If cos x = 1 3 , π < x < 3 π 2 , find the values of sin x 2 and tan x 2

cos x = 1 3 , π < x < 3 π 2 x is in I I I r d Quadrant. We have 2 sin 2 x 2 = 1 − cos x = 1 - 1 3 = 2 3 ∴ sin 2 x 2 = 1 3 ∴ sin x 2 = 1 3 . . . 1 π 2 < x 2 < 3 π 4 2 cos 2 x 2 = 1 + cos x = 1 + 1 3 = 4 3 ∴ cos x 2 = - 2 3 ( cos is -negative in 2 n d Quadrant ) . . . 2 From 1 and 2 tan x 2 = 1 3 - 2 3 = - 1 2 ∴ tan x 2 = − 1 2 = - 1 2

If cos x = 1 3 , π < x < 3 π 2 , find the values of sin x 2 and tan x 2

cos x = 1 3 , π < x < 3 π 2 x is in I I I r d Quadrant. We have 2 sin 2 x 2 = 1 − cos x = 1 - 1 3 = 2 3 ∴ sin 2 x 2 = 1 3 ∴ sin x 2 = 1 3 . . . 1 π 2 < x 2 < 3 π 4 2 cos 2 x 2 = 1 + cos x = 1 + 1 3 = 4 3 ∴ cos x 2 = - 2 3 ( cos is -negative in 2 n d Quadrant ) . . . 2 From 1 and 2 tan x 2 = 1 3 - 2 3 = - 1 2 ∴ tan x 2 = − 1 2 = - 1 2

Prove that: 1 + cos π 8 1 + cos 3 π 8 1 + cos 5 π 8 1 + cos 7 π 8 = 1 8 .

To prove that: 1 + cos π 8 1 + cos 3 π 8 1 + cos 5 π 8 1 + cos 7 π 8 = 1 8 . We know, cos 5 π 8 = cos π 2 + π 8 = - sin π 8 cos 3 π 8 = cos π 2 - π 8 = sin π 8 cos 7 π 8 = cos π - π 8 = - cos π 8 So, L.H.S. = 1 + cos π 8 1 - cos π 8 1 + sin π 8 1 - sin π 8 = 1 - cos 2 π 8 1 - sin 2 π 8 = 1 - 1 + cos π 4 2 1 - 1 - cos π 4 2 ∵ cos 2 A = 2 cos 2 A - 1 = 1 - 2 sin 2 A = 1 4 1 - 1 2 1 + 1 2 = 1 8 Hence, proved.

Show that cot 142 1 2 ° = 2 + 3 - 2 - 6

To Prove :cot 142 1 ° 2 = 2 + 3 - 2 - 6 Proof : ⇒ cot 142 1 ° 2 = cot ( 90 ° + 52 . 5 ° ) As it lies in second quadrant therefore a negative sign will also come.So: ⇒ Cot 142 1 2 ° = - tan 52 . 5 ° ⇒ cot 142 1 2 ° = - tan ( 45 ° + 7 . 5 ° ) Now, in right-hand side we will apply the formula of tan ( A + B ) = tan A + tan B 1 - tan A tan B ⇒ Cot 142 1 2 ° = - tan 45 ° + tan 7 . 5 ° 1 - tan 45 ° . tan 7 . 5 ° ⇒ Cot 142 1 2 ° = - 1 + tan 7 . 5 ° 1 - tan 7 . 5 ° ⇒ Cot 142 1 2 ° = - 1 + Sin 7 . 5 ° Cos 7 . 5 ° 1 - Sin 7 . 5 ° Cos 7 . 5 ° ⇒ Cot 142 1 2 ° = - Cos 7 . 5 ° + Sin 7 . 5 ° Cos 7 . 5 ° - Sin 7 . 5 ° By rationalising the right-hand side we get: ⇒ Cot 142 1 2 ° = - Cos 7 . 5 ° + Sin 7 . 5 ° Cos 7 . 5 ° - Sin 7 . 5 ° × Cos 7 . 5 ° + Sin 7 . 5 ° Cos 7 . 5 ° + Sin 7 . 5 ° ⇒ Cot 142 1 2 ° = - ( Cos 7 . 5 ° + Sin 7 . 5 ° ) 2 Cos 2 7 . 5 ° - Sin 2 7 . 5 ° ⇒ Cot 142 1 2 ° = - Cos 2 7 . 5 ° + Sin 2 7 . 5 ° - 2 Sin 7 . 5 ° C o s 7 . 5 ° Cos 2 7 . 5 ° - Sin 2 7 . 5 ° As we know that Sin 2 θ + Cos 2 θ = 1 Sin 2 θ = 2 Sin θ . Cos θ Cos 2 θ = Cos 2 θ - Sin 2 θ By using these identities we get: ⇒ Cot 142 1 2 ° = - 1 + Sin 15 ° Cos 15 ° ⇒ Cot 142 1 2 ° = - 1 + Sin ( 45 ° - 30 ° ) Cos ( 45 ° - 30 ° ) ⇒ Cot 142 1 2 ° = - 1 + Sin 45 ° . Cos 30 ° - Cos 45 ° . Sin 30 ° Cos 45 ° . Cos 30 ° + Sin 45 ° . Sin 30 ° ⇒ Cot 142 1 2 ° = - 1 + 1 2 . 3 2 - 1 2 . 1 2 1 2 . 3 2 + 1 2 . 1 2 ⇒ Cot 142 1 2 ° = - 2 2 + 3 - 1 3 + 1 Now, by rationalising right-hand side we get: ⇒ Cot 142 1 2 ° = - 2 2 + 3 - 1 3 + 1 × 3 - 1 3 - 1 ⇒ Cot 142 1 2 ° = - 2 6 - 2 2 + 3 - 3 - 3 + 1 3 - 1 ⇒ Cot 142 1 2 ° = - 2 ( 6 - 2 + 2 - 3 ) 2 ⇒ Cot 142 1 2 ° = 2 + 3 - 2 - 6

