J & K CET Engineering J and K - CET Engineering Solved Paper-2009

  • question_answer
    \[\sin \,\,{{47}^{o}}+\sin {{61}^{o}}-\sin {{11}^{o}}-\sin {{25}^{o}}\] is equal to

    A)  \[\sin \,{{7}^{o}}\]          

    B)  \[\cos \,{{7}^{o}}\]

    C)  \[\sin \,{{36}^{o}}\]         

    D)  \[\cos \,{{36}^{o}}\]

    Correct Answer: B

    Solution :

    \[\sin {{47}^{o}}-sin{{25}^{o}}+\sin {{61}^{o}}-\sin {{11}^{o}}\] \[=2\,\,\cos \,{{36}^{o}}\,\sin {{11}^{o}}+2\cos \,{{36}^{o}}\,\sin {{25}^{o}}\] \[=2\,\,\cos \,{{36}^{o}}\,[\sin {{11}^{o}}+\,\sin {{25}^{o}}]\] \[=2\,\,\cos \,{{36}^{o}}\left[ 2\,\sin \,\left( \frac{{{25}^{o}}+{{11}^{o}}}{2} \right)\,\cos \,\left( \frac{{{25}^{o}}-{{11}^{o}}}{2} \right) \right]\] \[=4\,\cos \,{{36}^{o}}\,\sin {{18}^{o}}\,\cos {{7}^{o}}\] \[=4\left( \frac{\sqrt{5}+1}{4} \right)\,\left( \frac{\sqrt{5}-1}{4} \right)\,\cos \,{{7}^{o}}\] \[=\frac{5-1}{4}\,\cos \,{{7}^{o}}\] \[=\,\cos \,{{7}^{o}}\]


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