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 .

If 0 < θ < π 8 , show that 2 + 2 + 2 + 2 cos 4 θ = 2 cos θ 2 .

We have, 2 + 2 + 2 + 2 cos 4 θ = 2 + 2 + 2 ( 1 + cos 4 θ ) = 2 + 2 + 2 · 2 cos 2 2 θ ∵ cos 2 A = 2 cos 2 A - 1 = 2 + 2 + 2 cos 2 θ = 2 + 2 ( 1 + cos 2 θ ) = 2 + 2 · 2 cos 2 θ = 2 + 2 cos θ = 2 ( 1 + cos θ ) = 2 · 2 cos 2 θ 2 = 2 cos θ 2 Hence, proved that 2 + 2 + 2 + 2 cos 4 θ = 2 cos θ 2 .

If a ≤ cos θ + 3 2 sin θ + π 4 + 6 ≤ b , then find the largest value of a and smallest value of b .

Given, that a ≤ cos θ + 3 2 sin θ + π 4 + 6 ≤ b .....(1) Consider, cos θ + 3 2 sin θ + π 4 + 6 = cos θ + 3 2 sin θ cos π 4 + cos θ sin π 4 + 6 = cos θ + 3 2 sin θ 1 2 + cos θ 1 2 + 6 = cos θ + 3 sin θ + 3 cos θ + 6 = 4 cos θ + 3 sin θ + 6 We know that minimum value and maximum value of A cos θ + B sin θ + C is given by C - A 2 + B 2 and C + A 2 + B 2 . So, minimum value of 4 cos θ + 3 sin θ + 6 = 6 - 4 2 + 3 2 = 6 - 25 = 6 - 5 = 1 and maximum value of 4 cos θ + 3 sin θ + 6 = 6 + 4 2 + 3 2 = 6 + 25 = 6 + 5 = 11 ∴ 1 ≤ 4 cos θ + 3 sin θ + 6 ≤ 11 So, on comparing to eq(1), we get The largest value of a is 1 . The smallest value of b is 11 .