frac{sin 3theta + sin 5theta + sin 7theta + s | vedclass

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

Question

(C) To solve the expression $\frac{\sin 3	heta + \sin 5	heta + \sin 7	heta + \sin 9	heta}{\cos 3	heta + \cos 5	heta + \cos 7	heta + \cos 9	het

Vedclass · Accepted Answer

$	an 6	heta$

Vedclass · Answer

(C) To solve the expression $\frac{\sin 3	heta + \sin 5	heta + \sin 7	heta + \sin 9	heta}{\cos 3	heta + \cos 5	heta + \cos 7	heta + \cos 9	heta}$,we group the terms in the numerator and denominator:Numerator: $(\sin 9	heta + \sin 3	heta) + (\sin 7	heta + \sin 5	heta)$Denominator: $(\cos 9	heta + \cos 3	heta) + (\cos 7	heta + \cos 5	heta)$Using the sum-to-product formulas $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$ and $\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:Numerator $= 2 \sin 6	heta \cos 3	heta + 2 \sin 6	heta \cos 	heta = 2 \sin 6	heta (\cos 3	heta + \cos 	heta)$Denominator $= 2 \cos 6	heta \cos 3	heta + 2 \cos 6	heta \cos 	heta = 2 \cos 6	heta (\cos 3	heta + \cos 	heta)$Dividing the numerator by the denominator:$\frac{2 \sin 6	heta (\cos 3	heta + \cos 	heta)}{2 \cos 6	heta (\cos 3	heta + \cos 	heta)} = \frac{\sin 6	heta}{\cos 6	heta} = 	an 6	heta$.

$\frac{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta}{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta} = $

Similar Questions

The value of $\sin^{2} \theta + \sin^{2}(\theta + 60^{\circ}) + \sin^{2}(\theta - 60^{\circ}) =$

The value of $\cot 70^\circ + 4\cos 70^\circ$ is

If $\frac{2\sin \alpha}{1 + \cos \alpha + \sin \alpha} = y,$ then $\frac{1 - \cos \alpha + \sin \alpha}{1 + \sin \alpha} = $

In triangle $ABC$,the value of $\sin 2A + \sin 2B + \sin 2C$ is equal to

The equation $(a + b)^2 = 4ab \sin^2 \theta$ is possible only when

Vedclass Test Series

Exam Paper Generator

Online Exam Module