If sin left( x + frac{4pi}{9} right) = a and | vedclass

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

Question

(D) Given,$\sin \left( x + \frac{4\pi}{9} \right) = a$. Note that $\frac{4\pi}{9} = 80^{\circ}$.We need to find $\cos \left( x + \frac{7\pi}{9} \right

Vedclass · Accepted Answer

$\frac{-\sqrt{1 - a^2} - a\sqrt{3}}{2}$

Vedclass · Answer

(D) Given,$\sin \left( x + \frac{4\pi}{9} \right) = a$. Note that $\frac{4\pi}{9} = 80^{\circ}$.We need to find $\cos \left( x + \frac{7\pi}{9} \right)$. Note that $\frac{7\pi}{9} = 140^{\circ}$.We can write $\cos \left( x + 140^{\circ} \right) = \cos \left( (x + 80^{\circ}) + 60^{\circ} \right)$.Using the identity $\cos(A + B) = \cos A \cos B - \sin A \sin B$:$\cos \left( (x + 80^{\circ}) + 60^{\circ} \right) = \cos(x + 80^{\circ}) \cos 60^{\circ} - \sin(x + 80^{\circ}) \sin 60^{\circ}$.Since $\frac{\pi}{9} In this interval,$\cos(x + 80^{\circ})$ is negative,so $\cos(x + 80^{\circ}) = -\sqrt{1 - \sin^2(x + 80^{\circ})} = -\sqrt{1 - a^2}$.Substituting the values: $\cos(x + 140^{\circ}) = (-\sqrt{1 - a^2}) \cdot \frac{1}{2} - a \cdot \frac{\sqrt{3}}{2} = \frac{-\sqrt{1 - a^2} - a\sqrt{3}}{2}$.

If $\sin \left( x + \frac{4\pi}{9} \right) = a$ and $\frac{\pi}{9} < x < \frac{\pi}{3}$,then $\cos \left( x + \frac{7\pi}{9} \right)$ equals:

Similar Questions

If $\tan A = \frac{1 - \cos B}{\sin B}$,find $\tan 2A$ in terms of $\tan B$ and show that:

The value of $x,$ which satisfies the equation $2 \operatorname{cosec}^{2} 30^{\circ} + x \sin^{2} 60^{\circ} - \frac{3}{4} \tan^{2} 30^{\circ} = 10$ is

$\frac{{\sin \theta }}{{1 - \cot \theta }} + \frac{{\cos \theta }}{{1 - \tan \theta }} = $

If $\tan A + \cot A = 4,$ then ${\tan ^4}A + {\cot ^4}A$ is equal to

The sum of all values of $\theta \in (0, \frac{\pi}{2})$ satisfying $\sin^2 2\theta + \cos^4 2\theta = \frac{3}{4}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module