Question

. Evaluate : cos60˚cos30˚- sin60˚sin30˚

Answer 1

$\bf{We \: know \: that:}\\\\\star \: \sf{cos \: 60^{\circ} = \dfrac{1}{2}}\\\\\\\star \: \sf{cos \: 30^{\circ} = \dfrac{\sqrt{3}}{2}}\\\\\\\star \: \sf{sin \: 60^{\circ} = \dfrac{\sqrt{3}}{2}}\\\\\\\star \: \sf{sin \: 30^{\circ} = \dfrac{1}{2}}$

Substitute the above values in the given question.

$\sf{cos \: 60^{\circ}cos30^{\circ}- sin60^{\circ}sin30^{\circ}}\\\\\\\Rightarrow \: \sf{\bigg(\dfrac{1}{2} \times \dfrac{\sqrt{3}}{2}\bigg) - \bigg(\dfrac{\sqrt{3}}{2} \times \dfrac{1}{2}\bigg)}\\\\\\\\\Rightarrow \: \sf{\dfrac{\sqrt{3}}{4} - \dfrac{\sqrt{3}}{4}}\\\\\\\Rightarrow \sf{0}\\\\\\\therefore \large{\boxed{\boxed{\bf{cos \: 60^{\circ}cos30^{\circ}- sin60^{\circ}sin30^{\circ} = 0}}}}$

Know more:

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$