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Question: Find the value of \({\sin ^2}{5^ \circ } + {\sin ^2}{10^ \circ } + {\sin ^2}{15^ \circ } + ... + {\s...

Find the value of sin⁡25∘+sin⁡210∘+sin⁡215∘+...+sin⁡290∘{\sin ^2}{5^ \circ } + {\sin ^2}{10^ \circ } + {\sin ^2}{15^ \circ } + ... + {\sin ^2}{90^ \circ }
A) 172\dfrac{{17}}{2}
B) 192\dfrac{{19}}{2}
C) 212\dfrac{{21}}{2}
D) 232\dfrac{{23}}{2}

Explanation

Solution

At first we need to combine the terms in order in which we can write sin⁡2(90∘−θ)=cos⁡2θ{\sin ^2}({90^ \circ } - \theta ) = {\cos ^2}\theta and by using the identity sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 we get the value of sin⁡25∘+sin⁡210∘+sin⁡215∘+...+sin⁡290∘{\sin ^2}{5^ \circ } + {\sin ^2}{10^ \circ } + {\sin ^2}{15^ \circ } + ... + {\sin ^2}{90^ \circ }

Complete step by step solution:
We can combine the given terms as follows
⇒(sin⁡25∘+sin⁡285∘)+(sin⁡210∘+sin⁡280∘)+(sin⁡215∘+sin⁡275∘)+...+(sin⁡240∘+sin⁡250∘)+sin⁡245∘+sin⁡290∘\Rightarrow ({\sin ^2}{5^ \circ } + {\sin ^2}{85^ \circ }) + ({\sin ^2}{10^ \circ } + {\sin ^2}{80^ \circ }) + ({\sin ^2}{15^ \circ } + {\sin ^2}{75^ \circ }) + ... + ({\sin ^2}{40^ \circ } + {\sin ^2}{50^ \circ }) + {\sin ^2}{45^ \circ } + {\sin ^2}{90^ \circ }
Now sin⁡2θ\sin ^{2} \theta can be written has sin⁡2(90∘−θ)=cos⁡2θ{\sin ^2}({90^ \circ } - \theta ) = {\cos ^2}\theta
⇒(sin⁡25∘+sin⁡2(90∘−5))+(sin⁡210∘+sin⁡2(90∘−10))+...+(sin⁡240∘+sin⁡2(90∘−40))+sin⁡245∘+sin⁡290∘\Rightarrow ({\sin ^2}{5^ \circ } + {\sin ^2}({90^ \circ } - 5)) + ({\sin ^2}{10^ \circ } + {\sin ^2}({90^ \circ } - 10)) + ... + ({\sin ^2}{40^ \circ } + {\sin ^2}({90^ \circ } - 40)) + {\sin ^2}{45^ \circ } + {\sin ^2}{90^ \circ }
⇒(sin⁡25∘+cos⁡25∘)+(sin⁡210∘+cos⁡210∘)+(sin⁡215∘+cos⁡215∘)+...+(sin⁡240∘+cos⁡240∘)+sin⁡245∘+sin⁡290∘\Rightarrow ({\sin ^2}{5^ \circ } + {\cos ^2}{5^ \circ }) + ({\sin ^2}{10^ \circ } + {\cos ^2}{10^ \circ }) + ({\sin ^2}{15^ \circ } + {\cos ^2}{15^ \circ }) + ... + ({\sin ^2}{40^ \circ } + {\cos ^2}{40^ \circ }) + {\sin ^2}{45^ \circ } + {\sin ^2}{90^ \circ }
We know that sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
We get,

⇒1+1+1+1+1+1+1+1+sin⁡245∘+sin⁡290∘ ⇒8+(12)2+1 ⇒9+12=18+12=192  \Rightarrow 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + {\sin ^2}{45^ \circ } + {\sin ^2}{90^ \circ } \\\ \Rightarrow 8 + {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} + 1 \\\ \Rightarrow 9 + \dfrac{1}{2} = \dfrac{{18 + 1}}{2} = \dfrac{{19}}{2} \\\

Therefore the correct option is (B).

Note:
An equation is called an identity when it is true for all values of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angles involved.