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Magazine Chapter 4: Problem 38
Find exact expressions for the indicated quantities, given that $$ \cos \frac{\pi}{12}=\frac{\sqrt{2+\sqrt{3}}}{2} \text { and } \sin \frac{\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2} $$ [These values for \(\cos \frac{\pi}{12}\) and \(\sin \frac{\pi}{8}\) will be derived.] $$ \tan \left(-\frac{3 \pi}{8}\right) $$
Short Answer
Expert verified
The exact expression for \(\tan\left(-\frac{3\pi}{8}\right)\) is \(\frac{-1}{\sqrt{2}+\sqrt{2-\sqrt{3}}}\).
Step by step solution
01
Express the given angle as a subtraction of known angles
To be able to use the given values for \(\cos\frac{\pi}{12}\) and \(\sin\frac{\pi}{8}\), rewrite the angle \(-\frac{3\pi}{8}\) as a subtraction of \(\frac{\pi}{12}\) and \(\frac{\pi}{8}\) or their multiples:
\[
-\frac{3\pi}{8} = - \frac{\pi}{8} - \frac{\pi}{12}.
\]
02
Compute the tangents of the known angles
Using the given values of \(\cos\frac{\pi}{12}\) and \(\sin\frac{\pi}{8}\), as well as the Pythagorean identity, compute the tangents of the angles \(\frac{\pi}{12}\) and \(\frac{\pi}{8}\):
Recall the Pythagorean identity: \[\sin^2{\theta}+\cos^2{\theta}=1\]
and the definition of tangent: \[\tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}}.\]
First, for \(\frac{\pi}{12}\):
\[
\sin^2{\frac{\pi}{12}} = 1 - \cos^2{\frac{\pi}{12}} = 1 - \left(\frac{\sqrt{2+\sqrt{3}}}{2}\right)^2 = \frac{2-\sqrt{3}}{2}.
\]
So,
\[
\sin{\frac{\pi}{12}}=\sqrt{\frac{2-\sqrt{3}}{2}}.
\]
Thus,
\[
\tan{\frac{\pi}{12}} = \frac{\sqrt{\frac{2-\sqrt{3}}{2}}}{\frac{\sqrt{2+\sqrt{3}}}{2}} = \sqrt{2-\sqrt{3}}.
\]
Next, for \(\frac{\pi}{-8}\), note that \(\cos{\frac{\pi}{8}}=\sin{\frac{\pi}{2}-\frac{\pi}{8}}=\sin{\frac{3*\pi}{8}}\), making:
\[
\sin \frac{\pi}{-8} = \sin \left(\frac{\pi}{2}-\frac{3\pi}{8} \right) = \sin \frac{\pi}{8},
\]
\[
\cos \frac{\pi}{-8} = -\left(1 - \sin^2{\frac{\pi}{8}} \right)= -\left(1 - \frac{2 - \sqrt{2}}{2} \right)= -\frac{\sqrt{2}}{2}.
\]
Hence,
\[
\tan{\frac{\pi}{-8}} = \frac{\sin{\frac{\pi}{-8}}}{\cos{\frac{\pi}{-8}}} = \sqrt{2}.
\]
03
Apply the angle subtraction formula for the tangent
Now apply the angle subtraction formula for the tangent:
\[
\tan{(\alpha-\beta)} = \frac{\tan{\alpha}-\tan{\beta}}{1+\tan{\alpha}\tan{\beta}}.
\]
In our case, \(\alpha = -\frac{\pi}{8}\) and \(\beta = \frac{\pi}{12}\), so:
\[
\tan\left(-\frac{3\pi}{8}\right) = \frac{\tan{\frac{\pi}{-8}}-\tan{\frac{\pi}{12}}}{1+\tan{\frac{\pi}{-8}}\tan{\frac{\pi}{12}}} = \frac{\sqrt{2}-\sqrt{2-\sqrt{3}}}{1+\sqrt{2}\sqrt{2-\sqrt{3}}}.
