CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    \[si{{n}^{2}}5{}^\circ +si{{n}^{2}}10{}^\circ +si{{n}^{2}}15{}^\circ ...+si{{n}^{2}}90{}^\circ \]is equal to:

    A)  \[8\frac{1}{2}\]                               

    B)  9

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

    D)         \[4\frac{1}{2}\]

    E)  0

    Correct Answer: C

    Solution :

    \[si{{n}^{2}}5{}^\circ +si{{n}^{2}}10{}^\circ \text{ +}si{{n}^{2}}15{}^\circ +...+si{{n}^{2}}90{}^\circ \] \[=si{{n}^{2}}5{}^\circ +si{{n}^{2}}10{}^\circ \text{ +}........\text{+}si{{n}^{2}}45{}^\circ +...\] \[+si{{n}^{2}}80{}^\circ +si{{n}^{2}}85{}^\circ \text{ +}si{{n}^{2}}90{}^\circ \] \[=si{{n}^{2}}5{}^\circ +si{{n}^{2}}10{}^\circ \text{ +}....\text{+}\frac{1}{2}+....+{{\cos }^{2}}10{}^\circ \]                                                 \[+{{\cos }^{2}}5{}^\circ +{{\sin }^{2}}90{}^\circ \] \[=({{\sin }^{2}}5{}^\circ +{{\cos }^{2}}5{}^\circ )+({{\sin }^{2}}10{}^\circ +{{\cos }^{2}}10{}^\circ )+\]                 \[.....+({{\sin }^{2}}40{}^\circ +{{\cos }^{2}}40{}^\circ )+{{\sin }^{2}}45{}^\circ \]                                                                 \[+{{\sin }^{2}}90{}^\circ \] \[=1+1+1+1+1+1+1+1+\frac{1}{2}+1\] \[=9\frac{1}{2}\]


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