A) \[2x\]
B) \[x\]
C) \[{{x}^{2}}\]
D) \[-x\]
Correct Answer: D
Solution :
As given that\[\sin {{52}^{o}}=\cos 2\alpha =x\] \[\therefore \]\[\cos ({{90}^{o}}-{{52}^{o}})=\cos 2\alpha =x\] \[\Rightarrow \] \[\cos {{38}^{o}}=\cos 2\alpha =x\] \[\Rightarrow \] \[2\alpha ={{38}^{o}}\Rightarrow \alpha =\frac{38}{2}={{19}^{o}}\] \[\therefore \] \[\beta ={{180}^{o}}-({{52}^{o}}+{{19}^{o}})\] \[\beta ={{109}^{o}}\] \[\therefore \] \[\cos 2\beta =\cos 2\times ({{109}^{o}})\] \[=\cos {{218}^{o}}=\cos (180+{{38}^{o}})\] \[=-\cos {{38}^{o}}\] \[\Rightarrow \] \[\cos 2\beta =-x\]
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