Question

If $\theta = \frac{\pi}{15}$, then value of $\frac{(\sin 5\theta + \cos 5\theta) + (\sin 7\theta - \cos 7\theta)}{(\sin 8\theta + \cos 8\theta) + (\sin 10\theta - \cos 10\theta)}$ is equal to -

• A. -1
• B. $\frac{\sqrt{3}-1}{2\sqrt{2}}$
• C. 1
• D. $\frac{\sqrt{3}+1}{2\sqrt{2}}$

Answer 1

Answer: (C)

1

Answer 2

The problem asks us to find the value of a given trigonometric expression when $\theta = \frac{\pi}{15}$.

Let the given expression be $E$: $E = \frac{(\sin 5\theta + \cos 5\theta) + (\sin 7\theta - \cos 7\theta)}{(\sin 8\theta + \cos 8\theta) + (\sin 10\theta - \cos 10\theta)}$

First, let's simplify the numerator ($N$) and the denominator ($D$) separately.

Step 1: Simplify the Numerator (N) $N = (\sin 5\theta + \cos 5\theta) + (\sin 7\theta - \cos 7\theta)$ Rearrange the terms: $N = (\sin 5\theta + \sin 7\theta) + (\cos 5\theta - \cos 7\theta)$ Apply the sum-to-product formulas: $\sin A + \sin B = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$ $\cos A - \cos B = -2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$

For the first part $(\sin 5\theta + \sin 7\theta)$: $A = 5\theta, B = 7\theta$ $\frac{A+B}{2} = \frac{5\theta+7\theta}{2} = 6\theta$ $\frac{A-B}{2} = \frac{5\theta-7\theta}{2} = -\theta$ So, $\sin 5\theta + \sin 7\theta = 2 \sin(6\theta) \cos(-\theta) = 2 \sin(6\theta) \cos(\theta)$ (since $\cos(-x) = \cos x$)

For the second part $(\cos 5\theta - \cos 7\theta)$: $A = 5\theta, B = 7\theta$ $\frac{A+B}{2} = 6\theta$ $\frac{A-B}{2} = -\theta$ So, $\cos 5\theta - \cos 7\theta = -2 \sin(6\theta) \sin(-\theta) = -2 \sin(6\theta) (-\sin\theta) = 2 \sin(6\theta) \sin(\theta)$ (since $\sin(-x) = -\sin x$)

Substitute these back into $N$: $N = 2 \sin(6\theta) \cos(\theta) + 2 \sin(6\theta) \sin(\theta)$ Factor out $2 \sin(6\theta)$: $N = 2 \sin(6\theta) (\cos(\theta) + \sin(\theta))$

Step 2: Simplify the Denominator (D) $D = (\sin 8\theta + \cos 8\theta) + (\sin 10\theta - \cos 10\theta)$ Rearrange the terms: $D = (\sin 8\theta + \sin 10\theta) + (\cos 8\theta - \cos 10\theta)$ Apply the sum-to-product formulas:

For the first part $(\sin 8\theta + \sin 10\theta)$: $A = 8\theta, B = 10\theta$ $\frac{A+B}{2} = \frac{8\theta+10\theta}{2} = 9\theta$ $\frac{A-B}{2} = \frac{8\theta-10\theta}{2} = -\theta$ So, $\sin 8\theta + \sin 10\theta = 2 \sin(9\theta) \cos(-\theta) = 2 \sin(9\theta) \cos(\theta)$

For the second part $(\cos 8\theta - \cos 10\theta)$: $A = 8\theta, B = 10\theta$ $\frac{A+B}{2} = 9\theta$ $\frac{A-B}{2} = -\theta$ So, $\cos 8\theta - \cos 10\theta = -2 \sin(9\theta) \sin(-\theta) = -2 \sin(9\theta) (-\sin\theta) = 2 \sin(9\theta) \sin(\theta)$

Substitute these back into $D$: $D = 2 \sin(9\theta) \cos(\theta) + 2 \sin(9\theta) \sin(\theta)$ Factor out $2 \sin(9\theta)$: $D = 2 \sin(9\theta) (\cos(\theta) + \sin(\theta))$

Step 3: Evaluate the Expression E Now, substitute the simplified $N$ and $D$ back into $E$: $E = \frac{2 \sin(6\theta) (\cos(\theta) + \sin(\theta))}{2 \sin(9\theta) (\cos(\theta) + \sin(\theta))}$ Given $\theta = \frac{\pi}{15}$, $\cos(\theta) + \sin(\theta) = \cos(\frac{\pi}{15}) + \sin(\frac{\pi}{15})$. Since $\frac{\pi}{15}$ is in the first quadrant, both $\cos(\frac{\pi}{15})$ and $\sin(\frac{\pi}{15})$ are positive, so their sum is not zero. Therefore, we can cancel the common term $(\cos(\theta) + \sin(\theta))$: $E = \frac{\sin(6\theta)}{\sin(9\theta)}$

Step 4: Substitute the value of $\theta$ Given $\theta = \frac{\pi}{15}$. Calculate $6\theta$ and $9\theta$: $6\theta = 6 \times \frac{\pi}{15} = \frac{2\pi}{5}$ $9\theta = 9 \times \frac{\pi}{15} = \frac{3\pi}{5}$

Substitute these values into $E$: $E = \frac{\sin(\frac{2\pi}{5})}{\sin(\frac{3\pi}{5})}$ We know that $\sin(\pi - x) = \sin x$. Here, $\frac{3\pi}{5} = \pi - \frac{2\pi}{5}$. So, $\sin(\frac{3\pi}{5}) = \sin(\pi - \frac{2\pi}{5}) = \sin(\frac{2\pi}{5})$.

Substitute this back into the expression for $E$: $E = \frac{\sin(\frac{2\pi}{5})}{\sin(\frac{2\pi}{5})}$ Since $\frac{2\pi}{5} = 72^\circ$, $\sin(\frac{2\pi}{5}) \neq 0$. Therefore, $E = 1$

The final answer is $\boxed{1}$.

Explanation of the solution:

1. Simplify Numerator: Group terms $(\sin 5\theta + \sin 7\theta)$ and $(\cos 5\theta - \cos 7\theta)$. Apply sum-to-product formulas $\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$ and $\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$. This yields $N = 2 \sin(6\theta)(\cos\theta + \sin\theta)$.
2. Simplify Denominator: Similarly, group terms $(\sin 8\theta + \sin 10\theta)$ and $(\cos 8\theta - \cos 10\theta)$. Apply sum-to-product formulas. This yields $D = 2 \sin(9\theta)(\cos\theta + \sin\theta)$.
3. Simplify Expression: Divide the simplified numerator by the simplified denominator. The common factor $2(\cos\theta + \sin\theta)$ cancels out, leaving $E = \frac{\sin(6\theta)}{\sin(9\theta)}$.
4. Substitute $\theta$: Substitute $\theta = \frac{\pi}{15}$ into the simplified expression. This gives $6\theta = \frac{2\pi}{5}$ and $9\theta = \frac{3\pi}{5}$. So, $E = \frac{\sin(\frac{2\pi}{5})}{\sin(\frac{3\pi}{5})}$.
5. Final Evaluation: Use the identity $\sin(\pi - x) = \sin x$. Since $\frac{3\pi}{5} = \pi - \frac{2\pi}{5}$, we have $\sin(\frac{3\pi}{5}) = \sin(\frac{2\pi}{5})$. Therefore, $E = \frac{\sin(\frac{2\pi}{5})}{\sin(\frac{2\pi}{5})} = 1$.