Question

cos2π/3+sin5π/6 evalute the following​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider,

$\rm :\longmapsto\:cos\bigg(\dfrac{2\pi}{3} \bigg)$

$\: \: \rm= \: \: \:cos\bigg(\pi - \dfrac{\pi}{3} \bigg)$

$\: \: \rm= \: \: - \:cos\bigg( \dfrac{\pi}{3} \bigg)$

$\: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: cos(\pi - x) = - \: cosx\bigg \}}$

$\: \: \rm= \: \: - \: \dfrac{1}{2}$

$\bf :\longmapsto\:cos\bigg(\dfrac{2\pi}{3} \bigg) = - \: \dfrac{1}{2}$

Consider,

$\rm :\longmapsto\:sin\bigg(\dfrac{5\pi}{6} \bigg)$

$\: \: \rm= \: \:sin\bigg(\pi - \dfrac{\pi}{6} \bigg)$

$\: \: \rm= \: \:sin\bigg( \dfrac{\pi}{6} \bigg)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \:sin(\pi - x) = sinx \bigg \}}$

$\: \: \rm= \: \:\dfrac{1}{2}$

$\bf :\longmapsto\:sin\bigg(\dfrac{5\pi}{6} \bigg) = \dfrac{1}{2}$

Hence,

$\rm :\longmapsto\:cos\bigg(\dfrac{2\pi}{3} \bigg) + sin\bigg(\dfrac{5\pi}{6} \bigg)$

$\: \: \rm= \: \:\dfrac{1}{2} - \dfrac{1}{2}$

$\: \: \rm= \: \:0$

$\bf :\implies\:cos\bigg(\dfrac{2\pi}{3} \bigg) + sin\bigg(\dfrac{5\pi}{6} \bigg) = 0$

Additional Information :-

Sign of Trigonometric ratios in Quadrants

sin (90°-θ)  =  cos θ

cos (90°-θ)  =  sin θ

tan (90°-θ)  =  cot θ

csc (90°-θ)  =  sec θ

sec (90°-θ)  =  csc θ

cot (90°-θ)  =  tan θ

sin (90°+θ)  =  cos θ

cos (90°+θ)  =  -sin θ

tan (90°+θ)  =  -cot θ

csc (90°+θ)  =  sec θ

sec (90°+θ)  =  -csc θ

cot (90°+θ)  =  -tan θ

sin (180°-θ)  =  sin θ

cos (180°-θ)  =  -cos θ

tan (180°-θ)  =  -tan θ

csc (180°-θ)  =  csc θ

sec (180°-θ)  =  -sec θ

cot (180°-θ)  =  -cot θ

sin (180°+θ)  =  -sin θ

cos (180°+θ)  =  -cos θ

tan (180°+θ)  =  tan θ

csc (180°+θ)  =  -csc θ

sec (180°+θ)  =  -sec θ

cot (180°+θ)  =  cot θ

sin (270°-θ)  =  -cos θ

cos (270°-θ)  =  -sin θ

tan (270°-θ)  =  cot θ

csc (270°-θ)  =  -sec θ

sec (270°-θ)  =  -csc θ

cot (270°-θ)  =  tan θ

sin (270°+θ)  =  -cos θ

cos (270°+θ)  =  sin θ

tan (270°+θ)  =  -cot θ

csc (270°+θ)  =  -sec θ

sec (270°+θ)  =  cos θ

cot (270°+θ)  =  -tan θ

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$