Question

Тригонометрия 1. 2cos115° + 5sin205° \ 3cos65°

2. (cos20° + cos40°)^2 + (sin20°-sin40°)^2

3. Дано: sin a = 0.6 pi\2 < a < pi

Вычислить: cos a, sin2a, cos2a

Answer 1

Пошаговое объяснение:

1).

2cos115°=2×cos(180°-65°)=2×(-cos65°)=-2cos65°

5sin205°=5×sin(270°-65°)=5×(-cos65°)=-5cos65°

-2cos65°-5cos65°=-7cos65°

$\frac{2cos {115}^{0} + 5sin {205}^{0} }{3cos {65}^{0}} = \frac{ - 7cos {65}^{0} }{3cos {65}^{0} } = - \frac{7}{3}$

2).

${(cos {20}^{0} + cos {40}^{0}) }^{2} + {(sin {20}^{0} - sin {20}^{0}) }^{2} =$

$= {cos}^{2} {20}^{0} + 2cos {20}^{0} \times cos {40}^{0} + {cos}^{2} {40}^{0} + {sin}^{2} {20}^{0} - 2sin {20}^{0} \times sin {40}^{0} + {sin}^{2} {40}^{0} = ( {cos}^{2} {20}^{0} + {sin}^{2} {20}^{0}) + 2 \times (cos {20}^{0}cos {40}^{0} - sin {20}^{0} sin {40}^{0} ) + ( {cos}^{2} {40}^{0} + {sin}^{2} {20}^{0} ) = 1 + 2 \times cos( {20}^{0} + {40}^{0}) + 1 =$

=2+2×cos60°=2+2×(1/2)=2+1=3