Q.The value of $\sin\dfrac{\pi}{10}\,\sin\dfrac{13\pi}{10}$ is
(A) $\dfrac{1}{2}$
(B) $-\dfrac{1}{2}$
(C) $-\dfrac{1}{4}$
(D) $1$

The key idea here is to use trigonometric identities to simplify the given expression, specifically angle transformations and standard trigonometric values.

First, we transform the angle $\dfrac{13\pi}{10}$:

Now, substitute this back into the expression:

We know the standard values for $\sin 18^\circ$ and $\sin 54^\circ$:

$\sin\dfrac{\pi}{10} = \sin 18^\circ = \dfrac{\sqrt{5}-1}{4}$ …

This problem tests your understanding of trigonometric identities for angle reduction and your knowledge of specific trigonometric values for angles like $18^\circ$ and $36^\circ$. The key is to recognize that the given angles are related and can be simplified to a form where standard values can be applied.

1. Simplify the second angle using angle reduction. The angle $\dfrac{13\pi}{10}$ is greater than $\pi$. We can express it as a sum involving $\pi$:

Now, we use the trigonometric identity for sine of an angle in the third quadrant:
> [!FORMULA]
> $\sin(\pi + \theta) = -\sin\theta$

Applying this identity:

> [!WARNING]
> When using angle reduction formulas like $\sin(\pi + \theta) = -\sin\theta$, always pay close attention to the quadrant of the original angle to correctly determine the sign of the reduced expression. $\frac{13\pi}{10}$ is in the third quadrant, where sine is negative.

2. Rewrite the original expression.

Substitute the simplified term back into the original expression:

3. Convert angles to degrees for easier recognition. It is often helpful to convert radian measures to degrees, especially when dealing with common angles like $18^\circ$ or $36^\circ$.

So the expression becomes:

4. Relate $\sin 54^\circ$ to a complementary angle.

   We know that $\sin\theta = \cos(90^\circ - \theta)$.

   $\sin(90^\circ - \theta) = \cos\theta$

   Applying this to $\sin 54^\circ$:

The expression now simplifies to:

5. Recall the standard values for $\sin 18^\circ$ and $\cos 36^\circ$.

   These are fundamental trigonometric values that are often required in competitive exams.

   Important

   $\sin 18^\circ = \dfrac{\sqrt{5}-1}{4}$

   $\cos 36^\circ = \dfrac{\sqrt{5}+1}{4}$

   ›Proof

   Derivation of $\sin 18^\circ$ and $\cos 36^\circ$

   Let $\theta = 18^\circ$. Then $5\theta = 90^\circ$.

   We can write $2\theta = 90^\circ - 3\theta$.

   Taking sine on both sides:

   $\sin(2\theta) = \sin(90^\circ - 3\theta)$

   $\sin(2\theta) = \cos(3\theta)$

   Using the double and triple angle formulas:

   $2\sin\theta\cos\theta = 4\cos^3\theta - 3\cos\theta$

   Since $\theta = 18^\circ$, $\cos\theta \neq 0$, so we can divide by $\cos\theta$:

   $2\sin\theta = 4\cos^2\theta - 3$

   Substitute $\cos^2\theta = 1 - \sin^2\theta$:

   $2\sin\theta = 4(1 - \sin^2\theta) - 3$

   $2\sin\theta = 4 - 4\sin^2\theta - 3$

   $4\sin^2\theta + 2\sin\theta - 1 = 0$

   Let $x = \sin\theta = \sin 18^\circ$. This is a quadratic equation:

   $4x^2 + 2x - 1 = 0$

   Using the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

   $x = \dfrac{-2 \pm \sqrt{2^2 - 4(4)(-1)}}{2(4)}$ …