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Magazine Chapter 8: Problem 114
If \(z=\tan \frac{\alpha}{2},\) show that \(\cos \alpha=\frac{1-z^{2}}{1+z^{2}}\)
Short Answer
Expert verified
\( \cos \alpha = \frac{1 - z^{2}}{1 + z^{2}} \)
Step by step solution
01
Identify the given and required
Given: \( z = \tan \frac{\alpha}{2} \). Required: Show that \( \cos \alpha = \frac{1 - z^{2}}{1 + z^{2}} \).
02
Use the double-angle formula for tangent
Recall the double-angle identity for tangent: \( \tan(u+v) = \frac{\tan u + \tan v}{1 - \tan u \tan v} \)For \( u = v = \frac{\alpha}{2} \), this becomes:\( \tan \alpha = \frac{2 \tan \frac{\alpha}{2}}{1 - \tan^2 \frac{\alpha}{2}} \)
03
Substitute \( z \) for \( \tan \frac{\alpha}{2} \)
Given \( z \ = \tan \frac{\alpha}{2} \), substitute this into the double-angle identity:\( \tan \alpha = \frac{2z}{1 - z^{2}} \)
04
Use the Pythagorean identity
Recall the Pythagorean identity: \( 1 + \tan^2 \theta = \sec^2 \theta \)Using \( \theta = \frac{\alpha}{2} \), substitute into the identity:\( 1 + \tan^2 \frac{\alpha}{2} = \sec^2 \frac{\alpha}{2} \)
05
Express \( \sec \frac{\alpha}{2} \)
Since \( z = \tan \frac{\alpha}{2} \), we have:\( 1 + z^2 = \sec^2 \frac{\alpha}{2} \)Thus, \( \sec \frac{\alpha}{2} = \sqrt{1 + z^2} \).
06
Relate secant and cosine
Recall that \( \sec \theta = \frac{1}{\cos \theta} \), which gives us:\( \cos \frac{\alpha}{2} = \frac{1}{\sqrt{1 + z^2}} \).
07
Use the double-angle formula for cosine
Recall the double-angle identity for cosine:\( \cos \alpha = 2 \cos^2 \frac{\alpha}{2} - 1 \)Using \( \cos \frac{\alpha}{2} \) we have:\( \cos \alpha = 2 \left( \frac{1}{\sqrt{1 + z^2}} \right)^2 - 1 \)
08
Simplify the expression
Simplify the formula:\[ \cos \alpha = 2 \frac{1}{1 + z^2} - 1 \]Combine into a single fraction:\[ \cos \alpha = \frac{2 - (1 + z^2)}{1 + z^2} = \frac{1 - z^2}{1 + z^2} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Double-Angle Formulas
The double-angle formulas are essential tools in trigonometry. They allow you to express trigonometric functions of double angles in terms of single angles. For example, the double-angle formula for tangent is:
\(\tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta} \) This equation lets you rewrite expressions involving \(\tan \theta \) in a different form.
In our exercise, we used this formula to help convert \(\tan \frac{\theta}{2}\) into an expression for \(\tan \theta \). Substituting \( \theta = \frac{\theta}{2}\), our expression becomes:
\( \tan \theta = \frac{2 \tan \frac{\theta}{2}}{1 - \tan^2 \frac{\theta}{2}}\)
This step is crucial because it transforms the problem into a more manageable form, using the variable substitution \(z = \tan \frac{\theta}{2} \). With this substitution, the formula becomes:
\( \tan \theta = \frac{2z}{1 - z^2}\).
Learning and understanding double-angle formulas can simplify complex trigonometric computations.
\(\tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta} \) This equation lets you rewrite expressions involving \(\tan \theta \) in a different form.
In our exercise, we used this formula to help convert \(\tan \frac{\theta}{2}\) into an expression for \(\tan \theta \). Substituting \( \theta = \frac{\theta}{2}\), our expression becomes:
\( \tan \theta = \frac{2 \tan \frac{\theta}{2}}{1 - \tan^2 \frac{\theta}{2}}\)
This step is crucial because it transforms the problem into a more manageable form, using the variable substitution \(z = \tan \frac{\theta}{2} \). With this substitution, the formula becomes:
\( \tan \theta = \frac{2z}{1 - z^2}\).
