Problem 81 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \frac{\sin A+\sin 3 A+\sin 5 A+\sin 7 A}{\cos A+\cos 3 A+\cos 5 A+\cos 7 A}=\tan 4 A $$

Step-by-Step Solution

Answer

To prove the given equation \(\frac{\sin A+\sin 3 A+\sin 5 A+\sin 7 A}{\cos A+\cos 3 A+\cos 5 A+\cos 7 A}=\tan 4 A\), we apply the sine and cosine addition and subtraction formulas. We first group the sine and cosine terms separately and then apply the formulas. After substituting the simplified terms back into the equation and canceling out common factors, we arrive at the result \(\frac{\sin 4 A}{\cos 4 A}\), which is equal to \(\tan 4 A\).

Step 1: Group sine terms and cosine terms

First, let's group the sine and cosine terms separately. \[ \frac{\sin A+\sin 3 A+\sin 5 A+\sin 7 A}{\cos A+\cos 3 A+\cos 5 A+\cos 7 A}=\frac{\sin A + \sin 7 A + \sin{3A} + \sin{5A}}{\cos A + \cos 7 A + \cos{3A} + \cos{5A}} \]

Step 2: Apply sine and cosine addition and subtraction formulas

We can apply the sine addition formula to the numerators \(\sin A + \sin 7 A\) and \(\sin 3 A + \sin 5 A\): \[ \sin A + \sin 7 A = 2\sin \frac{A + 7 A}{2}\cos \frac{A - 7 A}{2} = 2\sin 4 A \cos(-3 A) \] \[ \sin 3 A + \sin 5 A = 2\sin \frac{3 A + 5 A}{2}\cos \frac{3 A - 5 A}{2} = 2\sin 4 A \cos(- A) \] Similarly, we apply the cosine addition formula to the denominators \(\cos A + \cos 7 A\) and \(\cos 3 A + \cos 5 A\): \[ \cos A + \cos 7 A = 2\cos \frac{A + 7 A}{2}\cos \frac{A - 7 A}{2} = 2\cos 4 A \cos(-3 A) \] \[ \cos 3 A + \cos 5 A = 2\cos \frac{3 A + 5 A}{2}\cos \frac{3 A - 5 A}{2} = 2\cos 4 A \cos(- A) \]

Step 3: Substitute back into the equation and simplify

Substitute the simplified sine and cosine terms back into the equation: \[ \frac{2\sin 4 A \cos (-3 A) + 2\sin 4 A \cos(-A)}{2\cos 4 A \cos(-3 A) + 2\cos 4 A \cos(-A)} = \frac{\sin 4 A(\cos(-3 A) + \cos(-A))}{\cos 4 A (\cos(-3 A) + \cos(-A))} \] Now, we can cancel the \((\cos(-3 A) + \cos(-A))\) term both from the numerator and denominator: \[ \frac{\sin 4 A}{\cos 4 A} \]

Step 4: Convert to tangent

Finally, we convert the result to tangent using the formula \(\tan(x) = \frac{\sin x}{\cos x}\): \[ \frac{\sin 4 A}{\cos 4 A} = \tan 4 A \] Now we have shown that the given equation holds true: \[ \frac{\sin A+\sin 3 A+\sin 5 A+\sin 7 A}{\cos A+\cos 3 A+\cos 5 A+\cos 7 A} = \tan 4 A \]

Key Concepts

Sine Addition Formula

Understanding the sine addition formula is essential when analyzing trigonometric expressions. It's a tool to simplify expressions involving the sine of a sum or difference. In its most basic form, the formula is given by:
\[\begin{equation}\begin{aligned}\text{For the sum,}\ \sin(A+B) &= \sin A \cos B + \cos A \sin B. \end{aligned} \end{equation}\]
For the difference, it’s:
\[\begin{equation}\begin{aligned}\sin(A-B) &= \sin A \cos B - \cos A \sin B. \end{aligned} \end{equation}\]
This identity allows us to combine two sine terms with different arguments into a single expression. It's particularly useful in problems like our textbook exercise, where we encounter sums of sine functions with different angles, making it simpler to solve complex trigonometric equations. The cool thing is that we can use this to transform the numerator of our given problem, grouping sine terms in pairs, which then reveal a pattern that we can simplify even further.

Cosine Addition Formula

Just like the sine addition formula, the cosine addition formula is a powerful tool in trigonometry. It helps in breaking down the cosine of a sum or difference of two angles into a product involving cosines and sines of individual angles. The formula states that:
\[\begin{equation}\begin{aligned}\text{For the sum,}\ \cos(A+B) &= \cos A \cos B - \sin A \sin B. \end{aligned} \end{equation}\]
And for the difference:
\[\begin{equation}\begin{aligned}\cos(A-B) &= \cos A \cos B + \sin A \sin B. \end{aligned} \end{equation}\]
In our exercise, this is strategically applied to the denominator. This technique helps in revealing the underlying structure of the trigonometric expression, which can be manipulated to bring the equation into a form that is easily solvable. By pairing cosine terms just like sine terms, it contributes to the simplification of the overall equation, which is vital in pursuing a clearer path to the solution.

Tangent Function

The tangent function combines both sine and cosine and can be considered as the ratio of sine to cosine for a given angle. The formula is straightforward:
\[\begin{equation}\tan A = \frac{\sin A}{\cos A}\end{equation}\]
In our exercise, the final step involves converting the simplified sine and cosine ratios into a tangent. This step merges both the numerator and denominator into a form that is recognized as the tangent function. It's interesting to note how the complexities of trigonometry can be resolved by understanding and applying such simple relationships between trigonometric functions. This understanding not only led to the solution but also validated the initial trigonometric identity. Students often find comfort in knowing that by mastering the basic identities, more intricate equations become easier to tackle. This relationship is key to solving many trigonometry problems efficiently and is a shining example of the interconnectedness of trigonometric functions.