simplify sin 2 θ 1 + cos 2 θ .

Given, sin 2 θ 1 + cos 2 θ = 2 sin θ cos θ 1 + 2 cos 2 θ - 1 ∵ sin 2 A = 2 sin A cos A & cos 2 A = 2 cos 2 A - 1 = 2 sin θ cos θ 2 cos 2 θ = sin θ cos θ = tan θ Therefore, sin 2 θ 1 + cos 2 θ = tan θ .

simplify sin 2 θ 1 + cos 2 θ .

Given, sin 2 θ 1 + cos 2 θ = 2 sin θ cos θ 1 + 2 cos 2 θ - 1 ∵ sin 2 A = 2 sin A cos A & cos 2 A = 2 cos 2 A - 1 = 2 sin θ cos θ 2 cos 2 θ = sin θ cos θ = tan θ Therefore, sin 2 θ 1 + cos 2 θ = tan θ .

simplify 3 cos θ + cos 3 θ 3 sin θ - sin 3 θ .

Given, 3 cos θ + cos 3 θ 3 sin θ - sin 3 θ = 3 cos θ + 4 cos 3 θ - 3 cos θ 3 sin θ - 3 sin θ - 4 sin 3 θ = 3 cos θ + 4 cos 3 θ - 3 cos θ 3 sin θ - 3 sin θ + 4 sin 3 θ = 4 cos 3 θ 4 sin 3 θ = cot 3 θ Therefore, 3 cos θ + cos 3 θ 3 sin θ - sin 3 θ = cot 3 θ .

Express sin 4 θ sin θ in terms of cos 3 θ and cos θ .

Given, sin 4 θ sin θ = sin 2 2 θ sin θ = 2 sin 2 θ cos 2 θ sin θ ∵ sin 2 A = 2 sin A cos A = 2 2 sin θ cos θ 2 cos 2 θ - 1 sin θ = 4 cos θ 2 cos 2 θ - 1 ∴ sin 4 θ sin θ = 8 cos 3 θ - 4 cos θ Hence, the given expression is expressed in terms of cos 3 θ and cos θ .

Express 1 - cos θ + sin θ 1 + cos θ + sin θ in terms of tan θ 2 .

Given, 1 - cos θ + sin θ 1 + cos θ + sin θ = 1 - 1 - 2 sin 2 θ 2 + 2 sin θ 2 cos θ 2 1 + 2 cos 2 θ 2 - 1 + 2 sin θ 2 cos θ 2 ∵ sin 2 A = 2 sin A cos A & cos 2 A = 2 cos 2 A - 1 = 1 - 2 sin 2 A = 2 sin 2 θ 2 + 2 sin θ 2 cos θ 2 2 cos 2 θ 2 + 2 sin θ 2 cos θ 2 = 2 sin θ 2 sin θ 2 + cos θ 2 2 cos θ 2 cos θ 2 + sin θ 2 = sin θ 2 cos θ 2 = tan θ 2 ∴ 1 - cos θ + sin θ 1 + cos θ + sin θ = tan θ 2 . Hence the given expression is expressed in terms of tan θ 2

If sin α = 3 5 , where π 2 < α < π , evaluate cos 3 α and tan 2 α .

We have, sin α = 3 5 and π 2 < α < π . So, α lies in 2 nd quadrant. Now, cos α = 1 - sin 2 α = 1 - 9 25 = 16 25 = ± 4 5 So, cos α = - 4 5 ∵ cos x is negative in 2 nd quadrant Now, cos 3 α = 4 cos 3 α - 3 cos α = 4 - 4 5 3 - 3 - 4 5 = 4 - 64 125 + 12 5 = - 256 + 300 125 = 44 125 ∴ cos 3 α = 44 125 . We know that tan α = sin α cos α = 3 5 - 4 5 = - 3 4 Now, tan 2 α = 2 tan α 1 - tan 2 α = 2 - 3 4 1 - - 3 4 2 = - 6 4 1 - 9 16 = - 6 4 7 16 ∴ tan 2 α = - 24 7 .