sin 50^circ - sin 70^circ + sin 10^circ = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Using the formula $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:$\sin 50^\circ - \sin 70^\circ + \sin 10^\circ$$= 2 \cos \frac{50^\c

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Using the formula $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:$\sin 50^\circ - \sin 70^\circ + \sin 10^\circ$$= 2 \cos \frac{50^\circ + 70^\circ}{2} \sin \frac{50^\circ - 70^\circ}{2} + \sin 10^\circ$$= 2 \cos 60^\circ \sin(-10^\circ) + \sin 10^\circ$$= -2 \cos 60^\circ \sin 10^\circ + \sin 10^\circ$$= -2 	imes \frac{1}{2} 	imes \sin 10^\circ + \sin 10^\circ$$= -\sin 10^\circ + \sin 10^\circ = 0.$

$\sin 50^\circ - \sin 70^\circ + \sin 10^\circ = $

Similar Questions

Which of the following have the same value?
$(a)$ $\sin 120^{\circ}$
$(b)$ $\cos 930^{\circ}$
$(c)$ $\tan 840^{\circ}$
$(d)$ $\cot (-1110^{\circ})$

$\cos 36^{\circ} - \cos 72^{\circ}$ is equal to

If $\sin x = -\frac{24}{25},$ then the value of $\tan x$ is

Find the angle in radian through which a pendulum swings if its length is $75\, cm$ and the tip describes an arc of length $15\, cm$.

The value of $\frac{1+\tanh x}{1-\tanh x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module