`sin^(2)24^(@)-cos^(2)84^(@)=?`

Options

A. `(sqrt(5)-1)/(8)`
B. `(sqrt(5)+1)/(8)`
C. `(sqrt(5)-1)/(4)`
D. None of these

Answer

To solve the problem \( \sin^2 24^\circ - \cos^2 84^\circ \), we can follow these steps:

### Step 1: Rewrite \( \cos^2 84^\circ \)
Using the co-function identity, we know that:
\[ \cos(90^\circ - \theta) = \sin(\theta) \]
Thus, we can rewrite \( \cos 84^\circ \) as:
\[ \cos 84^\circ = \sin(90^\circ - 84^\circ) = \sin 6^\circ \]
So, we have:
\[ \cos^2 84^\circ = \sin^2 6^\circ \]

### Step 2: Substitute into the expression
Now we substitute this back into the original expression:
\[ \sin^2 24^\circ - \cos^2 84^\circ = \sin^2 24^\circ - \sin^2 6^\circ \]

### Step 3: Apply the difference of squares identity
We can use the identity \( a^2 - b^2 = (a - b)(a + b) \):
\[ \sin^2 24^\circ - \sin^2 6^\circ = (\sin 24^\circ - \sin 6^\circ)(\sin 24^\circ + \sin 6^\circ) \]

### Step 4: Use the sum and difference identities for sine
Now we apply the sine addition and subtraction identities:
\[ \sin a - \sin b = 2 \sin\left(\frac{a-b}{2}\right) \cos\left(\frac{a+b}{2}\right) \]
\[ \sin a + \sin b = 2 \sin\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) \]
For \( a = 24^\circ \) and \( b = 6^\circ \):
- \( \sin 24^\circ - \sin 6^\circ = 2 \sin\left(\frac{24^\circ - 6^\circ}{2}\right) \cos\left(\frac{24^\circ + 6^\circ}{2}\right) = 2 \sin(9^\circ) \cos(15^\circ) \)
- \( \sin 24^\circ + \sin 6^\circ = 2 \sin\left(\frac{24^\circ + 6^\circ}{2}\right) \cos\left(\frac{24^\circ - 6^\circ}{2}\right) = 2 \sin(15^\circ) \cos(9^\circ) \)

### Step 5: Combine the results
Now substituting these results back into our expression:
\[ \sin^2 24^\circ - \sin^2 6^\circ = (2 \sin 9^\circ \cos 15^\circ)(2 \sin 15^\circ \cos 9^\circ) \]
This simplifies to:
\[ 4 \sin 9^\circ \cos 15^\circ \sin 15^\circ \cos 9^\circ \]

### Step 6: Use the double angle identity
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ = 4 \cdot \frac{1}{2} \sin 18^\circ \cdot \frac{1}{2} \sin 30^\circ = \sin 18^\circ \cdot \frac{1}{2} \]
Since \( \sin 30^\circ = \frac{1}{2} \), we have:
\[ = \sin 18^\circ \cdot \frac{1}{2} \]

### Step 7: Final Value
The value of \( \sin 30^\circ = \frac{1}{2} \) and we know \( \sin 18^\circ = \frac{\sqrt{5}-1}{4} \):
\[ \sin 18^\circ \cdot \frac{1}{2} = \frac{\sqrt{5}-1}{8} \]

Thus, the final answer is:
\[ \frac{\sqrt{5}-1}{8} \]