Question

The value of \[ \sin^6 15^\circ + \cos^6 15^\circ \text{ is equal to:} \]

• A. \( \frac{13}{16} \)
• B. \( \frac{11}{16} \)
• C. \( \frac{9}{16} \)
• D. \( \frac{7}{16} \)
• E. \( \frac{5}{16} \)

Hint:

When evaluating powers of trigonometric functions, use known identities and simplifications to break down the terms. For example, use the identity \( \sin^2 x + \cos^2 x = 1 \) to simplify many expressions.

Answer 1

The Correct Option is A

We are asked to find the value of \( \sin^6{15^\circ} + \cos^6{15^\circ} \). 
We can simplify this expression using algebraic identities.
First, recall the identity: \[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \] Let \( a = \sin^2{15^\circ} \) and \( b = \cos^2{15^\circ} \). 
Thus, the expression becomes: \[ \sin^6{15^\circ} + \cos^6{15^\circ} = (\sin^2{15^\circ} + \cos^2{15^\circ}) \left( (\sin^2{15^\circ})^2 - \sin^2{15^\circ} \cos^2{15^\circ} + (\cos^2{15^\circ})^2 \right) \] Since \( \sin^2{15^\circ} + \cos^2{15^\circ} = 1 \) (the Pythagorean identity), we are left with: \[ \sin^6{15^\circ} + \cos^6{15^\circ} = 1 \times \left( \sin^4{15^\circ} - \sin^2{15^\circ} \cos^2{15^\circ} + \cos^4{15^\circ} \right) \]
Next, simplify the expression inside the parentheses. Notice that \( \sin^4{15^\circ} + \cos^4{15^\circ} = (\sin^2{15^\circ} + \cos^2{15^\circ})^2 - 2\sin^2{15^\circ} \cos^2{15^\circ} \). 
Since \( \sin^2{15^\circ} + \cos^2{15^\circ} = 1 \), this simplifies to: \[ \sin^4{15^\circ} + \cos^4{15^\circ} = 1 - 2\sin^2{15^\circ} \cos^2{15^\circ} \] 
Thus, the original expression becomes: \[ \sin^6{15^\circ} + \cos^6{15^\circ} = 1 - 3\sin^2{15^\circ} \cos^2{15^\circ} \] Now, we need to compute \( \sin^2{15^\circ} \cos^2{15^\circ} \). 
Using the double angle identity, we know that: \[ \sin{30^\circ} = 2 \sin{15^\circ} \cos{15^\circ} \] Since \( \sin{30^\circ} = \frac{1}{2} \), we have: \[ 2 \sin{15^\circ} \cos{15^\circ} = \frac{1}{2} \quad \Rightarrow \quad \sin{15^\circ} \cos{15^\circ} = \frac{1}{4} \] 
Therefore: \[ \sin^2{15^\circ} \cos^2{15^\circ} = \left( \frac{1}{4} \right)^2 = \frac{1}{16} \] Substitute this into the equation for \( \sin^6{15^\circ} + \cos^6{15^\circ} \): \[ \sin^6{15^\circ} + \cos^6{15^\circ} = 1 - 3 \times \frac{1}{16} = 1 - \frac{3}{16} = \frac{16}{16} - \frac{3}{16} = \frac{13}{16} \] 
Thus, the correct answer is option (A), \( \frac{13}{16} \).