Question

cos 570° sin 510° + sin(-330°).cos(-390°)
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Answer 1

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LHS =cos (570)sin (510) + sin (- 330)cos (- 390)

= cos (570) sin (510) + [ – sin (330) ]cos (390) [ because sin( – x ) = – sin x and cos( – x ) = cos x ]

= cos (570)sin(510) – sin (330)

= cos (90 * 6 + 30) sin (90 * 5 + 60) – sin (90 * 3 + 60) cos (90 * 4 + 30)

= – cos (30) cos (60) – [ – cos (60) ] cos (30)

= – cos (30) cos (60) + cos (30) sin (60)

= 0

Hope it's Helpful.....:)

Answer 2

cos2θ + 3 sin θ + 3 sin2θ = 0

1 – sin2θ + 3 sin θ + 3 sin2θ = 0

2 sin2θ + 3 sin θ + 1 = 0

2sin2θ + 2 sin θ + sin θ + 1 = 0

2 sin θ (sin θ +1 ) + 1 (sin θ + 1) = 0

(2 sin θ + 1) (sin θ + 1) = 0

2 sin θ + 1 = 0, sin θ + 1 = 0

sin θ = -½, sin θ = -1

$cos2θ + 3 sin θ + 3 sin2θ = 01 – sin2θ + 3 sin θ + 3 sin2θ = 02 sin2θ + 3 sin θ + 1 = 02sin2θ + 2 sin θ + sin θ + 1 = 02 sin θ (sin θ +1 ) + 1 (sin θ + 1) = 0(2 sin θ + 1) (sin θ + 1) = 02 sin θ + 1 = 0, sin θ + 1 = 0sin θ = -½, sin θ = -1$