewrite $$ \frac{23 \pi}{12} $$ as: \[ \frac{23 \pi}{12} = \frac{12 \pi}{12} + \frac{11 \pi}{12} = \pi + \frac{11 \pi}{12} \] Since $$ \tan(\pi + x) = \tan(x) $$, we have: \[ \tan \frac{23 \pi}{12} = \tan \frac{11 \pi}{12} \] Next, we can express $$ \frac{11 \pi}{12} $$ as: \[ \frac{11 \pi}{12} = \frac{6 \pi}{12} + \frac{5 \pi}{12} = \frac{\pi}{2} + \frac{5 \pi}{12} \] Since $$ \tan\left(\frac{\pi}{2} + x\right) $$ is undefined, we can instead use the identity for $$ \tan\left(\pi - x\right) $$: \[ \tan \frac{11 \pi}{12} = -\tan \frac{\pi}{12} \] Now, we need to calculate $$ \tan \frac{\pi}{12} $$. We can use the half-angle formula: \[ \tan \frac{\pi}{12} = \tan \left(\frac{\pi/6}{2}\right) = \frac{1 - \cos \frac{\pi}{6}}{\sin \frac{\pi}{6}} \] Calculating $$ \tan \frac{\pi}{12} $$ will give us the final result. Let's compute $$ \tan \frac{\pi}{12} $$ first. The values are: - $$ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} $$ - $$ \sin \frac{\pi}{6} = \frac{1}{2} $$ Now, substituting these values into the formula: \[ \tan \frac{\pi}{12} = \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}} = 2\left(1 - \frac{\sqrt{3}}{2}\right) = 2 - \sqrt{3} \] Thus, we have: \[ \tan \frac{11 \pi}{12} = -\tan \frac{\pi}{12} = -\left(2 - \sqrt{3}\right) \] Now, let's summarize the calculations and find $$ \tan \frac{23 \pi}{12} $$: \[ \tan \frac{23 \pi}{12} = -\left(2 - \sqrt{3}\right) \] So, the final answer is: \[ \tan \frac{23 \pi}{12} = \sqrt{3} - 2 \] $$ \tan \frac{23 \pi}{12} = \sqrt{3} - 2 $$

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Home Question Bank Math Trigonometry

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Ask by Zimmerman Lawson. in Canada

Mar 24,2025

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To solve for

, we first rewrite the angle. The angle can be expressed as:

Since

, we have:

Next, we know that

Using the tangent addition formula, we find:

Therefore,

Next, we need to compute

. We can express

as the sum of

and

Using the cotangent addition formula, we have:

Plug in

(where

) and

(where

Multiplying numerator and denominator by

Thus we have:

We can also rationalize this last result. Therefore:

Evaluating this, we find:

Thus,

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