If sin x + cos x = a,where a in [-sqrt{2}, sq | vedclass

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

Question

(D) Given $\sin x + \cos x = a$. Squaring both sides,we get $1 + 2 \sin x \cos x = a^2$,so $\sin x \cos x = \frac{a^2 - 1}{2}$.The sum is $S = \sum_{n

Vedclass · Accepted Answer

$\frac{2(1 + a - a^2)}{(a - 1)^2}$

Vedclass · Answer

(D) Given $\sin x + \cos x = a$. Squaring both sides,we get $1 + 2 \sin x \cos x = a^2$,so $\sin x \cos x = \frac{a^2 - 1}{2}$.The sum is $S = \sum_{n=1}^{\infty} \sin^n x + \sum_{n=1}^{\infty} \cos^n x = \frac{\sin x}{1 - \sin x} + \frac{\cos x}{1 - \cos x}$.$S = \frac{\sin x(1 - \cos x) + \cos x(1 - \sin x)}{(1 - \sin x)(1 - \cos x)} = \frac{(\sin x + \cos x) - 2 \sin x \cos x}{1 - (\sin x + \cos x) + \sin x \cos x}$.Substituting $\sin x + \cos x = a$ and $\sin x \cos x = \frac{a^2 - 1}{2}$:$S = \frac{a - (a^2 - 1)}{1 - a + \frac{a^2 - 1}{2}} = \frac{a - a^2 + 1}{\frac{2 - 2a + a^2 - 1}{2}} = \frac{2(1 + a - a^2)}{a^2 - 2a + 1} = \frac{2(1 + a - a^2)}{(a - 1)^2}$.

If $\sin x + \cos x = a$,where $a \in [-\sqrt{2}, \sqrt{2}] - \{-1, 1\}$,then $\sum_{n=1}^{\infty} (\sin^n x + \cos^n x)$ is equal to -

Similar Questions

If $\tan \theta \cdot \tan \left(120^{\circ}-\theta\right) \tan \left(120^{\circ}+\theta\right)=\frac{1}{\sqrt{3}}$,then $\theta$ is equal to

If $y = (1 + \tan A)(1 - \tan B)$ where $A - B = \frac{\pi}{4}$,then $(y + 1)^{y + 1}$ is equal to

The value of the series $\cos 12^{\circ} + \cos 84^{\circ} + \cos 132^{\circ} + \cos 156^{\circ}$ is

Let $S = \{\theta \in [0, 2\pi] : 8^{2 \sin^2 \theta} + 8^{2 \cos^2 \theta} = 16\}$. Then $n(S) + \sum_{\theta \in S} \left(\sec \left(\frac{\pi}{4} + 2\theta\right) \operatorname{cosec} \left(\frac{\pi}{4} + 2\theta\right)\right)$ is equal to.

The value of the $\sin 1^{\circ} + \sin 2^{\circ} + \ldots + \sin 359^{\circ}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module