Understanding the Trigonometric Product

The question asks us to find the value of the product of three sine functions with specific angles: $\sin 12^\circ \sin 48^\circ \sin 54^\circ$. This involves using trigonometric identities to simplify the expression and arrive at a numerical value.

Key Trigonometric Identities

To solve this problem, we will use the following standard trigonometric identities:

• Product-to-Sum Formula: $\sin A \sin B = \dfrac{1}{2} [\cos(A-B) - \cos(A+B)]$
• Co-function Identity: $\sin(90^\circ - \theta) = \cos \theta$
• Difference of Squares: $a^2 - b^2 = (a-b)(a+b)$
• Known Value: $\cos 36^\circ = \dfrac{\sqrt{5} + 1}{4}$
• Known Value: $\cos 60^\circ = \dfrac{1}{2}$

Step-by-Step Solution

Let the expression be $P$. So, $P = \sin 12^\circ \sin 48^\circ \sin 54^\circ$. We will evaluate this step by step:

Step 1: Apply Product-to-Sum Formula

First, let's combine the first two terms, $\sin 12^\circ \sin 48^\circ$, using the product-to-sum formula $\sin A \sin B = \dfrac{1}{2} [\cos(A-B) - \cos(A+B)]$. Here, $A = 48^\circ$ and $B = 12^\circ$.

$$ \sin 48^\circ \sin 12^\circ = \dfrac{1}{2} [\cos(48^\circ - 12^\circ) - \cos(48^\circ + 12^\circ)] $$

$$ \sin 48^\circ \sin 12^\circ = \dfrac{1}{2} [\cos 36^\circ - \cos 60^\circ] $$

Step 2: Substitute Known Values

We know that $\cos 60^\circ = \dfrac{1}{2}$. Substituting this value:

$$ \sin 48^\circ \sin 12^\circ = \dfrac{1}{2} \left[\cos 36^\circ - \dfrac{1}{2}\right] $$

Step 3: Substitute back into the expression P

Now, substitute this result back into the original expression for P:

$$ P = \left( \dfrac{1}{2} \left[\cos 36^\circ - \dfrac{1}{2}\right] \right) \sin 54^\circ $$

Step 4: Use Co-function Identity

We can simplify $\sin 54^\circ$ using the co-function identity $\sin(90^\circ - \theta) = \cos \theta$.

$$ \sin 54^\circ = \sin(90^\circ - 36^\circ) = \cos 36^\circ $$

Substitute this into the expression for P:

$$ P = \left( \dfrac{1}{2} \left[\cos 36^\circ - \dfrac{1}{2}\right] \right) \cos 36^\circ $$

$$ P = \dfrac{1}{2} \left[\cos^2 36^\circ - \dfrac{1}{2} \cos 36^\circ\right] $$

Step 5: Substitute the Value of cos 36°

The exact value of $\cos 36^\circ$ is $\dfrac{\sqrt{5} + 1}{4}$. Substitute this value:

$$ P = \dfrac{1}{2} \left[ \left(\dfrac{\sqrt{5} + 1}{4}\right)^2 - \dfrac{1}{2} \left(\dfrac{\sqrt{5} + 1}{4}\right) \right] $$

Step 6: Simplify the Expression

Let's simplify the terms inside the bracket:

Calculate $\left(\dfrac{\sqrt{5} + 1}{4}\right)^2$: $$ \left(\dfrac{\sqrt{5} + 1}{4}\right)^2 = \dfrac{(\sqrt{5})^2 + 2(\sqrt{5})(1) + 1^2}{4^2} = \dfrac{5 + 2\sqrt{5} + 1}{16} = \dfrac{6 + 2\sqrt{5}}{16} = \dfrac{3 + \sqrt{5}}{8} $$

Calculate $\dfrac{1}{2} \left(\dfrac{\sqrt{5} + 1}{4}\right)$: $$ \dfrac{1}{2} \left(\dfrac{\sqrt{5} + 1}{4}\right) = \dfrac{\sqrt{5} + 1}{8} $$

Now substitute these back into the expression for P:

$$ P = \dfrac{1}{2} \left[ \dfrac{3 + \sqrt{5}}{8} - \dfrac{\sqrt{5} + 1}{8} \right] $$

$$ P = \dfrac{1}{2} \left[ \dfrac{(3 + \sqrt{5}) - (\sqrt{5} + 1)}{8} \right] $$

$$ P = \dfrac{1}{2} \left[ \dfrac{3 + \sqrt{5} - \sqrt{5} - 1}{8} \right] $$

$$ P = \dfrac{1}{2} \left[ \dfrac{2}{8} \right] $$

$$ P = \dfrac{1}{2} \times \dfrac{1}{4} $$

Step 7: Final Calculation

Multiply the fractions to get the final value:

$$ P = \dfrac{1}{8} $$

Result

The value of $\sin 12^\circ \sin 48^\circ \sin 54^\circ$ is $\dfrac{1}{8}$.