Решить с развёрнутым ответом

Question 1

Question 2

1.558
а) Используем формулу понижения степени:
${cos}^{2}x = \frac{1 + cos2x}{2} \\ 1 + cos2x = 2 {cos}^{2} x \\ \\ 1 + cos120 = 2 {cos}^{2}60 = 2 \times {( \frac{1}{2}) }^{2} = 2 \times \frac{1}{4} = \frac{1}{2}$
б)
${sin}^{2} x = \frac{1 - cos2x}{2} \\ 1 - cos2x = 2 {sin}^{2} x \\ \\ 1 - cos120 = 2 {sin}^{2} 60 = 2 {( \frac{ \sqrt{3} }{2} )}^{2} = 2 \times \frac{3}{4} = \frac{3}{2}$
1.559.
а) Используем формулу косинуса двойного угла:
$cos {}^{2} a - {sin}^{2} a = cos2a$
${cos}^{2} 67.5 - {sin}^{2} 67.5 = cos(2 \times 67.5) = cos135 = cos(180 - 45) = - cos45 = - \frac{ \sqrt{2} }{2}$
б)
${sin}^{2} 75 - {cos}^{2} 75 = - ( {cos}^{2} 75 - {sin}^{2} 75) = - cos(2 \times 75) = - cos150 = - cos(180 - 30) = - ( - cos30) = cos30 = \frac{ \sqrt{3} }{2}$
1.560
а)
${( {cos}^{2} \frac{\pi}{8} - {sin}^{2} \frac{\pi}{8}) }^{2} = {(cos(2 \times \frac{\pi}{8})) }^{2} = {(cos \frac{\pi}{4}) }^{2} = {( \frac{ \sqrt{2} }{2} )}^{2} = \frac{2}{4} = \frac{1}{2}$
б) Используем формулу синуса двойного угла
$sin2 \alpha = 2sin \alpha cos \alpha$
${(2sin \frac{\pi}{12}cos \frac{\pi}{12}) }^{ - 1} = \frac{1}{2sin \frac{\pi}{12} cos \frac{\pi}{12} } = \frac{1}{sin(2 \times \frac{\pi}{12}) } = \frac{1}{sin\frac{\pi}{6} } = \frac{1}{ \frac{1}{2} } = 2$