A) \[\frac{3}{4}\]
B) \[-\frac{3}{4}\]
C) \[\frac{1}{16}\]
D) \[\frac{1}{4}\]
Correct Answer: A
Solution :
\[\cos \left( \frac{1}{2}{{\cos }^{-1}}\frac{1}{8} \right)=x\] (let) \[\Rightarrow \]\[{{\cos }^{-1}}\frac{1}{8}=2{{\cos }^{-1}}x\] \[={{\cos }^{-1}}(2{{x}^{2}}-1)\] \[\Rightarrow \]\[\frac{1}{8}=2{{x}^{2}}-1\] \[\Rightarrow \]\[2{{x}^{2}}=\frac{1}{8}+1=\frac{9}{8}\] \[\Rightarrow \]\[{{x}^{2}}=\frac{9}{16}\] \[\Rightarrow \]\[x=\sqrt{\frac{9}{16}}=\frac{3}{4}\] \[[\because \,0\le x\le 1]\]
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