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

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given $\sin \left( x + \frac{4\pi}{9} \right) = a$. We need to find $\cos \left( x + \frac{7\pi}{9} \right)$. Note that $\frac{7\pi}{9} = \frac{4\pi}{9} + \frac{3\pi}{9} = \frac{4\pi}{9} + \frac{\pi}{3}$. So,$\cos \left( x + \frac{7\pi}{9} \right) = \cos \left( \left( x + \frac{4\pi}{9} \right) + \frac{\pi}{3} \right)$. Using the formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$: $\cos \left( \left( x + \frac{4\pi}{9} \right) + \frac{\pi}{3} \right) = \cos \left( x + \frac{4\pi}{9} \right) \cos \frac{\pi}{3} - \sin \left( x + \frac{4\pi}{9} \right) \sin \frac{\pi}{3}$. Given $\frac{\pi}{9} In this interval,the cosine function is negative. Since $\sin \left( x + \frac{4\pi}{9} \right) = a$,then $\cos \left( x + \frac{4\pi}{9} \right) = -\sqrt{1 - a^2}$. Substituting the values: $-\sqrt{1 - a^2} \cdot \frac{1}{2} - a \cdot \frac{\sqrt{3}}{2} = \frac{-\sqrt{1 - a^2} - a\sqrt{3}}{2}$.

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

Prove that $3 \sin \frac{\pi}{6} \sec \frac{\pi}{3} - 4 \sin \frac{5 \pi}{6} \cot \frac{\pi}{4} = 1$.

The values of $x$ in $\left(0, \frac{\pi}{2}\right)$ satisfying the equation $\sin x \cos x = \frac{1}{4}$ are $.......$

Convert $6$ radians into degree measure.

Which of the following is true?

$\sin ^2 \frac{2 \pi}{3}+\cos ^2 \frac{5 \pi}{6}-\tan ^2 \frac{3 \pi}{4}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module