frac{sin A+sin 7 A+sin 13 A}{cos A+cos 7 A+co | vedclass

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

Question

(C) Given expression: $\frac{\sin A+\sin 7 A+\sin 13 A}{\cos A+\cos 7 A+\cos 13 A}$ Group the terms with $A$ and $13A$: $= \frac{(\sin 13 A+\sin A)+\s

Vedclass · Accepted Answer

$	an 7 A$

Vedclass · Answer

(C) Given expression: $\frac{\sin A+\sin 7 A+\sin 13 A}{\cos A+\cos 7 A+\cos 13 A}$ Group the terms with $A$ and $13A$: $= \frac{(\sin 13 A+\sin A)+\sin 7 A}{(\cos 13 A+\cos A)+\cos 7 A}$ Using sum-to-product formulas $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$ and $\cos C + \cos D = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})$: $= \frac{2 \sin(\frac{13A+A}{2}) \cos(\frac{13A-A}{2}) + \sin 7 A}{2 \cos(\frac{13A+A}{2}) \cos(\frac{13A-A}{2}) + \cos 7 A}$ $= \frac{2 \sin 7 A \cos 6 A + \sin 7 A}{2 \cos 7 A \cos 6 A + \cos 7 A}$ Factor out $\sin 7 A$ from the numerator and $\cos 7 A$ from the denominator: $= \frac{\sin 7 A(2 \cos 6 A + 1)}{\cos 7 A(2 \cos 6 A + 1)}$ $= \frac{\sin 7 A}{\cos 7 A} = 	an 7 A$

$\frac{\sin A+\sin 7 A+\sin 13 A}{\cos A+\cos 7 A+\cos 13 A} =$

Similar Questions

If $\frac{x}{\cos \alpha} = \frac{y}{\cos \left(\frac{2 \pi}{3} - \alpha\right)} = \frac{z}{\cos \left(\frac{2 \pi}{3} + \alpha\right)}$,then the value of $(x + y + z)$ is equal to

If $\tan A = -\frac{1}{2}$ and $\tan B = -\frac{1}{3},$ then $A + B = $

$\frac{\sinh(x+y) + \sinh(x-y)}{\cosh(x+y) - \cosh(x-y)} = $

Prove that $\cos 2x \cos \frac{x}{2} - \cos 3x \cos \frac{9x}{2} = \sin 5x \sin \frac{5x}{2}$

Prove that: $\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x$

Vedclass Test Series

Exam Paper Generator

Online Exam Module