JEE Main & Advanced Mathematics Trigonometric Identities Question Bank Self Evaluation Test - Trigonometric Function

  • question_answer
    The value of \[{{\sin }^{2}}5{}^\circ +{{\sin }^{2}}10{}^\circ +{{\sin }^{2}}15{}^\circ +{{\sin }^{2}}\]\[20{}^\circ +.....+{{\sin }^{2}}90{}^\circ \] is

    A) 7                     

    B) 8    

    C) 9                     

    D) \[\frac{19}{2}\]

    Correct Answer: D

    Solution :

    \[{{\sin }^{2}}5{}^\circ +{{\sin }^{2}}10{}^\circ +{{\sin }^{2}}15{}^\circ +.....+......+.....\] \[{{\sin }^{2}}75{}^\circ +{{\sin }^{2}}80{}^\circ +{{\sin }^{2}}85{}^\circ +{{\sin }^{2}}90{}^\circ \] \[\Rightarrow \,\,{{\sin }^{2}}5{}^\circ +{{\sin }^{2}}10{}^\circ +{{\sin }^{2}}15{}^\circ +.....{{\sin }^{2}}\] \[(90-15{}^\circ )+{{\sin }^{2}}(90-10)+si{{n}^{2}}(90-5)+1\] \[\Rightarrow \,\,{{\sin }^{2}}5+{{\sin }^{2}}10+{{\sin }^{2}}15{}^\circ +......+{{\cos }^{2}}15{}^\circ \]\[+\,\,{{\cos }^{2}}10{}^\circ +{{\cos }^{2}}5{}^\circ +1\] \[\Rightarrow \,\,\,\,\,(1+1+1+....8\,times)+si{{n}^{2}}45{}^\circ +1\] \[\Rightarrow \,\,8+\frac{1}{2}+1=\frac{19}{2}\]


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