Question

Sin(2x-30)-sin(2x-60) Дать в виде Добутка

Answer 1

Обыкновенная разность синусов

sin(2x-30)-sin(2x-60)=2sin((2x-30-2x+60)/2)cos((2x-30+2x-60)/2)=
=2sin(15)cos(2x-45)

Answer 2

I can help you with your math question. Here is the detailed solution:

Using the formula for the difference of two sines, we have:

$$\sin(2x-30)-\sin(2x-60) = 2\cos(\frac{2x-30+2x-60}{2})\sin(\frac{2x-30-2x+60}{2})$$

Simplifying the expression, we get:

$$\sin(2x-30)-\sin(2x-60) = 2\cos(2x-45)\sin(15)$$

Using the values of $\cos(2x-45)$ and $\sin(15)$, we can write the final answer as:

$$\sin(2x-30)-\sin(2x-60) = \sqrt{2}\cos(2x-45)\frac{\sqrt{6}-\sqrt{2}}{4}$$

This is the product form of the expression. I hope this helps.Ответь+подробно.+Sin(2x-30)-sin(2x-60)++Дать+в+виде+Добутка