\]
04
Simplify the expression
Finally, simplify the expression for the tangent of \(-\frac{3\pi}{8}\):
\[
\tan\left(-\frac{3\pi}{8}\right) = \frac{\sqrt{2}-\sqrt{2-\sqrt{3}}}{1+\sqrt{2}\sqrt{2-\sqrt{3}}} = \frac{(\sqrt{2} - \sqrt{2-\sqrt{3}})(\sqrt{2} + \sqrt{2-\sqrt{3}})}{(1+\sqrt{2}\sqrt{2-\sqrt{3}})(\sqrt{2} + \sqrt{2-\sqrt{3}})} = \frac{-1}{\sqrt{2}+\sqrt{2-\sqrt{3}}}.
\]
Thus, the exact expression for the tangent of \(-\frac{3\pi}{8}\) is \(\frac{-1}{\sqrt{2}+\sqrt{2-\sqrt{3}}}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Angle Subtraction Formula
The angle subtraction formula for tangent is a powerful tool. It helps in evaluating trigonometric functions at angles that are not directly available using basic trigonometric values. Given two angles, \(\alpha\) and \(\beta\), the angle subtraction formula for tangent is expressed as: \[ \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \] This formula is particularly useful in scenarios where you are given angles that can be expressed as the difference of two familiar angles. For example, if we need to find \(\tan\left(-\frac{3\pi}{8}\right)\), and know the tangents of angles \(\alpha = -\frac{\pi}{8}\) and \(\beta = \frac{\pi}{12}\), you can directly use this formula. Simply calculate the tangents for these known angles. Then substitute them into the formula to find your result.
Pythagorean Identity
The Pythagorean identity is a fundamental relation in trigonometry. It states that: \[ \sin^2 \theta + \cos^2 \theta = 1 \] This equation holds true for any angle \(\theta\) and is crucial for deriving other trigonometric identities. It allows us to compute either \(\sin \theta\) or \(\cos \theta\) if one is known. In the context of our exercise, we used this identity to find \(\sin\frac{\pi}{12}\). Given \(\cos\frac{\pi}{12}\), we rearranged the equation to get: \[ \sin^2\frac{\pi}{12} = 1 - \cos^2\frac{\pi}{12} \] This calculation is key to finding the tangent later, as \(\tan\theta = \frac{\sin \theta}{\cos \theta}\). By knowing both sine and cosine values, evaluating tangent becomes simple.
Exact Trigonometric Values
Exact trigonometric values are the values of sine, cosine, and tangent for specific angles. They are often represented using radical expressions. These values are crucial in solving trigonometric problems precisely without numerical approximations. Angles like \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), and \(\frac{\pi}{3}\) have well-known exact trigonometric values. In this exercise, we dealt with slightly more complex angles such as \(\frac{\pi}{12}\) and \(\frac{\pi}{8}\). Understanding these exact values helps in building a strong foundation in trigonometry. For instance, the exact value for \(\cos\frac{\pi}{12}\) is \(\frac{\sqrt{2+\sqrt{3}}}{2}\). These radical forms are derived from transforming known angles and using trigonometric identities. Having a grasp of exact values simplifies the trigonometric computations significantly.
Tangent Function
The tangent function, denoted as \(\tan\), is one of the primary trigonometric functions. It is defined as the ratio of \(\sin\theta\) to \(\cos\theta\): \[ \tan\theta = \frac{\sin\theta}{\cos\theta} \] This ratio is a fundamental relationship in trigonometry. It helps in solving problems involving right triangles and angles. The tangent function is periodic with a period of \(\pi\), meaning it repeats its values in every interval of \(\pi\). Understanding the tangent function is crucial for this exercise. By using the definitions and properties of both sine and cosine, we can find the tangent of any angle for which sine and cosine are known. In the exercise, we calculated \(\tan\frac{\pi}{12}\) and \(\tan\frac{\pi}{-8}\) using their sine and cosine values, and then applied the angle subtraction formula.
