What is the value of \(\frac{3}{2}\left(\frac{\cos 39°}{\sin 51°}\right)-\sqrt{\sin^2 39°+\sin^2 51°}\) ?
\(\frac{1}{2}\)
Notice the two angles are complementary: \(39°+51°=90°\), which lets every ratio be reduced to a single angle.
Using \(\sin(90°-\theta)=\cos\theta\), we get \(\sin 51°=\sin(90°-39°)=\cos 39°\).
Therefore the first fraction is \(\dfrac{\cos 39°}{\sin 51°}=\dfrac{\cos 39°}{\cos 39°}=1\), and the first term equals \(\dfrac{3}{2}\times1=\dfrac{3}{2}\).
For the radical, again \(\sin 51°=\cos 39°\), so \(\sin^2 39°+\sin^2 51°=\sin^2 39°+\cos^2 39°\).
By the Pythagorean identity \(\sin^2\theta+\cos^2\theta=1\), this sum is \(1\), and \(\sqrt{1}=1\).
Combine the two parts: \(\dfrac{3}{2}-1=\dfrac{1}{2}\). The key concepts are complementary-angle conversion and the Pythagorean identity.
The value of the expression is 1/2.
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