Prove that sin 2 48 ° - cos 2 12 ° = - 5 + 1 8

L . H . S = sin 2 48 ° − cos 2 12 ° = cos 2 42 ° − cos 2 12 ° = cos 42 ° + cos 12 ° cos 42 ° − cos 12 ° = - 2 cos 27 ° cos 15 ° 2 sin 27 ° sin 15 ° = − 2 sin 27 ° cos 27 ° 2 sin 15 ° cos 15 ° = − sin 54 ° sin 30 ° = - 5 + 1 4 · 1 2 = − 5 + 1 8 Hence, proved.

Prove that 4 sin 24 ° + cos 6 ° = 3 + 15

To Prove , 4(sin24 ° + cos 6 ° ) = 3 + 15 Proof: L.H.S. 4 ( sin 24 ° + cos 6 ° ) = 4 [ sin 24 ° + cos ( 90 ° - 84 ° ) ] = 4 [ sin 24 ° + sin 84 ° ] As we know that ⇒ sin A + sin B = 2 sin ( A + B 2 ) cos ( A - B 2 ) L.H.S = 4 [ 2 sin ( 84 + 24 2 ) ° cos ( 84 - 24 2 ) ° ] = 8 sin ( 108 2 ) ° · cos ( 60 2 ) ° = 8 sin 54 ° . cos 30 ° = 8 sin 54 ° × 3 2 = 4 3 sin 54 ° As we know that sin 3 θ = 3 sin θ - 4 sin 3 θ So, LH.S. 4 3 [ 3 sin18 ° - 4 sin 3 18 ° ] = 4 3 [ 3 ( 5 - 1 4 ) - 4 ( ( 5 - 1 4 ) 3 ) ] = 4 3 [ 3 5 - 3 4 - ( 5 5 - 1 + 3 5 - 15 ) 16 ] = 4 3 [ 3 5 - 3 4 - { 3 5 - 16 16 } ] = 4 3 ( 12 5 - 12 - 8 5 + 16 16 ) = 4 3 ( 12 5 - 12 - 8 5 + 16 16 ) = 4 3 [ 4 5 + 4 16 ] = 4 3 [ 4 5 + 4 16 ] = 16 3 [ 5 + 1 16 ] = 3 + 15

Prove that: sin 6 ° sin 42 ° sin 66 ° sin 78 ° = 1 16 .

To prove, sin 6 ° sin 42 ° sin 66 ° sin 78 ° = 1 16 LHS = sin 6 ° sin 42 ° sin 66 ° sin 78 ° = sin 6 ° sin 90 ° - 48 ° sin 90 ° - 24 ° sin 90 ° - 12 ° = sin 6 ° cos 48 ° cos 24 ° cos 12 ° ∵ sin 90 ° - θ = cos θ = 1 2 cos 6 ° 2 sin 6 ° cos 6 ° cos 12 ° cos 24 ° cos 48 ° = 1 4 cos 6 ° 2 sin 12 ° cos 12 ° cos 24 ° cos 48 ° ∵ 2 sin x cos x = sin 2 x = 1 4 cos 6 ° 2 sin 24 ° cos 24 ° cos 48 ° = 1 16 cos 6 ° 2 sin 48 ° cos 48 ° = sin 96 ° 16 cos 6 ° = sin 90 ° + 6 ° 16 cos 6 ° = cos 6 ° 16 cos 6 ° ∵ sin ( 90 ° + θ ) = cos θ = 1 16 = RHS Hence, proved.

Prove that sin π 5 sin 2 π 5 sin 3 π 5 sin 4 π 5 = 5 16 .

By LHS, sin π 5 sin 2 π 5 sin 3 π 5 sin 4 π 5 = sin 36 ° sin 72 ° sin 108 ° sin 144 ° = sin 36 ° sin 72 ° sin 72 ° sin 36 ° ∵ sin π - θ = sin θ = sin 2 36 ° sin 2 72 ° = sin 2 36 ° cos 2 18 ° ∵ sin π 2 - θ = cos θ = 5 - 5 8 × 5 + 5 8 = 25 - 5 64 ∵ a + b a - b = a 2 - b 2 = 20 64 = 5 16 = RHS