Question

Evaluate sin(-690°)cos(-300°)+cos(-750°)sin(-240°)

Answer 1

We have
sin(-690°) cos(-300°) + cos(-750°) sin(-240°)
Lets take each part seperate, so we have
sin(-690°) = -sin(360 x 2 - 30) = -sin(-30)    [ sin(360 - x) = -sin(x) ]

= -sin(-30) = sin(30) = 1/2    [ sin(-x) = -sin(x) ]
cos(-300°) = cos(300)  [ cos(-x) = cos(x) ]
cos(300) = cos(360 - 60)   [ cos(x) = cos(360 - x) ]
= cos(60) = 1/2

cos(-750) = cos(750)   [ cos(-x) = cos(x) ]
= cos(360 x 2 + 30) = cos(30) = √3/2  [ cos(x) = cos(360 + x) ]
sin(-240) = -sin(240)  [ sin(-x) = -sin(x) ]
= -sin(180 + 60) = sin(60) = √3/2  [ sin(180 + x) = -sin(x) ]

Putting all together, we get
= 1/2 x 1/2 + √3/2 x √3/2
= 1/4 + 3/4
= 1
Answer is 1

Answer 2

Step-by-step explanation:

sin (-690).cos(-300)+cos750.sin240