JEE MATHEMATICS MCQ: TRANSFORMATIONS

1 Easy

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Analyzing +4 | -1

\alpha ,\,\;\beta ,\,\;\gamma ,\,\,\delta

are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity

k

, then the value of

4\sin \,\,\frac{\alpha }{2} + 3\sin \,\,\frac{\beta }{2} + 2\sin \,\,\frac{\gamma }{2} + \sin \,\,\frac{\delta }{2}

is equal to

View answer and explanation

Correct option: B

2\,\sqrt {1 + k}

Option B (

2\,\sqrt {1 + k}

) is correct because by using the properties of sine and trigonometric identities, we can simplify the expression to

2\sqrt{1+k}

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 \;