Question

Sin (-15°) cos 75° + cos 15° sin 75°;соs(a-b)-2sina•sinb.a-alfa b-bettaпожалуйсто ​

Answer 1

Решение на фото(формулы написаны как объяснение)

Answer 2

To calculate the expression $\sin(-15^\circ) \cos(75^\circ) + \cos(15^\circ) \sin(75^\circ)$, we can use trigonometric identities and properties.

First, let's evaluate $\sin(-15^\circ)$ and $\cos(75^\circ)$ using the properties of trigonometric functions:

1. $\sin(-15^\circ) = -\sin(15^\circ)$ (since sine is an odd function)
2. $\cos(75^\circ) = \cos(45^\circ + 30^\circ)$ (using the sum-to-product trigonometric identity) $= \cos(45^\circ) \cos(30^\circ) - \sin(45^\circ) \sin(30^\circ)$ $= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \times \frac{1}{2}$ $= \frac{\sqrt{6} - \sqrt{2}}{4}$

Next, let's evaluate $\cos(15^\circ)$ and $\sin(75^\circ)$:

3. $\cos(15^\circ) = \cos(45^\circ - 30^\circ)$ (using the difference-to-product trigonometric identity) $= \cos(45^\circ) \cos(30^\circ) + \sin(45^\circ) \sin(30^\circ)$ $= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2}$ $= \frac{\sqrt{6} + \sqrt{2}}{4}$

4. $\sin(75^\circ) = \sin(45^\circ + 30^\circ)$ (using the sum-to-product trigonometric identity) $= \sin(45^\circ) \cos(30^\circ) + \cos(45^\circ) \sin(30^\circ)$ $= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2}$ $= \frac{\sqrt{6} + \sqrt{2}}{4}$