If \(2\sin\theta-1=0\), then find the value of \(2\cos^2\theta-\sin^3\theta\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\frac{11}{8}\)
The equation gives \(\sin\theta\) directly, and \(\cos^2\theta\) then follows from the identity \(\sin^2\theta+\cos^2\theta=1\); no angle needs to be found.
From the given equation, \(\sin\theta=\frac{1}{2}\)
From the identity, \(\cos^2\theta=1-\frac{1}{4}=\frac{3}{4}\), so \(2\cos^2\theta=\frac{3}{2}\)
Cube of the sine: \(\sin^3\theta=\left(\frac{1}{2}\right)^3=\frac{1}{8}\)
Required value: \(2\cos^2\theta-\sin^3\theta=\frac{3}{2}-\frac{1}{8}=\frac{12-1}{8}=\frac{11}{8}\)