Prove that cos 48 ° · cos 12 ° = 3 + 5 8

To Prove: cos 48 ° · cos 12 ° = 3 + 5 8 L.H.S. = cos 48 ° · cos 12 ° = 1 2 2 cos 48 ° · cos 12 ° Apply the formula, 2 cos A cos B = cos A + B + cos A - B ⇒ L.H.S. = 1 2 cos 48 ° + 12 ° + cos 48 ° - 12 ° = 1 2 cos 60 ° + cos 36 ° = 1 2 1 2 + 5 + 1 4 ∵ cos 36 ° = 5 + 1 4 = 1 2 3 + 5 4 = 3 + 5 8 = R.H.S. Hence Proved.

Prove that cos 48 ° · cos 12 ° = 3 + 5 8

To Prove: cos 48 ° · cos 12 ° = 3 + 5 8 L.H.S. = cos 48 ° · cos 12 ° = 1 2 2 cos 48 ° · cos 12 ° Apply the formula, 2 cos A cos B = cos A + B + cos A - B ⇒ L.H.S. = 1 2 cos 48 ° + 12 ° + cos 48 ° - 12 ° = 1 2 cos 60 ° + cos 36 ° = 1 2 1 2 + 5 + 1 4 ∵ cos 36 ° = 5 + 1 4 = 1 2 3 + 5 4 = 3 + 5 8 = R.H.S. Hence Proved.

Prove that cos θ + cos 2 π 3 + θ + cos 4 π 3 + θ = 0

To Prove: cos θ + cos 2 π 3 + θ + cos 4 π 3 + θ = 0 L.H.S. = cos θ + cos 2 π 3 + θ + cos 4 π 3 + θ We know that cos C + cos D = 2 cos C + D 2 cos C - D 2 ⇒ L.H.S. = cos θ + 2 cos 2 π 3 + 4 π 3 + 2 θ 2 cos 2 π 3 - 4 π 3 2 = cos θ + 2 cos π + θ cos - 2 π 6 = cos θ - 2 cos θ · cos π 3 ∵ cos - θ = cos θ and cos π + θ = - cos θ = cos θ - 2 cos θ · 1 2 = cos θ - cos θ = 0 = R.H.S. Hence Proved.

Prove that sin 2 α - π 4 + sin 2 α + π 12 - sin 2 α - π 12 = 1 2

To Prove: sin 2 α - π 4 + sin 2 α + π 12 - sin 2 α - π 12 = 1 2 L.H.S. = sin 2 α - π 4 + sin 2 α + π 12 - sin 2 α - π 12 Using the identity, a 2 - b 2 = a - b a + b , we have L.H.S. = sin 2 α - π 4 + sin α + π 12 - sin α - π 12 sin α + π 12 + sin α - π 12 Apply the formulae, sin C - sin D = 2 cos C + D 2 · sin C - D 2 and sin C + sin D = 2 sin C + D 2 · cos C - D 2 = sin 2 α - π 4 + 2 cos 2 α 2 sin π 12 · 2 sin 2 α 2 cos π 12 = sin 2 α - π 4 + 2 sin α cos α 2 sin π 12 cos π 12 = sin 2 α - π 4 + sin 2 α sin 2 π 12 = sin 2 α - π 4 + sin 2 α sin π 6 L.H.S. = sin 2 α - π 4 + sin 2 α 1 2 We know that sin 2 θ = 1 - cos 2 θ 2 = 1 - cos 2 α - π 4 2 + 1 2 sin 2 α = 1 - cos 2 α - π 2 2 + 1 2 sin 2 α = 1 - cos - π 2 - 2 α 2 + 1 2 sin 2 α = 1 - sin 2 α 2 + 1 2 sin 2 α ∵ cos 90 ° - θ = sin θ and cos - θ = cos θ = 1 2 Hence Proved.

If sin x + sin y = 1 4 and cos x + cos y = 1 3 , then show that tan x + y 2 = 3 4 .

