Chapter 5: Problem 64
Verify each identity. $$2 \tan \frac{\alpha}{2}=\frac{\sin ^{2} \alpha+1-\cos ^{2} \alpha}{\sin \alpha(1+\cos \alpha)}$$
Short Answer
Expert verified
The provided trigonometric identity is verified.
Step by step solution
01
Convert given equation to sines and cosines
Starting with the left-hand side (LHS) of the given equation which is \(2 \tan \frac{\alpha}{2}\). As we know that \( \tan x = \frac{\sin x}{\cos x} \), it can be rewritten as \(\frac{2 \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}\)
02
Use the double-angle formulas
Now, we have \(\frac{2 \sin \frac{\alpha}{2}}{1 - 2 \sin^{2} \frac{\alpha}{2}}\) after using the cosine double-angle formula: \(\cos^2 x = 1 - 2\sin^2 x\).
03
Simplify further
The LHS can be further simplified to \(\frac{\sin \alpha}{1 + \cos \alpha}\). This is done by using the double angle formulas again: \(\sin 2x = 2\sin x \cos x\) and replacing \(\sin \frac{\alpha}{2}\) with \(\frac{\sin \alpha}{1 + \cos \alpha}\). The identifier \(\cos \frac{\alpha}{2}\) in the denominator is replaced with \(\sqrt{\frac{1 + \cos \alpha}{2}}\) using the half-angle identity, giving us the desired form of \(\frac{\sin \alpha}{1 + \cos \alpha}\).
04
Work on the right-hand side (RHS) of the equation
The right-hand side (RHS) of the equation is \(\frac{\sin^{2}\alpha+1-\cos^{2}\alpha}{\sin \alpha (1+\cos\alpha)}\). This can be shortened to \(\frac{\sin^{2}\alpha +1 - (1-\sin^{2}\alpha)}{\sin \alpha (1+\cos \alpha)}\), and even further it can be simplified to \(\frac{\sin \alpha}{1 + \cos \alpha}\), which is equal to the LHS.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Half-Angle Identities
Half-angle identities are essential tools when dealing with trigonometric expressions. These identities help in expressing trigonometric functions involving half of an angle. One of the most common half-angle identities is for cosine, given by:\[ \cos \left(\frac{\alpha}{2}\right) = \sqrt{\frac{1 + \cos \alpha}{2}} \]This identity simplifies various expressions, especially when dealing with complex trigonometric problems. For example, in the provided solution, evaluating \(\cos \frac{\alpha}{2}\) in the denominator was simplified using this identity.
- These identities allow you to rewrite trigonometric functions in terms of the whole angle.
- Using the square root in the formula can sometimes introduce "+" or "-" signs. However, the context of the problem tells us which one to use based on the quadrant of the angle.
Double-Angle Formulas
Double-angle formulas are another powerful tool in trigonometry. They express trigonometric functions of double angles in terms of single angles. A well-known example is the formulas for sine and cosine:
- \( \sin 2\theta = 2\sin \theta \cos \theta \)
- \( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \)
- Double-angle formulas often rearrange terms, simplifying otherwise complicated trigonometric identities.
- Knowing these formulas by heart accelerates the resolution process in verifying identities or solving problems.
Simplifying Trigonometric Expressions
Simplifying trigonometric expressions often involves using a combination of identities to transform a complex expression into a simpler, more recognizable form. In our original problem, the expression was simplified from a complicated fraction to a more manageable form:
- The initial expression involved \( \sin \frac{\alpha}{2} \) and \( \cos \frac{\alpha}{2} \), which were expressed using known identities to simplify the fractions further.
- Recognizing familiar forms such as \( \frac{\sin \alpha}{1 + \cos \alpha} \) is crucial in identifying and verifying trigonometric identities.
- Effective use of these identities dramatically simplifies proofs and verifications of identities.
- Breaking down expressions and substituting them with known simplifications saves time and computation effort.