A) \[\sin \,15{}^\circ \]
B) \[\cos 15{}^\circ \]
C) \[sin\,15{}^\circ \,\cos 15{}^\circ \]
D) \[sin\,15{}^\circ \,\cos 75{}^\circ \]
Correct Answer: C
Solution :
we know that sin \[15{}^\circ =\frac{\sqrt{3-1}}{2\sqrt{2}}\](irrational)\[\cos \,15{}^\circ =\frac{\sqrt{3+1}}{2\sqrt{2}}\](irrational)\[\begin{align} & \sin \,15{}^\circ \,\cos 15{}^\circ =\frac{1}{2}(2\sin 15{}^\circ cos15{}^\circ ) \\ & =\frac{1}{2}\sin 30{}^\circ =\frac{1}{4}(rational) \\ \end{align}\] |
\[\begin{align} & \sin 15{}^\circ \cos 75{}^\circ =\sin {}^\circ 15{}^\circ \cos (90-15{}^\circ ) \\ & \\ \end{align}\]\[=\sin 15{}^\circ \sin \,15{}^\circ ={{\sin }^{2}}15{}^\circ \] |
\[=\frac{1}{2}(1+\cos 30{}^\circ )\] |
\[=\frac{1}{2}\left( 1-\frac{\sqrt{3}}{2} \right)\]Irrational |
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