Given, sin x + sin y = 1 4 and cos x + cos y = 1 3 To Prove: tan x + y 2 = 3 4 Now consider, sin x + sin y = 1 4 ⇒ 2 sin x + y 2 cos x - y 2 = 1 4 → 1 And, cos x + cos y = 1 3 ⇒ 2 cos x + y 2 cos x - y 2 = 1 3 → 2 Dividing 1 by 2 , we get 2 sin x + y 2 cos x - y 2 2 cos x + y 2 cos x - y 2 = 1 4 1 3 ⇒ sin x + y 2 cos x + y 2 = 3 4 ⇒ tan x + y 2 = 3 4 Hence Proved.

If sin x + sin y = 1 4 and cos x + cos y = 1 3 , then show that cot ( x + y ) = 7 24 .

Given, sin x + sin y = 1 4 and cos x + cos y = 1 3 To Prove: cot x + y = 7 24 Now consider, sin x + sin y = 1 4 ⇒ 2 sin x + y 2 cos x - y 2 = 1 4 → 1 And, cos x + cos y = 1 3 ⇒ 2 cos x + y 2 cos x - y 2 = 1 3 → 2 Dividing 1 by 2 , we get 2 sin x + y 2 cos x - y 2 2 cos x + y 2 cos x - y 2 = 1 4 1 3 ⇒ sin x + y 2 cos x + y 2 = 3 4 ⇒ tan x + y 2 = 3 4 → 3 Now, we know that tan 2 A = 2 tan A 1 - tan 2 A Thus, tan x + y = 2 tan x + y 2 1 - tan 2 x + y 2 = 2 3 4 1 - 3 4 2 [From 3 ] = 3 2 1 - 9 16 = 3 2 × 16 7 = 24 7 Now, we know that cot θ = 1 tan θ ⇒ cot x + y = 1 tan x + y = 1 24 7 Hence, cot x + y = 7 24 Hence Proved.

If neither A - π 12 nor A - 5 π 12 is an integral multiple of π , prove that cot π 12 - A + tan π 12 + A = 4 cos 2 A 1 - 2 sin 2 A .

To Prove: cot π 12 - A + tan π 12 + A = 4 cos 2 A 1 - 2 sin 2 A L.H.S. = cot π 12 - A + tan π 12 + A = cos π 12 - A sin π 12 - A + sin π 12 + A cos π 12 + A = cos π 12 - A cos π 12 + A + sin π 12 + A sin π 12 - A sin π 2 - A cos π 12 + A We know that cos ( A + B ) cos ( A - B ) = cos 2 B - sin 2 A and sin ( A + B ) sin ( A - B ) = sin 2 A - sin 2 B = cos 2 A - sin 2 π 12 + sin 2 π 12 - sin 2 A sin π 2 - A cos π 12 + A = 2 cos 2 A - sin 2 A 2 sin π 12 - A cos π 12 + A L.H.S. = 2 cos 2 A - sin 2 A 2 cos π 12 + A sin π 12 - A Apply, 2 cos C sin D = sin ( C + D ) - sin ( C - D ) = 2 cos 2 A - sin 2 A sin 2 π 12 - sin ( 2 A ) = 2 cos 2 A sin π 6 - sin ( 2 A ) = 2 cos 2 A 1 2 - sin 2 A = 4 cos 2 A 1 - 2 sin 2 A Hence Proved.

If cos x + cos y = 4 5 and cos x - cos y = 2 7 , find the value of 14 tan x - y 2 + 5 cot x + y 2 .

Given cos x + cos y = 4 5 2 cos x + y 2 cos x - y 2 = 4 5 . . . . . 1 cos x - cos y = 2 7 - 2 sin x - y 2 sin x + y 2 = 2 7 . . . . 2 ⇒ 2 cos x + y 2 · cos x - y 2 - 2 sin ( x - y 2 ) sin x + y 2 = 4 5 2 7 ⇒ - cot x + 4 2 · cot x - y 2 = 14 5 ⇒ - 5 cot x + y 2 = 14 tan x - y 2 ∴ 14 tan x - y 2 + 5 cot x + y 2 = 0