Question

1)Вычислите sin^4(3π/2− 2α) , если cos(π − 4α) = −1/3.

2)Упростите выражения sin 50 + 8 sin 10sin 50sin 70( в градусах)

Answer 1

1) $\sin^4( \frac{3\pi}{2} - 2\alpha) = S$

$\cos(\pi - 4\alpha) = -\frac{1}{3}$

Используем формулу косинуса двойного аргумента:

$\cos(2x) = \cos^2(x) - \sin^2(x) = 1 - \sin^2(x) - \sin^2(x) = 1 - 2\sin^2(x)$

тогда

$\sin^2(x) = \frac{1 - \cos(2x)}{2}$

тогда

$S = (\sin^2(\frac{3\pi}{2} - 2\alpha))^2 = (\frac{1 - \cos(3\pi - 4\alpha)}{2})^2 =$

$= (\frac{1 - \cos(\pi - 4\alpha)}{2})^2 = (\frac{1 - (-\frac{1}{3})}{2})^2 =$

$= (\frac{1+\frac{1}{3}}{2})^2 = (\frac{3+1}{2\cdot 3})^2 = (\frac{4}{6})^2 =$

$= (\frac{2}{3})^2 = \frac{4}{9}$

2) $\sin(50^\circ) + 8\sin(10^\circ)\sin(50^\circ)\sin(70^\circ) =$

$= \sin(50^\circ)\cdot ( 1 + 8\sin(10^\circ)\sin(90^\circ - 20^\circ) ) =$

$= \sin(50^\circ)\cdot (1 + 8\sin(10^\circ)\cos(20^\circ) ) =$

$= \sin(50^\circ)\cdot (1 + \frac{8\sin(10^\circ)\cos(10^\circ)\cos(20^\circ)}{\cos(10^\circ)}) = V$

Используем формулу синуса двойного аргумента:

$\sin(2x) = 2\sin(x)\cos(x)$

$V = \sin(50^\circ)\cdot (1+ \frac{4\sin(20^\circ)\cos(20^\circ)}{\cos(10^\circ)}) =$

$= \sin(50^\circ)\cdot (1 + \frac{2\sin(40^\circ)}{\cos(10^\circ)}) =$

$= \sin(50^\circ) + \frac{2\sin(50^\circ)\sin(40^\circ)}{\cos(10^\circ)} = W$

Используем формулу "произведение синусов":

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

$W = \sin(50^\circ) + \frac{\cos(50^\circ - 40^\circ) - \cos(50^\circ + 40^\circ)}{\cos(10^\circ)} =$

$= \sin(50^\circ) + \frac{\cos(10^\circ) - \cos(90^\circ)}{\cos(10^\circ)} =$

$= \sin(50^\circ) + \frac{\cos(10^\circ) - 0}{\cos(10^\circ)} =$

$= \sin(50^\circ) + 1$.