An angle \(\theta\) lying strictly between \(0^\circ\) and \(90^\circ\) is chosen so that \(\sin\theta+\cos\theta=\displaystyle\frac{17}{13}\). Work out the product \(\sin\theta\cos\theta\).
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{60}{169}\)
Squaring \(\sin\theta+\cos\theta\) gives \(\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta\), and the first two terms add up to 1. The product is then found directly.
\((\sin\theta+\cos\theta)^2=\displaystyle\frac{289}{169}\)
\(1+2\sin\theta\cos\theta=\displaystyle\frac{289}{169}\), so \(2\sin\theta\cos\theta=\displaystyle\frac{120}{169}\)
\(\sin\theta\cos\theta=\displaystyle\frac{60}{169}\)
Such an angle exists: \(\sin\theta=\frac{5}{13}\), \(\cos\theta=\frac{12}{13}\) (or the other way round) fit the condition.