Question:
If
\[
\sin\theta+\cos\theta=\sqrt{2}\cos\alpha,
\]
where \(0<\theta<\frac{\pi}{2}\), then the value of
\[
\sin2\theta
\]
is:
If
\[
\sin\theta+\cos\theta=\sqrt{2}\cos\alpha,
\]
where \(0<\theta<\frac{\pi}{2}\), then the value of
\[
\sin2\theta
\]
is:
Show Hint
Always remember
\[
(\sin\theta+\cos\theta)^2
=
1+\sin2\theta.
\]
This identity appears frequently in trigonometric simplification problems.
Updated On: Jun 10, 2026
- \(\cos2\alpha\)
- \(\sin2\alpha\)
- \(1-\cos2\alpha\)
- \(2\cos^2\alpha-1\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Concept:
The expression
\[
\sin\theta+\cos\theta
\]
can often be simplified using the identity
\[
(\sin\theta+\cos\theta)^2
=
\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta.
\]
Using
\[
\sin^2\theta+\cos^2\theta=1
\]
and
\[
\sin2\theta=2\sin\theta\cos\theta,
\]
we can connect the given expression directly to the required quantity.
Step 1: Square the given equation. Given, \[ \sin\theta+\cos\theta = \sqrt2\cos\alpha. \] Squaring both sides, \[ (\sin\theta+\cos\theta)^2 = 2\cos^2\alpha. \]
Step 2: Expand the left-hand side. \[ \sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta = 2\cos^2\alpha. \] Using \[ \sin^2\theta+\cos^2\theta=1, \] we get \[ 1+\sin2\theta = 2\cos^2\alpha. \]
Step 3: Isolate \(\sin2\theta\). \[ \sin2\theta = 2\cos^2\alpha-1. \]
Step 4: Final Conclusion. \[ \boxed{\sin2\theta=2\cos^2\alpha-1} \] Hence the correct answer is \[ \boxed{\text{Option (D)}}. \]
Step 1: Square the given equation. Given, \[ \sin\theta+\cos\theta = \sqrt2\cos\alpha. \] Squaring both sides, \[ (\sin\theta+\cos\theta)^2 = 2\cos^2\alpha. \]
Step 2: Expand the left-hand side. \[ \sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta = 2\cos^2\alpha. \] Using \[ \sin^2\theta+\cos^2\theta=1, \] we get \[ 1+\sin2\theta = 2\cos^2\alpha. \]
Step 3: Isolate \(\sin2\theta\). \[ \sin2\theta = 2\cos^2\alpha-1. \]
Step 4: Final Conclusion. \[ \boxed{\sin2\theta=2\cos^2\alpha-1} \] Hence the correct answer is \[ \boxed{\text{Option (D)}}. \]
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