\(\sin^2 5^\circ+\sin^2 10^\circ+\sin^2 15^\circ+\ldots+\sin^2 90^\circ\) value of:
\(9\dfrac{1}{2}\)
The series is \(\sin^2(5k^\circ)\) for \(k=1,2,\ldots,18\) (angles \(5^\circ,10^\circ,\ldots,90^\circ\)), i.e. 18 terms.
Use the pairing identity \(\sin^2\theta+\sin^2(90^\circ-\theta)=\sin^2\theta+\cos^2\theta=1\). Pair the terms: \((5^\circ,85^\circ),(10^\circ,80^\circ),(15^\circ,75^\circ),(20^\circ,70^\circ),(25^\circ,65^\circ),(30^\circ,60^\circ),(35^\circ,55^\circ),(40^\circ,50^\circ)\) — that is 8 pairs, each summing to 1.
The remaining unpaired terms are \(45^\circ\) and \(90^\circ\): \(\sin^2 45^\circ=\dfrac12\) and \(\sin^2 90^\circ=1\).
Total sum \(=8(1)+\dfrac12+1=9\dfrac12\).
Similar Questions
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Important Questions from Trigonometric Identities
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If \(\sin \left( {A - B} \right) = \frac{1}{2}\) and \(\cos \left( {A + B} \right) = \frac{1}{2}\) , where A > B > 0° and A + B is an acute angle, then the value of A is:
Find the value of $\cos 10^\circ \times \cos 30^\circ \times \cos 50^\circ \times \cos 70^\circ$
