Problem 21 - Algebra and Trigonometry with Analytic Geometry

Question

Verify the identity. $$ \frac{\sin u-\sin v}{\sin u+\sin v}=\frac{\tan \frac{1}{2}(u-v)}{\tan \frac{1}{2}(u+v)} $$

Step-by-Step Solution

Answer

The identity is verified: both sides are equal.

Step 1: Express Left Side in Terms of Tan Half-Angle Formula

Using the sine difference identity, express \( \sin u - \sin v \) and \( \sin u + \sin v \) with half-angle identities. By the half-angle identity, \( \sin u - \sin v = 2 \cos\left(\frac{u+v}{2}\right) \sin\left(\frac{u-v}{2}\right) \) and \( \sin u + \sin v = 2 \sin\left(\frac{u+v}{2}\right) \cos\left(\frac{u-v}{2}\right) \).

Step 2: Simplify the Expression

Substitute these identities into the original expression on the left side: \[ \frac{2 \cos\left(\frac{u+v}{2}\right) \sin\left(\frac{u-v}{2}\right)}{2 \sin\left(\frac{u+v}{2}\right) \cos\left(\frac{u-v}{2}\right)} \]The '2's cancel out, leaving:\[ \frac{\cos\left(\frac{u+v}{2}\right)}{\sin\left(\frac{u+v}{2}\right)} \cdot \frac{\sin\left(\frac{u-v}{2}\right)}{\cos\left(\frac{u-v}{2}\right)} \]

Step 3: Convert Fraction to Tangent

Recognize that \( \frac{\cos\left(\frac{u+v}{2}\right)}{\sin\left(\frac{u+v}{2}\right)} = \cot\left(\frac{u+v}{2}\right) \) and \( \frac{\sin\left(\frac{u-v}{2}\right)}{\cos\left(\frac{u-v}{2}\right)} = \tan\left(\frac{u-v}{2}\right) \). Hence, the expression \[ \frac{\tan\left(\frac{u-v}{2}\right)}{\tan\left(\frac{u+v}{2}\right)} \] matches the right-hand side.

Step 4: Conclusion

Both sides of the equation are identical as we have shown both to be equal to \( \frac{\tan\left(\frac{u-v}{2}\right)}{\tan\left(\frac{u+v}{2}\right)} \). Therefore, the given trigonometric identity is verified.

Key Concepts

Half-Angle Identities

When working with trigonometric identities, knowing the half-angle identities can be very helpful. These identities relate the trigonometric functions of half an angle to the functions of the angle itself. In the exercise, the half-angle identities are used to break down more complex expressions for simplification. The basic forms of these identities are as follows:

• For sine: \( \sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{2}} \)
• For cosine: \( \cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}} \)
• For tangent: \( \tan\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \)

Using these identities, we can express differences and sums in terms of half-angles, which allows easier manipulation of trigonometric equations. In our problem, the half-angle form helped to express both \( \sin u - \sin v \) and \( \sin u + \sin v \), simplifying the initial problem significantly.

Sine Difference Identity

Another powerful tool in trigonometry is the sine difference identity. The identity states:

• \( \sin(u - v) = \sin u \cos v - \cos u \sin v \)

This identity allows us to simplify expressions where sine appears as a difference between two angles. In the exercise, we don't directly see the sine difference identity in its traditional form, but it plays a role in transformations related to half-angle identities. It helps relate the expression \( \sin u - \sin v \) to a form that can be simplified using those identities. Mastering this can simplify many trigonometric equations and proofs.

Tangent Function

The tangent function is fundamental to trigonometry, often representing the ratio of the sine and cosine of an angle. Formally, it is given by:

• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)

In the provided exercise, tangent plays a crucial role in transforming expressions into simpler terms. Once half-angle identities have simplified the expression, we can convert it to a tangent form. Recognizing that:

• \( \tan\left(\frac{u - v}{2}\right) = \frac{\sin\left(\frac{u - v}{2}\right)}{\cos\left(\frac{u - v}{2}\right)} \)

This format directly matches the right side of our exercise's given identity. Understanding the behavior of the tangent function allows for these transformations, which lie at the heart of verifying trigonometric identities.

Cotangent Function

As the reciprocal of the tangent function, the cotangent function is also a key player in trigonometry. It is defined as:

• \( \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} \)

In our exercise, the cotangent appearance emerges naturally when we express parts of the problem in different formats. After simplifying the expression using half-angle identities, we note:

• \( \frac{\cos\left(\frac{u + v}{2}\right)}{\sin\left(\frac{u + v}{2}\right)} = \cot\left(\frac{u + v}{2}\right) \)

This transformation is crucial, as it allows the original equation to realign with the target form given. Understanding cotangent's interplay with tangent reveals opposite yet complementary perspectives in trigonometric problem-solving.