A) \[sin\text{ }47{}^\circ \]
B) \[\text{cos }47{}^\circ \]
C) \[\text{2}\,\sin 47{}^\circ \]
D) \[\text{2}\,\cos 47{}^\circ \]
Correct Answer: B
Solution :
\[{{\sin }^{2}}66\frac{1{}^\circ }{2}-{{\sin }^{2}}23\frac{1{}^\circ }{2}\] \[={{\left[ \sin \left( 90{}^\circ -23\frac{1{}^\circ }{2} \right) \right]}^{2}}-{{\sin }^{2}}23\frac{1{}^\circ }{2}\] \[={{\cos }^{2}}23\frac{1{}^\circ }{2}-{{\sin }^{2}}23\frac{1{}^\circ }{2}\] \[=\cos 2\left( 23\frac{1{}^\circ }{2} \right)=\cos 47{}^\circ \] \[(\because \,\cos 2A=co{{s}^{2}}A-{{\sin }^{2}}A)\] \[=\cos \left[ 2\times \left( \frac{47}{2} \right) \right]=\cos 47{}^\circ \]
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