Question

Известно, что sin∠А + cos∠А = 0,5. 
Найдите sin∠А ∙cos∠А.
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Answer 1

Ответ:

$\sin(\angle A) + \cos(\angle A) = \frac{1}{2} \\ {( \sin(\angle A) + \cos(\angle A) }^{2} = {( \frac{1}{2}) }^{2} \\ \sin^{2} (\angle A) + 2 \sin(\angle A) \cos(\angle A) + \cos^{2} (\angle A) = \frac{1}{4}$

Тригонометрическое тождество: $sin^2(\angle A)+cos^2(\angle A)=1$

Поэтому:

$1 + 2 \sin(\angle A) \cos(\angle A) = \frac{1}{4} \\ 2 \sin(\angle A) \cos(\angle A) = \frac{1}{4} - 1 \\ 2 \sin(\angle A) \cos(\angle A) = - \frac{3}{4} \\ \sin(\angle A) \cos(\angle A) = - \frac{3}{8}$

Answer 2

sin∠А + cos∠А = 0,5

(sin∠А + cos∠А)² = 0,25
(sin∠А + cos∠А)² - 2·sin∠А ∙cos∠А= sin²∠А + 2·sin∠А ∙cos∠A + cos²∠А - 2·sin∠А ∙cos∠А = sin²∠А + cos²∠А = 1
0,25 - 2·sin∠А ∙cos∠А = 1
2·sin∠А ∙cos∠А = -0.75

sin∠А ∙cos∠А=-0.375