Question

sin 480⁰ cos 690⁰+cos 780⁰ sin1050⁰​

Answer 1

Answer:

sin 480° (cos 690°) + cos 780° sin1050°

First solving first part then second and then adding both of them

sin 480° (cos 690°)

• sin(2π - x) = -sinx

• and cos (2π -x) = cos x

• using 4π

• cos (4π - x) = cos x

So

sin ( 360 - 480) cos ( 720-690)

- sin( -120) cos(30)

• sin(-x) = -sin x and sin (π/2 + x) = cos x

sin120° cos 30

sin ( π/2 +30) cos (30)

cos(30) cos(30)

cos²(30)

Now

cos 780⁰ sin1050⁰

Using

• Using sin (2π - x) = -sin (x)
• Using sin (2π - x) = -sin (x) so sin ( 6π - x) = - sin x

• Using sin (2π - x) = -sin (x) so sin ( 6π - x) = - sin xcos (2π - x) = cos x

Now

cos 780 sin 1050

cos ( 720 -780) sin (6(180) - 1050)

cos ( 720 - 780) sin(1080 - 1050)

cos (-60) [-sin (30)]

• Using cos(-x) = cos x

-cos(60) sin(30)

Now

adding them

cos²(30) - cos(60) sin(30)

• cos 30 = √3/2

• sin 30 = 1/2

• cos 60 = 1/2

=(√3/2)² - (1/2)(1/2)

= 3/4-1/4

= 2/4

= 1/2