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AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 22 of 160

Marks: +1, -0

\cos^2

\cos^2

\sin^2

\sin^2

35° + cos 15° - cos 5° sin 35° =

Solution:

Consider the expression,

\cos^{2} 5^{\circ} - \cos^{2} 15^{\circ} - \sin^{2} 15^{\circ} + \sin^{2} 35^{\circ} + \cos^{2} 15^{\circ} \sin 15^{\circ} - \cos 5^{\circ} \sin 35^{\circ}

= \cos 5^{\circ}(\cos 5^{\circ} - \sin 35^{\circ}) - \cos 15^{\circ}(\cos 15^{\circ} - \sin 15^{\circ}) + \sin^{2} 35^{\circ} - \sin^{2} 15^{\circ}

= \cos 5^{\circ}(\cos 5^{\circ} - \cos 55^{\circ}) - \cos 15^{\circ}(\cos 15^{\circ} - \cos 75^{\circ}) + \sin^{2} 50^{\circ} \sin^{2} 20^{\circ}

= \cos 5^{\circ}(2 \sin 30^{\circ} \sin 25^{\circ}) - \cos 15^{\circ}(2 \sin 45^{\circ} \sin 30^{\circ}) + \sin 50^{\circ} \sin 20^{\circ}

= \left( \begin{array}{c} -\;\; \frac{\cos 15^{\circ}}{\sqrt{2}} + \cos 5^{\circ} \sin 25^{\circ} \\ +\;\; \sin 50^{\circ} \sin 20^{\circ} \end{array} \right).

= \left( -\;\; \frac{\cos 15^{\circ}}{\sqrt{2}} + \;\; \frac{1}{2}(2 \cos 5^{\circ} \cos 65^{\circ}) + \;\; \frac{1}{2}(2 \sin 50^{\circ} \sin 20^{\circ}) \right)

= -\;\; \frac{\cos 15^{\circ}}{\sqrt{2}} + \;\; \frac{1}{2}[2 \cos 45^{\circ} \cos 15^{\circ}]

= -\;\; \frac{\cos 15^{\circ}}{\sqrt{2}} + \;\; \frac{\cos 15^{\circ}}{\sqrt{2}}

= 0

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