Question

Алгебра 10 клас 1004 та 1006

Answer 1

Решение.

Применяем формулы синуса суммы и разности углов, а также

косинуса суммы и разности углов :

 $\boldsymbol{cos(\alpha \pm \beta )=cos\alpha \cdot cos\beta \mp sin\alpha \cdot sin\beta }\\\\\boldsymbol{sin(\alpha \pm \beta )=sin\alpha \cdot cos\beta \pm cos\alpha \cdot sin\beta }$   .

$\bf 1004)\\\\cos58^\circ \cdot cos32^\circ -sin58^\circ \cdot sin32^\circ =cos(58^\circ +32^\circ )=cos90^\circ =0\\\\sin64^\circ \cdot sin19^\circ+cos64^\circ \cdot cos19^\circ =cos(64^\circ -19^\circ )=cos45^\circ =\dfrac{\sqrt2}{2}\\\\sin65^\circ \cdot cos55^\circ +cos65^\circ \cdot sin55^\circ =sin(65^\circ +55^\circ )=sin120^\circ =sin60^\circ =\dfrac{\sqrt3}{2}\\\\cos10^\circ \cdot cos20^\circ -sin20^\circ \cdot sin10^\circ =cos(10^\circ +20^\circ )=cos30^\circ =\dfrac{\sqrt3}{2}$  

$\bf 1006)\\\\\dfrac{sin53^\circ \cdot cos37^\circ +cos53^\circ \cdot sin37^\circ}{cos35^\circ \cdot cos65^\circ +sin35^\circ \cdot sin65^\circ }=\dfrac{sin90^\circ }{cos30^\circ }=\dfrac{1}{\frac{\sqrt3}{2}}=\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}\\\\\\\dfrac{cos67^\circ \cdot cos7^\circ +sin67^\circ \cdot sin7^\circ}{sin40^\circ \cdot cos50^\circ +cos40^\circ \cdot sin50^\circ }=\dfrac{cos60^\circ }{sin90^\circ }=\dfrac{\frac{1}{2}}{1}=\dfrac{1}{2}$

$\bf \dfrac{sin\dfrac{\pi }{12}\cdot cos\dfrac{5\pi }{12}+cos\dfrac{5\pi }{12}\cdot sin\dfrac{\pi }{12}}{cos\dfrac{\pi }{12}\cdot cos\dfrac{5\pi }{12}+sin\dfrac{\pi }{12}\cdot sin\dfrac{5\pi }{12}}=\dfrac{sin\dfrac{\pi }{2}}{cos\dfrac{\pi }{3}}=\dfrac{1}{\frac{\sqrt3}{2}}=\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}$  

$\bf \dfrac{cos\dfrac{2\pi }{15}\cdot cos\dfrac{\pi }{5}-sin\dfrac{2\pi }{15}\cdot sin\dfrac{\pi }{5}}{sin\dfrac{\pi }{5}\cdot cos\dfrac{\pi }{30}-cos\dfrac{\pi }{5}\cdot sin\dfrac{\pi }{30}}=\dfrac{cos\dfrac{\pi }{3}}{sin\dfrac{\pi }{6}}=\dfrac{\dfrac{\sqrt3}{2}}{\dfrac{1}{2}}=\sqrt3$