Which is equal to $$ \cos 15^{\circ} $$ ? $$ \sin 30^{\circ} \cos 75^{\circ} \cos 30^{\circ} \sin 75^{\circ} $$ Watch a vides Submit an example or

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1. We start with the expression
   $$
   \sin30^\circ \cos75^\circ \cos30^\circ \sin75^\circ.
   $$

2. Substitute the known values:
   $$
   \sin30^\circ = \frac{1}{2}, \quad \cos30^\circ = \frac{\sqrt{3}}{2}.
   $$

3. Notice that using complementary angles we have:
   $$
   \cos75^\circ = \sin(90^\circ - 75^\circ) = \sin15^\circ \quad \text{and} \quad \sin75^\circ = \cos(90^\circ - 75^\circ) = \cos15^\circ.
   $$
   Therefore, the expression becomes:
   $$
   \frac{1}{2} \cdot \sin15^\circ \cdot \frac{\sqrt{3}}{2} \cdot \cos15^\circ.
   $$

4. Group the constants and the trigonometric factors:
   $$
   \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4},
   $$
   so the product is:
   $$
   \frac{\sqrt{3}}{4}\, \sin15^\circ\cos15^\circ.
   $$

5. Use the double-angle identity for sine:
   $$
   \sin(2\theta) = 2\sin\theta\cos\theta.
   $$
   With $$\theta = 15^\circ$$ we have:
   $$
   2\sin15^\circ\cos15^\circ = \sin30^\circ = \frac{1}{2}.
   $$
   Thus,
   $$
   \sin15^\circ\cos15^\circ = \frac{1}{4}.
   $$

6. Substitute back into the expression:
   $$
   \frac{\sqrt{3}}{4} \cdot \frac{1}{4} = \frac{\sqrt{3}}{16}.
   $$

7. Therefore, the given product simplifies to:
   $$
   \sin30^\circ \cos75^\circ \cos30^\circ \sin75^\circ = \frac{\sqrt{3}}{16}.
   $$

8. We also recall that
   $$
   \cos15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4},
   $$
   which is approximately $$0.966$$. In contrast,
   $$\frac{\sqrt{3}}{16} \approx 0.108$$.

9. Thus, the expression $$\sin30^\circ \cos75^\circ \cos30^\circ \sin75^\circ$$ is not equal to $$\cos15^\circ$$.
The expression $$\sin30^\circ \cos75^\circ \cos30^\circ \sin75^\circ$$ simplifies to $$\frac{\sqrt{3}}{16}$$, which is not equal to $$\cos15^\circ$$.

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