Question

Cos square 15 ‐ cos square 75 = root 3 / 2

Answer 1

To prove ---> Cos² 15° - Cos² 75° = (√3 ) / 2

Proof --->

LHS = Cos² 15° - Cos² 75°

We have a formula , Cos²θ = 1 - Sin²θ , applying it here , we get

= 1 - Sin² 15° - ( 1 - Sin² 75° )

= 1 - Sin² 15° - 1 + Sin² 75°

= Sin² 75° - Sin² 15°

We have a formula ,

Sin²A - Sin²B = Sin ( A + B ) Sin ( A - B ) , applying it here, we get

= Sin ( 75° + 15° ) Sin ( 75° - 15° )

= Sin 90° Sin 60°

Putting Sin90° = 1 and Sin60° = √3 / 2 , we get

= 1 { ( √3 ) / 2 }

= √3 / 2 = RHS

Second method --->

LHS= Cos² 15° - Cos² 75°

= ( Cos 15° )² - ( Cos 75° )²

= ( Cos 15° + Cos 75° ) ( Cos 15° - Cos 75° )

= ( Cos 75° + Cos 15° ) ( Cos 15° - Cos 75° )

We have two formulee

(1) CosC + CosD = 2 Cos ( C+D/ 2 ) Cos ( C-D / 2 )

(2) CosC - CosD = 2 Sin ( C + D/ 2 ) Sin ( D- C / 2 )

Applying these formulee here we get

= 2 Cos( 75° + 15° / 2 ) Cos( 75° - 15°/2 )

2Sin( 15°+75°/2 ) Sin( 75° - 15° / 2 )

= 4 Cos(90°/2) Cos(60°/2) Sin (90°/2) Sin(60°/2)

= 4 Cos ( 45° ) Cos ( 30° ) Sin ( 45° ) Sin ( 30° )

= 4 ( 1 / √2 ) ( √3 / 2 ) ( 1 / √2 ) ( 1 / 2 )

= 4 × ( √3 / 8 )

= √3 / 2 = RHS

Third method --->

We have a formula

Cos2A = 2 Cos²A - 1

=> 2 Cos²A = 1 + Cos2A

Now taking , LHS = Cos² 15° - Cos² 75°

= 1/2 ( 2 Cos²15° - 2 Cos²75° )

Applying above formula , we get

= 1/2 { ( 1 + Cos30° ) - ( 1 + Cos150° ) }

= 1/2 ( 1 + Cos30° - 1 - Cos150° )

= 1/2 ( Cos30° - Cos150° )

We have a formula

Cosc - CosD = 2 Sin ( C + D / 2 ) Sin (D - C / 2 )

= 1/2 { 2 Sin (30° + 150° / 2 ) Sin ( 150° - 30° / 2 ) }

= Sin ( 180° / 2 ) Sin ( 120° / 2 )

= Sin ( 90° ) Sin ( 60° )

= ( 1 ) ( √3 / 2 )

= √3 / 2 = RHS