11th Class Mathematics Trigonometric Identities Question Bank Critical Thinking

  • question_answer \[1+\cos \,{{56}^{o}}+\cos \,{{58}^{o}}-\cos {{66}^{o}}=\] [IIT 1964]

    A) \[2\,\cos {{28}^{o}}\,\cos \,{{29}^{o}}\,\cos \,{{33}^{o}}\]

    B) \[4\,\cos {{28}^{o}}\,\cos \,{{29}^{o}}\,\cos \,{{33}^{o}}\]

    C) \[4\,\cos {{28}^{o}}\,\cos \,{{29}^{o}}\,\sin {{33}^{o}}\]

    D) \[2\,\cos {{28}^{o}}\,\cos \,{{29}^{o}}\,\sin \,{{33}^{o}}\]

    Correct Answer: C

    Solution :

    \[1+\cos 56{}^\circ +\cos 58{}^\circ -\cos 66{}^\circ \] \[=2{{\cos }^{2}}28{}^\circ +2\sin 62{}^\circ .\sin 4{}^\circ \] \[=2{{\cos }^{2}}28{}^\circ +2\cos 28{}^\circ .\sin 4{}^\circ \] \[=2\cos 28{}^\circ (\cos 28{}^\circ +\cos 86{}^\circ )\] \[=2\cos 28{}^\circ .2\cos 57{}^\circ \cos 29{}^\circ \] \[=4\cos 28{}^\circ \cos 29{}^\circ \sin 33{}^\circ \]. Aliter: Apply the conditional identity \[\cos A+\cos B-\cos C=-1+4\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2}\]\[[\because \,56{}^\circ +58{}^\circ +66{}^\circ =180{}^\circ ]\] We get the value of required expression equal to\[4\cos 28{}^\circ \cos 29{}^\circ \sin 33{}^\circ \].

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