Examveda If sin(θ + 18°) = cos60° (0° < θ < 90°), then the value of cos 5θ is? A. $$\frac{1}{2}$$C. $$\frac{1}{{\sqrt 2 }}$$D. 1Answer: Option A Solution (By Examveda Team) $$\eqalign{ & {\text{sin}}\left( {\theta + {{18}^ \circ }} \right) = \cos {60^ \circ }\left( {{0^ \circ } < \theta < {{90}^ \circ }} \right) \cr & \Rightarrow \theta + {18^ \circ } + {60^ \circ } = {90^ \circ } \cr & \Rightarrow \theta = {12^ \circ } \cr & \Rightarrow {\text{cos 5}}\theta {\text{ }} \cr & \Rightarrow {\text{cos }}{60^ \circ } \cr & \Rightarrow \frac{1}{2} \cr} $$ This Question Belongs to Arithmetic Ability >> Trigonometry
Solution (By Examveda Team) $$\eqalign{ & {\text{sin}}\left( {\theta + {{18}^ \circ }} \right) = \cos {60^ \circ }\left( {{0^ \circ } < \theta < {{90}^ \circ }} \right) \cr & \Rightarrow \theta + {18^ \circ } + {60^ \circ } = {90^ \circ } \cr & \Rightarrow \theta = {12^ \circ } \cr & \Rightarrow {\text{cos 5}}\theta {\text{ }} \cr & \Rightarrow {\text{cos }}{60^ \circ } \cr & \Rightarrow \frac{1}{2} \cr} $$
The equation $${\cos ^2}\theta $$ = $$\frac{{{{\left( {x + y} \right)}^2}}}{{4xy}}$$ is only possible when ? A. x = -yB. x > yC. x = yD. x < y View Answer
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