Q62 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics
If $\sin \theta-\cos \theta=\frac{1}{\sqrt{3}}$, then $\sin (2 \theta)+\cos (4 \theta)+\sin (6 \theta)=$
- A$\frac{37}{27}$
- B$\frac{-37}{27}$
- C$\frac{-43}{27}$
- Correct answerD$\frac{43}{27}$
Solution
We are given that
$$
\begin{aligned}
& \sin \theta-\cos \theta=\frac{1}{\sqrt{3}} \\
& \text { Now }(\sin \theta-\cos \theta)^2=\sin ^2 \theta+\cos ^2 \theta-\sin 2 \theta \\
& =\frac{1}{3}=1-\sin 2 \theta \Rightarrow \sin 2 \theta=\frac{3}{2} \\
& \therefore \cos 4 \theta=1-2 \sin ^2 2 \theta=1-2 \times \frac{4}{9} \\
& =1-\frac{8}{9}=\frac{1}{9} \Rightarrow \cos 4 \theta=\frac{1}{9} \\
& \Rightarrow \sin 6 \theta=\frac{22}{27}
\end{aligned}
$$
Now $\sin 2 \theta+\cos 4 \theta+\sin 6 \theta=\frac{2}{3}+\frac{1}{9}+\frac{22}{7}$
$$
\frac{18+3+22}{27}=\frac{22+22}{27}=\frac{43}{27}
$$
Hence, the correct answer is option: 'D'.