Question

find the value of sin7π/6 + cos 2π/3
​

Answer 1

Answer:

$\sin( \frac{7\pi}{6} ) + \cos( \frac{2\pi}{3} ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ = \sin(\pi + \frac{\pi}{6} ) + \cos( \pi - \frac{\pi}{3} ) \\ = \sin( \frac{\pi}{6} ) + \cos( - \frac{\pi}{3} ) \: \: \: \: \: \: \: \: \: \: \: \: \: \\ = \sin( \frac{\pi}{6} ) - \cos( \frac{\pi}{3} ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ = \frac{1}{2} - \frac{1}{2} = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Here's the answer to the question.

Answer 2

Answer:

sin(

6

7π

)+cos(

3

2π

)

=sin(π+

6

π

)+cos(π−

3

π

)

=sin(

6

π

)+cos(−

3

π

)

=sin(

6

π

)−cos(

3

π

)

=

2

1

−

2

1

=0