Question

ДАЮ 70 БАЛЛОВ
Найти sin(α+β), если известно, что cosα+cosβ=a и sinα+sinβ=b

Answer 1

Ответ:

$sin(\alpha+\beta) =\frac{2ab}{a^2+b^2}$

Объяснение:

$sin(\alpha+\beta)=sin\alpha cos\beta+cos\alpha sin\beta$

1.

$cos\alpha+cos\beta=a\ \ \ |()^2$

$cos^2\alpha+2cos\alpha cos\beta+cos^2\beta=a^2$

$sin\alpha+sin\beta=b\ \ \ |()^2$

$sin^2\alpha+2sin\alpha sin\beta+sin^2\beta=b^2$

$cos^2\alpha+2cos\alpha cos\beta+cos^2\beta+sin^2\alpha+2sin\alpha sin\beta+sin^2\beta=a^2+b^2$

$1+2cos\alpha cos\beta+2sin\alpha sin\beta+1=a^2+b^2$

$2cos\alpha cos\beta+2sin\alpha sin\beta=a^2+b^2-1-1$

$2(cos\alpha cos\beta+sin\alpha sin\beta)=a^2+b^2-2$

$2cos(\alpha -\beta)=a^2+b^2-2\ \ \ |:2$

$cos(\alpha -\beta)=\frac{a^2+b^2-2}{2}$

2.

$cos\alpha+cos\beta=a$

$sin\alpha+sin\beta=b$

$(cos\alpha+cos\beta)(sin\alpha+sin\beta)=ab$

$cos\alpha sin\alpha+cos\alpha sin\beta+cos\beta sin\alpha+cos\beta sin\beta=ab$

$(sin\alpha cos\beta+cos\alpha sin\beta)+cos\alpha sin\alpha+cos\beta sin\beta=ab$

$sin(\alpha+\beta)+cos\alpha sin\alpha+cos\beta sin\beta=ab$

$sin(\alpha+\beta)+cos\alpha sin\alpha+cos\beta sin\beta=ab\ \ \ |\cdot 2$

$2sin(\alpha+\beta)+2cos\alpha sin\alpha+2cos\beta sin\beta=2ab$

$2sin(\alpha+\beta)+sin(2\alpha)+sin(2\beta)=2ab$

$2sin(\alpha+\beta)+sin(2\alpha)+sin(2\beta)=2ab$

$2sin(\alpha+\beta)+2sin\frac{2\alpha+2\beta}{2}cos\frac{2\alpha-2\beta}{2}=2ab\ \ \ |:2$

$sin(\alpha+\beta)+sin\frac{2(\alpha+\beta)}{2}cos\frac{2(\alpha-\beta)}{2}=ab$

$sin(\alpha+\beta)+sin(\alpha+\beta)cos(\alpha-\beta)=ab$

$sin(\alpha+\beta) \left(1+cos(\alpha-\beta)\right) =ab$

3.

$sin(\alpha+\beta) \left(1+\frac{a^2+b^2-2}{2}\right) =ab$

$\frac{2+a^2+b^2-2}{2}sin(\alpha+\beta)  =ab$

$\frac{a^2+b^2}{2}sin(\alpha+\beta) =ab\ \ \ |:\frac{a^2+b^2}{2}$

$sin(\alpha+\beta) =\frac{ab}{\frac{a^2+b^2}{2}}$

$sin(\alpha+\beta) =\frac{2ab}{a^2+b^2}$