11th Class Mathematics Trigonometric Identities Question Bank Critical Thinking

  • question_answer The value of \[{{\sin }^{2}}{{5}^{o}}+{{\sin }^{2}}{{10}^{o}}+{{\sin }^{2}}{{15}^{o}}+...+\] \[{{\sin }^{2}}{{85}^{o}}+{{\sin }^{2}}{{90}^{o}}\] is  equal to [Karnataka CET 1999]

    A) 7

    B) 8

    C) 9

    D) \[9\frac{1}{2}\]

    Correct Answer: D

    Solution :

    Given expression is \[{{\sin }^{2}}{{5}^{o}}+{{\sin }^{2}}{{10}^{o}}+{{\sin }^{2}}{{15}^{o}}+.....+{{\sin }^{2}}{{85}^{o}}+{{\sin }^{2}}{{90}^{o}}.\] We know that \[\sin {{90}^{o}}=1\] or \[{{\sin }^{2}}{{90}^{o}}=1\]. Similarly, \[\sin {{45}^{o}}=\frac{1}{\sqrt{2}}\text{or}\,\text{si}{{\text{n}}^{\text{2}}}{{45}^{o}}=\frac{1}{2}\] and the angles are in A.P. of 18 terms. We also know that \[{{\sin }^{2}}{{85}^{o}}={{[\sin ({{90}^{o}}-{{5}^{o}})]}^{2}}\]\[={{\cos }^{2}}{{5}^{o}}.\] Therefore from the complementary rule, we find \[{{\sin }^{2}}{{5}^{o}}+{{\sin }^{2}}{{85}^{o}}={{\sin }^{2}}{{5}^{o}}+{{\cos }^{2}}{{5}^{o}}=1.\] Therefore, \[{{\sin }^{2}}{{5}^{o}}+{{\sin }^{2}}{{10}^{o}}+{{\sin }^{2}}{{15}^{o}}+...+{{\sin }^{2}}{{85}^{o}}+{{\sin }^{2}}{{90}^{o}}\] \[=(1+1+1+1+1+1+1+1)+1+\frac{1}{2}=9\frac{1}{2}\].


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