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 :
\[\sin {{15}^{o}}=\sin ({{45}^{o}}-{{30}^{o}})=\frac{\sqrt{3}-1}{2\sqrt{2}}=\]irrational \[\cos {{15}^{o}}=\cos ({{45}^{o}}-{{30}^{o}})=\frac{\sqrt{3}+1}{2\sqrt{2}}\]=irrational \[\therefore \,\,\,\sin {{15}^{o}}\cos {{15}^{o}}=\frac{1}{2}(2\sin {{15}^{o}}\cos {{15}^{o}})\] \[=\frac{1}{2}\sin {{30}^{o}}=\frac{1}{2}.\frac{1}{2}=\frac{1}{4}\]= rational \ \[\sin {{15}^{o}}\cos {{75}^{o}}=\sin {{15}^{o}}\sin {{15}^{o}}={{\sin }^{2}}{{15}^{o}}\] \[={{\left( \frac{\sqrt{3}-1}{2\sqrt{2}} \right)}^{2}}=\frac{4-2\sqrt{3}}{8}\]= irrationalYou need to login to perform this action.
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