Question

(k) sin (120° - A) + sin (120° - B) + sin (120° - C)​

Answer 1

Answer:

Step-by-step explanation:

Given :

To find : The value of sin (120° - A) + sin (120° - B) + sin (120° - C)​

Solution :

We know that,

sin (A - B) = sin A cos B - cos A sin B

Hence,

sin (120° - A) = sin 120° cos A + cos 120° sin A

sin (120° - B) = sin 120° cos B + cos 120° sin B

sin (120° - C) = sin 120° cos C + cos 120° sin C

________

To solve further,

We need to know the values of,

sin 120° & cos 120°

We know that,

sin (A + B) = sin A cos B + cos A sin B

sin 120° = sin (90 + 30)° = sin 90° cos 30° + cos 90° sin 30°

We know that,

sin 90° = 1,

cos 90° = 0,

cos 30° = $\frac{\sqrt{3}}{2}$,

sin 30° = $\frac{1}{2}$,

So,

sin 120° = sin 90° cos 30° + cos 90° sin 30°

⇒ $(1)(\frac{\sqrt{3}}{2}) + (0)(\frac{1}{2})$

⇒ $\frac{\sqrt{3}}{2}$

hence,

sin 120° = $\frac{\sqrt{3}}{2}$

We also know that,

sin² θ + cos² θ = 1

cos² θ = 1 - sin² θ

cos² 120° = 1 - sin² 120°

cos² 120° = 1 - $(\frac{\sqrt{3}}{2})^2$

cos² 120° = 1 - $\frac{3}{4}$

cos² 120° = $\frac{1}{4}$

cos 120° = $\sqrt{\frac{1}{4}}$ = - $\frac{1}{2}$

As cos θ < 0 , θ ∈ (90° , 270°)

Hence,

By substituting the values of cos 120° & sin 120°,

sin (120° - A) = sin 120° cos A + cos 120° sin A

sin (120° - B) = sin 120° cos B + cos 120° sin B

sin (120° - C) = sin 120° cos C + cos 120° sin C,

By adding all 3 equations,

sin (120° - A) + sin (120° - B) + sin (120° - C)

= sin 120°(cos A + cos B + cos C) + cos 120°(sin A + sin B + sin C)

= $\frac{\sqrt{3}}{2}(cos A + cos B + cos C) - \frac{1}{2}(sin A + sin B + sin C)$

= $\frac{\sqrt{3}(cos A + cos B + cos C) - (sin A + sin B + sin C)}{2}$