Question : If $\alpha$ and $\beta$ are positive acute angles, $\sin (4\alpha -\beta )=1$ and $\cos (2\alpha +\beta)=\frac{1}{2}$, then the value of $\sin (\alpha +2\beta)$ is:
Option 1: $0$
Option 2: $1$
Option 3:
$\frac{\sqrt{3}}{2}$
Option 4:
$\frac{1}{\sqrt{2}}$
New: SSC CHSL tier 1 answer key 2024 out | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
Correct Answer:
$\frac{1}{\sqrt{2}}$
Solution :
Given: $\sin (4\alpha -\beta )=1$ and $\cos (2\alpha +\beta)=\frac{1}{2}$
$\sin (4\alpha -\beta )=1$
⇒ $\sin (4\alpha -\beta )=\sin 90^{\circ}$
⇒ $4\alpha -\beta= 90^{\circ}$...................................... (1)
Now, $\cos (2\alpha +\beta)=\frac{1}{2}$
⇒ $\cos (2\alpha +\beta)=\cos 60^{\circ}$
⇒ $2\alpha +\beta=60^{\circ}$....................................(2)
Solving equations 1 and 2, we get:
$6\alpha=150^{\circ}$
⇒ $\alpha=25^{\circ}$
Putting the value of $\alpha$ in equation 1, we get,
$4×25^{\circ} -\beta= 90^{\circ}$
⇒ $\beta=10^{\circ}$
Thus, $\sin (\alpha +2\beta)=\sin (25^{\circ} +2×10^{\circ})$
$=\sin (25^{\circ} +20^{\circ})$
$=\sin45^{\circ}$
$=\frac{1}{\sqrt{2}}$
Hence, the correct answer is $\frac{1}{\sqrt{2}}$.
Related Questions
Know More about
Staff Selection Commission Combined High ...
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Get Updates Brochure
Your Staff Selection Commission Combined Higher Secondary Level Exam brochure has been successfully mailed to your registered email id “”.
