Problem 289 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Prove that \(\sin \theta \sec 3 \theta=\frac{1}{2}(\tan 3 \theta-\tan \theta)\) and hence find the sum to \(n\) terms of the series \(\sin \theta \sec 3 \theta+\sin 3 \theta \sec 3^{2} \theta+\sin 3^{2} \theta \sec 3^{3} \theta+\ldots \ldots \ldots \ldots \ldots .\)

Step-by-Step Solution

Answer

The sum to \(n\) terms of the series \(\sin \theta \sec 3 \theta+\sin 3 \theta \sec 3^{2} \theta+\sin 3^{2} \theta \sec 3^{3} \theta+\ldots \) is \(\frac{1}{2}(-\tan \theta + \tan 3^n \theta).\)

Step 1: Convert to Sine and Cosine

As the fundamental trigonometric functions are sine and cosine, convert the tangent and secant functions into sine and cosine for clarity and simplicity. From the fundamental trigonometric identities, we know \(\sec x = \frac{1}{\cos x} \) and \(\tan x = \frac{\sin x}{\cos x} \). So the given equation becomes: \(\sin \theta \cdot \frac{1}{\cos 3\theta} = \frac{1}{2} \left( \frac{\sin 3\theta}{\cos 3\theta} - \frac{\sin \theta}{\cos\theta} \right)\)

Step 2: Simplify and Rewrite

After simplification, the equation is \(\frac{\sin \theta}{\cos 3\theta} = \frac{1}{2} (\sin 3\theta \cos\theta - \sin \theta \cos 3\theta)\). To prove it, rewrite the right hand side as \(\frac{1}{2}(\sin 3\theta \cos\theta - 2 \sin \theta \cos 3\theta + \sin \theta \cos 3\theta)\), which simplifies to \(\frac{1}{2}[\sin 3\theta \cos\theta - 2 \sin \theta \cos 3\theta] + \frac{\sin \theta}{\cos 3\theta}\). This makes it easier to show that this is equal to the left hand side of the original equation.

Step 3: Apply Trigonometric Identity

Recall the identity \[\sin a \cos b = \frac{1}{2}[\sin(a+b) + \sin(a-b)]\]. Apply the same to part of the equation inside the braces and keep the remaining part the same. The equation will look like this: \[\frac{1}{2}\left[\frac{1}{2}(\sin(3\theta+\theta) + \sin(3\theta - \theta)) - 2 \sin \theta \cos 3\theta \right] + \frac{\sin \theta}{\cos 3\theta}\].

Step 4: Simplify

On simplifying, it becomes \(\frac{1}{2}[\sin 4\theta + \sin 2\theta - 2 \sin \theta \cos 3\theta] + \frac{\sin \theta}{\cos 3\theta}\). Finally, use the identity \[\sin a - 2 \sin b \cos c = \sin a - \sin 2c = \sin (a - 2c)\] to simplify the equation to \(\frac{1}{2}[\sin 4\theta + \sin 2\theta - \sin 2\theta] + \frac{\sin \theta}{\cos 3\theta}\).

Step 5: Prove the Identity

Now it's more clear that the equation becomes \(\frac{\sin \theta}{\cos 3\theta}\). So now it's apparent that the given identity holds true.

Step 6: Find the Sum of the Series

The given series is \(\sin \theta \sec 3 \theta + \sin 3 \theta \sec 3^{2} \theta + \sin 3^{2} \theta \sec 3^{3} \theta + ...) \). Replace each term using the proven identity. The series becomes a geometric series: \(\frac{1}{2}(\tan 3 \theta - \tan \theta) + \frac{1}{2}(\tan 3^{2} \theta - \tan 3 \theta) + \frac{1}{2}(\tan 3^{3} \theta - \tan 3^{2} \theta) + ...) \). To find the sum of this series up to n terms, note that it is a telescoping series. Most of the terms cancel out in the series, leaving just two terms for the sum up to the nth term, namely \(\frac{1}{2}(-\tan \theta + \tan 3^n \theta)\).

Key Concepts

Geometric Series

A geometric series is a sequence of terms where each term is found by multiplying the previous one by a fixed, non-zero number called the common ratio. These series have a common structure and can converge or diverge depending on the value of the common ratio.

• If the common ratio is between -1 and 1, the series converges.
• If it's greater than 1 or less than -1, the series diverges.

The sum of a geometric series with n terms can be expressed using the formula:\[S_n = a \frac{1 - r^n}{1 - r}\]where \(a\) is the first term, and \(r\) is the common ratio. Understanding geometric series helps in recognizing patterns and simplifying complex mathematical expressions.

Telescoping Series

A telescoping series is a type of series where many terms cancel out when simplified, making it easier to find the sum. Consider it as a series where each term "telescopes" into the next, revealing the sum quickly.

• In these series, intermediate terms cancel, leaving only the first and last terms.
• This series is particularly useful in simplifying complex sums.

For example, the series \(rac{1}{2}(\tan 3 \theta - \tan \theta) + \frac{1}{2}(\tan 3^2 \theta - \tan 3 \theta) + \ldots\) is telescoping because most terms cancel, leaving \(\frac{1}{2}(-\tan \theta + \tan 3^n \theta)\) as the sum.

Trigonometric Functions

Trigonometric functions are the building blocks of trigonometry, connecting angles and ratios of sides in right triangles. These include sine (\(\sin\)), cosine (\(\cos\)), and tangent (\(\tan\)).

• \(\sin\) and \(\cos\) function values range between -1 and 1.
• \(\tan\) varies between negative and positive infinity, with undefined points at odd multiples of \(\frac{\pi}{2}\).

Understanding these functions is vital as they are foundational for forming identities and relationships between angles and side lengths. They frequently appear in problems involving periodic phenomena, oscillations, and waves.

Sin and Cos Identities

Sine and cosine identities are formulas that relate different trigonometric functions to each other. These identities allow for the transformation and simplification of trigonometric expressions.
Some basic identities include:

• \(\sin^2 \theta + \cos^2 \theta = 1\)
• \(\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta\)
• \(\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta\)

Using these identities, more complex problems involving radians and degree measures can be tackled. They are especially useful in proving relations, like transforming \(\tan\) and \(\sec\), as seen in the given exercise.