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JEE MATHEMATICS MCQ: TRANSFORMATIONS

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The sum   S=sin  θ+sin⁡  2θ+...+sin⁡  nθ  \mathrm{\;S} = \mathrm{sin}\,\,\mathrm{\theta } + \sin \,\,2\mathrm{\theta } + ... + \sin \,\,n\mathrm{\theta \;} equals
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Correct option: A
sin⁡  12  (n+1)  θ  sin⁡  nθ2/sin⁡  θ2\sin \,\,\frac{1}{2}\,\,\left( {n + 1} \right)\,\,\mathrm{\theta }\,\,\sin \,\,\frac{{n\mathrm{\theta }}}{2}/\sin \,\,\frac{\mathrm{\theta }}{2}
Option A (sin⁡12(n+1)θsin⁡nθ2/sin⁡θ2\sin \frac{1}{2}\left( {n + 1} \right)\theta \sin \frac{{n\theta }}{2}/\sin \frac{{\theta }}{2}) is correct because it represents the sum of a sine series using the appropriate trigonometric formula.

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cos⁡  π11+cos⁡  3 π11+cos⁡  5 π11+cos⁡  7π11+cos⁡  9 π11\cos \,\,\frac{\pi }{{11}} + \cos \,\,\frac{{3\,\pi }}{{11}} + \cos \,\,\frac{{5\,\pi }}{{11}} + \cos \,\,\frac{{7\pi }}{{11}} + \cos \,\,\frac{{9\,\pi }}{{11}}, is If   cos⁡  x+cos⁡  y+cos⁡  α=0   and sin⁡x+sin⁡  y+sin⁡  α=0\;\cos \,\,x + \cos \,\,y + \cos \,\,\alpha = 0\;\,\mathrm{and}\,\sin x + \sin \,\,y + \sin \,\,\alpha = 0 then cot⁡(x+y2)  \cot \left( {\frac{{x + y}}{2}} \right)\; is equal to The period of the function f(x)=sin⁡(2x+36  π),f\left( x \right) = \sin \left( {\frac{{2x + 3}}{{6\;\pi }}} \right), is If 0<x,y<π0 < x,\mathrm{ }y < \pi and cos⁡x+cos⁡y−cos⁡(x+y)=32\cos x + \cos y - \cos (x + y) = \frac{3}{2}, then sin⁡x+cos⁡y\sin x + \cos y is equal to The value of   sin  36∘  sin⁡  72∘  sin⁡  108∘  sin⁡  144∘  \mathrm{\;sin}\,\,36^\circ \,\,\sin \,\,72^\circ \,\,\sin \,\,108^\circ \,\,\sin \,\,144^\circ \; is equal to
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