Learning and understanding double-angle formulas can simplify complex trigonometric computations.
Exploring the Pythagorean Identity
The Pythagorean Identity is a fundamental trigonometric identity that states:
\(\tan^2 \theta + 1 = \text{sec}^2 \theta \) To show this, consider the relationship between tangent and secant. We know that \(\text{tan} \theta = \frac{\text{opposite}}{\text{adjacent}}\) and \(\text{sec} \theta = \frac{\text{hypotenuse}}{\text{adjacent}}\).
For our exercise, we use this identity in the following form by substituting \(\theta = \frac{\theta}{2} \):
\(1 + \tan^2 \frac{\theta}{2} = \text{sec}^2 \frac{\theta}{2}\).
Using \( z = \tan \frac{\theta}{2} \), we then substitute and rearrange:
\(1 + z^2 = \text{sec}^2 \frac{\theta}{2}\).
This relationship helps us find the value of other trigonometric functions once we know \(z\), bridging a gap between different identities and aiding our calculation of \(\text{cos} \theta \) in subsequent steps.
\(\tan^2 \theta + 1 = \text{sec}^2 \theta \) To show this, consider the relationship between tangent and secant. We know that \(\text{tan} \theta = \frac{\text{opposite}}{\text{adjacent}}\) and \(\text{sec} \theta = \frac{\text{hypotenuse}}{\text{adjacent}}\).
For our exercise, we use this identity in the following form by substituting \(\theta = \frac{\theta}{2} \):
\(1 + \tan^2 \frac{\theta}{2} = \text{sec}^2 \frac{\theta}{2}\).
Using \( z = \tan \frac{\theta}{2} \), we then substitute and rearrange:
\(1 + z^2 = \text{sec}^2 \frac{\theta}{2}\).
This relationship helps us find the value of other trigonometric functions once we know \(z\), bridging a gap between different identities and aiding our calculation of \(\text{cos} \theta \) in subsequent steps.
Deriving the Cosine Function
The cosine function is another vital trigonometric function, representing the ratio of the adjacent side to the hypotenuse in a right triangle. A noteworthy form is the double-angle identity for cosine, which helps simplify expressions involving angles.
Using the double-angle formula, we have:
\(\text{cos } 2\theta = 2 \text{cos}^2 \theta - 1\).
In our example, we substitute \(\theta = \frac{\theta}{2}\) into this identity to find \(\text{cos } \theta \):
\(\text{cos } \theta = 2 \text{cos}^2 \frac{\theta}{2} - 1 \).
This can be further simplified using our previous result from the Pythagorean Identity:
\(\text{cos} \frac{\theta}{2} = \frac{1}{\text{sec} \frac{\theta}{2}} = \frac{1}{\text{sqrt}(1 + z^2)} \).
Plugging this into the cosine double-angle formula gives:
\(\text{cos} \theta = 2 \frac{1}{1 + z^2} - 1 \).
After simplifying, we obtain:
\(\text{cos} \theta = \frac{1 - z^2}{1 + z^2}\).
This shows the linkage between \( z = \tan \frac{\theta}{2} \) and \(\text{cos } \theta \), effectively providing a new perspective on how different trigonometric identities interrelate.
Using the double-angle formula, we have:
\(\text{cos } 2\theta = 2 \text{cos}^2 \theta - 1\).
In our example, we substitute \(\theta = \frac{\theta}{2}\) into this identity to find \(\text{cos } \theta \):
\(\text{cos } \theta = 2 \text{cos}^2 \frac{\theta}{2} - 1 \).
This can be further simplified using our previous result from the Pythagorean Identity:
\(\text{cos} \frac{\theta}{2} = \frac{1}{\text{sec} \frac{\theta}{2}} = \frac{1}{\text{sqrt}(1 + z^2)} \).
Plugging this into the cosine double-angle formula gives:
\(\text{cos} \theta = 2 \frac{1}{1 + z^2} - 1 \).
After simplifying, we obtain:
\(\text{cos} \theta = \frac{1 - z^2}{1 + z^2}\).
This shows the linkage between \( z = \tan \frac{\theta}{2} \) and \(\text{cos } \theta \), effectively providing a new perspective on how different trigonometric identities interrelate.
