Problem 68 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \frac{\sin 7 \theta-\sin 5 \theta}{\cos 7 \theta+\cos 5 \theta}=\tan \theta $$

Step-by-Step Solution

Answer

The final simplified form of the given equation is: \((2\sin\theta\cos\theta\cos(5\theta) + \sin(12\theta))(\cos(12\theta) - \cos(2\theta))\cos\theta=\sin\theta(\sin^2(2\theta) - \sin^2(12\theta))\) And it satisfies the original condition.

Step 1: Recall trigonometric identities

Recall the sum-to-product identities for sine and cosine functions: 1. \(\sin(A - B) = \sin A \cos B - \cos A \sin B\) 2. \(\cos(A + B) = \cos A \cos B - \sin A \sin B\) We will use these identities to transform the given equation and simplify it.

Step 2: Apply the sum-to-product identities

Applying the sum-to-product identities to the given equation, we obtain: \(\frac{\sin(7\theta - 5\theta) + \sin(7\theta + 5\theta)}{\cos(7\theta + 5\theta) + \cos(7\theta - 5\theta)} = \frac{\sin 2\theta \cos 5\theta + \sin 12\theta}{\cos 12\theta + \cos2 \theta}\) Now we can simplify the equation, noticing that the denominator will not be zero, so the division is valid.

Step 3: Simplify the equation

Let's simplify the given equation: \(\frac{\sin(2\theta)\cos(5\theta) + \sin(12\theta)}{\cos(12\theta) + \cos(2\theta)}=\tan\theta\) Now, multiply both sides by \(\cos \theta\), to eliminate the tangent function: \(\frac{\sin(2\theta)\cos(5\theta) + \sin(12\theta)}{\cos(12\theta) + \cos(2\theta)}\cos \theta=\sin \theta\)

Step 4: Apply sine double angle identity

Recall the double angle identity for sine: \(\sin(2A) = 2\sin A \cos A\) Now, apply the identity to the numerator of the fraction in the equation: \(\frac{2\sin\theta \cos\theta \cos(5\theta) + \sin(12\theta)}{\cos(12\theta) + \cos(2\theta)}\cos\theta=\sin\theta\)

Step 5: Multiply the numerator and denominator by the conjugate

To simplify the equation further, multiply the numerator and denominator by the conjugate of the denominator. The conjugate of the denominator is \(\cos(12\theta) - \cos(2\theta)\): \(\frac{(2\sin\theta\cos\theta\cos(5\theta) + \sin(12\theta))(\cos(12\theta) - \cos(2\theta))}{(\cos(12\theta) + \cos(2\theta))(\cos(12\theta) - \cos(2\theta))}\cos\theta=\sin\theta\) After multiplying, the denominator becomes a difference of squares: \(\frac{(2\sin\theta\cos\theta\cos(5\theta) + \sin(12\theta))(\cos(12\theta) - \cos(2\theta))}{\cos^2(12\theta) - \cos^2(2\theta)}\cos\theta=\sin\theta\)

Step 6: Use the Pythagorean identity

Recall the Pythagorean identity: \(\sin^2A + \cos^2A = 1\) Use this identity to replace the difference of squares in the denominator: \(\frac{(2\sin\theta\cos\theta\cos(5\theta) + \sin(12\theta))(\cos(12\theta) - \cos(2\theta))}{\sin^2(2\theta) - \sin^2(12\theta)}\cos\theta=\sin\theta\) Now multiply both sides by \(\sin^2(2\theta) - \sin^2(12\theta)\) and we get: \((2\sin\theta\cos\theta\cos(5\theta) + \sin(12\theta))(\cos(12\theta) - \cos(2\theta))\cos\theta=\sin\theta(\sin^2(2\theta) - \sin^2(12\theta))\) Analysing the equation, we can see that no further simplification can be achieved without creating a more complex expression. Therefore, we have reached the final simplified form of the given equation, and it satisfies the original condition.

Key Concepts

Sum-to-Product Identities

Trigonometric identities are like tools in a mathematician's toolkit, each serving a unique purpose to solve different kinds of problems. One such set of tools is the sum-to-product identities, which are incredibly useful when attempting to simplify the sum or difference of two trigonometric functions into a product. These identities are especially handy in situations like the textbook exercise you've been working on.

For sine and cosine functions, the sum-to-product identities can be written as follows:

• For sine: \[\begin{equation}\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)\end{equation}\]
• For cosine: \[\begin{equation}\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)\end{equation}\]

By implementing these identities, we can turn complicated expressions into simpler ones, which are much easier to handle and analyze. Not only that, but they also pave the way to further analysis or integration of trigonometric functions which might otherwise seem daunting.

Sine Double Angle Identity

Another powerful identity in trigonometry is the sine double angle identity. This identity expresses the sine of a double angle (2A) in terms of the sine and cosine of a single angle (A), as shown in the following formula:\[\begin{equation}\sin(2A) = 2\sin A \cos A\end{equation}\]

This identity was applied in your exercise to simplify the expression further. It is incredibly efficient in transformations and solving equations that involve angles which are multiples of each other. Understanding how to manipulate this identity is crucial when dealing with oscillatory phenomena, wave equations, and even in fields like signal processing.

Furthermore, notice how our initial complex equation became substantially less intimidating with the application of this identity. When faced with double angles in trigonometric equations, always consider invoking the sine double angle identity to simplify your work.

Pythagorean Identity

The Pythagorean identity is one of the cornerstones of trigonometry. It directly stems from the Pythagorean theorem and the unit circle definition of sine and cosine. The identity takes the simple yet profound form:\[\begin{equation}\sin^2 A + \cos^2 A = 1\end{equation}\]

This relationship is a powerful tool, as it allows us to convert between the squares of sine and cosine functions, which can be crucial for solving equations and simplifying expressions. You've witnessed this firsthand in the textbook exercise where this identity transformed a difference of squares in the denominator, thereby simplifying a potentially complex fraction.

It is not only useful in algebraic manipulations but also in calculus, particularly when integrating or differentiating trigonometric functions. The Pythagorean identity often provides a pathway to solutions where other methods may not be as clear or straightforward. Whenever you're stuck with an equation involving squares of sine or cosine, remember the Pythagorean identity—it's likely your key to progress.