Question

Дано: sin x - sin y = m ; cos x+cos y = n найти: sin (x-y) =? и cos(x-y)=?

Answer 1

Дано: sinx-siny=m; cosx+cosy=n. Найти: sin(x-y) и cos(x-y).
Решение:
1. Воспользуемся формулами разность синусов и сумма косинусов:
$sinx-siny=2sin \frac{x-y}{2}cos \frac{x+y}{2}=m; cosx+cosy=2cos \frac{x+y}{2}cos \frac{x-y}{2}=n.$
Заметим, что оба равенства содержат один и тот же член: $cos \frac{x+y}{2}$. Выразим его из обоих равенств:
$cos \frac{x+y}{2}= \frac{m}{2sin \frac{x-y}{2}};cos \frac{x+y}{2}= \frac{n}{2cos \frac{x-y}{2}}.$
В получившихся равенствах левые части равны, значит, равны и правые части:
$\frac{m}{2sin \frac{x-y}{2}}= \frac{n}{2cos \frac{x-y}{2}}$.
Преобразуем данное равенство:
$\frac{2sin \frac{x-y}{2}}{2cos \frac{x-y}{2}}= \frac{m}{n};$
$\frac{sin \frac{x-y}{2}}{cos \frac{x-y}{2}}= \frac{m}{n};$
$( \frac{sin \frac{x-y}{2}}{cos \frac{x-y}{2}})^{2}=( \frac{m}{n})^{2};$
$\frac{sin^{2} \frac{x-y}{2}}{cos^{2} \frac{x-y}{2}}= \frac{m^{2}}{n^{2}};$
Теперь используем формулы понижения степени синуса и косинуса:
$\frac{1-cos(x-y)}{2}: \frac{1+cos(x-y)}{2}= \frac{m^{2}}{n^{2}};$
Преобразуем данное равенство:
$\frac{1-cos(x-y)}{1+cos(x-y)}= \frac{m^{2}}{n^{2}};$
n²(1-cos(x-y))=m²(1+cos(x-y));
n²-n²cos(x-y)=m²+m²cos(x-y);
m²cos(x-y)+n²cos(x-y)=n²-m²;
cos(x-y)(m²+n²)=n²-m²;
$cos(x-y)= \frac{n^{2}-m^{2}}{m^{2}+n^{2}}.$
Используя основное тригонометрическое тождество, выразим sin(x-y):
$sin(x-y)= \sqrt{1-( \frac{n^{2}-m^{2}}{m^{2}+n^{2}})^{2}}.$
Ответ: $sin(x-y)= \sqrt{1-( \frac{n^{2}-m^{2}}{m^{2}+n^{2}})^{2}};cos(x-y)= \frac{n^{2}-m^{2}}{m^{2}+n^{2}}.$

Answer 2

Given Information:

We are given the following equations: - sin(x) - sin(y) = m - cos(x) * cos(y) = n

We need to find the values of sin(x-y) and cos(x-y).

Solution:

To find sin(x-y), we can use the trigonometric identity sin(x-y) = sin(x) * cos(y) - cos(x) * sin(y). Let's substitute the given values into this identity:

sin(x-y) = sin(x) * cos(y) - cos(x) * sin(y)

Substituting sin(x) - sin(y) = m and cos(x) * cos(y) = n:

sin(x-y) = m * cos(y) - n * sin(y) [[1]](https://arxiv.org/pdf/1907.01060)

Similarly, to find cos(x-y), we can use the trigonometric identity cos(x-y) = cos(x) * cos(y) + sin(x) * sin(y). Let's substitute the given values into this identity:

cos(x-y) = cos(x) * cos(y) + sin(x) * sin(y)

Substituting sin(x) - sin(y) = m and cos(x) * cos(y) = n:

cos(x-y) = n * cos(y) + m * sin(y) Therefore, the values of sin(x-y) and cos(x-y) are given by equations [[1]](https://arxiv.org/pdf/1907.01060) and respectively.

Answer:

- sin(x-y) = m * cos(y) - n * sin(y) - cos(x-y) = n * cos(y) + m * sin(y)

Please let me know if I can help you with anything else.