Question

sin 60 + theta minus sin 60 - theta is equal to sin theta

Answer 1

Question:-

Prove:-

$\sin(60 \degree+ \theta ) - \sin(60 \degree - \theta) = \sin \theta$

Required Formulae:-

$\sin(a + b) = \sin(a) \cos(b) + \cos(a) \sin(b)$

$\sin(a - b) = \sin(a) \cos(b) - \cos(a) \sin(b)$

Solution:-

To prove:-

$\sin(60 \degree+ \theta ) - \sin(60 \degree - \theta) = \sin \theta$

By applying formula,

$\sin(60 + \theta) = \sin60 cos \theta + \cos60 \sin \theta$

$= > \frac{ \sqrt{3} }{2} \cos \theta + \frac{1}{2} \sin \theta$

$= > \frac{ \sqrt{3} \cos \theta + \sin \theta}{2}$

$\sin(60 - \theta) = \sin60 \cos \theta - \cos60 \sin \theta$

$= > \frac{ \sqrt{3} }{2} \cos \theta - \frac{1}{2} \sin \theta$

$= > \frac{ \sqrt{3} \cos \theta - \sin \theta}{2}$

$\sin(60 + \theta) - \sin(60 - \theta)$

$\frac{ \sqrt{3} \cos \theta + \sin \theta}{2} - \frac{ \sqrt{3} \cos \theta - \sin \theta }{2}$

$\frac{ \sqrt{3} \cos \theta + \sin \theta - \sqrt{3} \cos \theta + \sin \theta }{2}$

$\frac{2 \sin \theta}{2}$

$\sin \theta$

Answer 2

Given:

A trigonometric equation Sin(60 + θ) - Sin( 60 - θ) = Sinθ.

To Find:

The proof of the given trigonometric equation.

Solution:

The given problem can be solved using the concepts of trigonometry.

1. The given trigonometric equation is Sin(60 + θ) - Sin( 60 - θ) = Sinθ.

2. According to the properties of trigonometry,

• Sin(A+B) = SinACosB + CosASinB,
• Sin(A-B) = SinACosB - CosASinB.

3. Consider the LHS of the equation and use the formulae mentioned above to expand the given trigonometric equation,

=> ( Sin60Cosθ + Cos60Sinθ) - ( Sin60Cosθ - Cos60Sinθ),

=> √(3/2)Cosθ + (1/2)Sinθ - √(3/2)Cosθ + (1/2)Sinθ,

=> 0 + (1/2)Sinθ + (1/2)Sinθ,

=> Sinθ = RHS.

Hence Proved.

Therefore, the given equation is proved.