JEE MATHEMATICS MCQ: TRANSFORMATIONS

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Analyzing +4 | -1

{a_{n + 1}} = \sqrt {\frac{1}{2}\,\,\left( {1 + {a_n}} \right)}

, then

\cos \,\,\left( {\frac{{\sqrt {1 - a_0^2} }}{{{a_1}\,{a_2}\,{a_3} \ldots \mathrm{to\;}\,\,\infty }}} \right) =

View answer and explanation

Correct option: C

{a_0}

Option C (

{a_0}

) is correct because the infinite product simplifies to a value that allows the cosine to be evaluated, resulting in

{a_0}

More Questions from TRANSFORMATIONS

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1 + \cos \,\,56^\circ + \cos \,\,58^\circ - \cos \,\,66^\circ = \lambda \cos \,\,28^\circ \,\,\cos \,\,29^\circ \,\,\sin \,\,33^\circ

\lambda =

\cos \,\,\frac{\pi }{{11}} + \cos \,\,\frac{{3\,\pi }}{{11}} + \cos \,\,\frac{{5\,\pi }}{{11}} + \cos \,\,\frac{{7\pi }}{{11}} + \cos \,\,\frac{{9\,\pi }}{{11}}

\;\cos \,\,x + \cos \,\,y + \cos \,\,\alpha = 0\;\,\mathrm{and}\,\sin x + \sin \,\,y + \sin \,\,\alpha = 0

\cot \left( {\frac{{x + y}}{2}} \right)\;

f\left( x \right) = \sin \left( {\frac{{2x + 3}}{{6\;\pi }}} \right),

0 < x,\mathrm{ }y < \pi

\cos x + \cos y - \cos (x + y) = \frac{3}{2}

\sin x + \cos y

\mathrm{\;sin}\,\,36^\circ \,\,\sin \,\,72^\circ \,\,\sin \,\,108^\circ \,\,\sin \,\,144^\circ \;