Question

Решить 2 и 3 задание.
Буду очень благодарна .

Answer 1

Пошаговое объяснение:

${ \sin(a) }^{2} + { \cos(a) }^{2} = 1 \\ { \sin(a) }^{2} = 1 - { \cos(a) }^{2} = 1 - {( - \frac{1}{2} ) }^{2} \\ = 1 - \frac{1}{4} = \frac{3}{4} \\ \sin(a) = \sqrt{ \frac{3}{4} } = \frac{ \sqrt{3} }{2} \\ \sin(2a) = 2 \times \sin(a) \times \cos(a) \\ = 2 \times \frac{ \sqrt{3} }{2} \times ( - \frac{1}{2} ) = - \frac{ \sqrt{3} }{2} \\ \tan(a) = \frac{ \sin(a) }{ \cos(a) } = \frac{ \frac{ \sqrt{3} }{2} }{ - \frac{1}{2} } = - \sqrt{3}$

${ \sin(a) }^{2} + { \cos(a) }^{2} = 1 \\ { \cos(a) }^{2} = 1 - { \sin(a) }^{2} = 1 - {( \frac{3}{5}) }^{2} \\ = 1 - \frac{9}{25} = \frac{16}{25} \\ \cos(a) = - \sqrt{ \frac{16}{25} } = - \frac{4}{5} \\ \sin(a + \frac{\pi}{3} ) = \sin(a) \times \cos( \frac{\pi}{3} ) \\ + \cos(a) \times \sin( \frac{\pi}{3} ) \\ = \frac{3}{5} \times \frac{1}{2} + ( - \frac{4}{5} ) \times \frac{ \sqrt{3} }{2} \\ = \frac{3}{10} + ( - \frac{4 \sqrt{3} }{10} ) = \frac{3 - 4 \sqrt{3} }{10}$

$( { \cot(a) }^{2} - { \tan(a) }^{2}) = \frac{4 \cos(2a) }{ { \sin(2a) }^{2} } \\( { \cot(a) }^{2} - { \tan(a) }^{2} ) = \frac{4 ({ \cos(a) }^{2} - { \sin(a) }^{2} ) }{ ({2 \times \sin(a) \times \cos(a) )}^{2} } = \\ \frac{4 ({ \cos(a) }^{2} - { \sin(a) }^{2} ) }{4 \times { \sin(a) }^{2} \times { \cos(a) }^{2} } = \frac{ { \cos(a) }^{2} - { \sin(a) }^{2} }{ { \sin(a) }^{2} \times { \cos(a) }^{2} } \\ = ( { \cot(a) }^{2} - { \tan(a) }^{2} )$