JEE MATHEMATICS MCQ: TRANSFORMATIONS

1 Easy

1 of 1

Application +4 | -1

n = 1,\;\,2,\,\;3,\,\; \ldots

, then

\cos \alpha \,\,\cos \,\,2\alpha \,\,\cos \,\,4\alpha \ldots \cos \,\,{2^{n - 1}}\alpha

is equal to

View answer and explanation

Correct option: D

\frac{{\sin \,\,{2^n}\alpha }}{{{2^n}\sin \alpha }}

Option D is correct because the given expression can be simplified using the identity

2\sin x \cos x = \sin 2x

repeatedly, leading to the result

\frac{{\sin \,{2^n}\alpha }}{{{{2^n}}\sin \,\alpha }}}

More Questions from TRANSFORMATIONS

Practice All Chapter MCQs →

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 \;