Question

2 sin15 * sin 75 Решения

Answer 1

$2sin15 \cdot sin75=\\\\=2 \cdot \frac{1}{2}(cos(-60)-cos90)=\\\\=2 \cdot \frac{1}{2} \cdot (\frac{1}{2}-0)=\\\\=\frac{1}{2}-0=\frac{1}{2}=0.5$

Answer 2

It looks like you're asking for the product of two trigonometric expressions: $2 \sin(15^\circ) \cdot \sin(75^\circ)$.

To simplify this expression, you can use the trigonometric identity:

$\sin(A) \cdot \sin(B) = \frac{1}{2} [\cos(A-B) - \cos(A+B)]$

In this case, $A = 15^\circ$ and $B = 75^\circ$, so the expression becomes:

$2 \sin(15^\circ) \cdot \sin(75^\circ) = 2 \cdot \frac{1}{2} [\cos(15^\circ - 75^\circ) - \cos(15^\circ + 75^\circ)]$

Simplify the angles:

$= \cos(-60^\circ) - \cos(90^\circ)$

Since $\cos(-60^\circ) = \cos(60^\circ)$ (cosine function is even):

$= \cos(60^\circ) - \cos(90^\circ)$

And $\cos(60^\circ) = \frac{1}{2}$:

$= \frac{1}{2} - 0$

Finally:

$= \frac{1}{2}$

So, $2 \sin(15^\circ) \cdot \sin(75^\circ) = \frac{1}{2}$.