If \(2\cos^2\theta-1=0\), where \(\theta\) is an acute angle, then find the value of \(2\sin^2\theta+\sin2\theta\).
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2
The given equation fixes \(\cos^2\theta\), and for an acute angle that fixes \(\theta\) itself; then the expression is worked out with known values.
\(2\cos^2\theta=1\), so \(\cos^2\theta=\frac{1}{2}\) and \(\cos\theta=\frac{1}{\sqrt{2}}\)
\(\theta\) is acute, so \(\theta=45^\circ\)
First term: \(\sin 45^\circ=\frac{1}{\sqrt{2}}\), so \(2\sin^2 45^\circ=2\times\frac{1}{2}=1\)
Second term: \(2\theta=90^\circ\), so \(\sin 2\theta=\sin 90^\circ=1\)
Value \(=1+1=\) 2