JEE Main & Advanced JEE Main Paper (Held On 26 May 2012)

  • question_answer
    The value of cos \[225{}^\circ \] + sin \[195{}^\circ \] is'.   JEE Main Online Paper (Held On 26-May-2012)  

    A) \[\frac{\sqrt{3}-1}{2\sqrt{2}}\]                  

    B) \[\frac{\sqrt{3}-1}{\sqrt{2}}\]

    C) \[-\frac{\sqrt{3}-1}{\sqrt{2}}\]  

    D) \[\frac{\sqrt{3}+1}{\sqrt{2}}\]

    Correct Answer: C

    Solution :

    Consider \[\cos {{225}^{o}}+\sin {{195}^{o}}\] \[=\cos ({{270}^{o}}-{{15}^{o}})+\sin ({{180}^{o}}+{{15}^{o}})\] \[=-\sin {{15}^{o}}-\sin {{15}^{o}}\] \[=-2\sin {{15}^{o}}=-2\left( \frac{\sqrt{3}-1}{2\sqrt{2}} \right)=-\left( \frac{\sqrt{3}-1}{\sqrt{2}} \right)\]


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