Question

Solve: \[{\sin ^2}\dfrac{{2\pi }}{3} + {\cos ^2}\dfrac{{5\pi }}{6} - {\tan ^2}\dfrac{{3\pi }}{4}\]

Answer 1

Hint: Here, we are given a trigonometric expression and we need to find the value of it. We will use \[\sin (\pi - x) = \sin x\], \[\cos (\pi - x) = - \cos x\] and \[\tan (\pi - x) = - \tan x\] . We will also the trigonometry ratios values too and so one should know those values to solve any question. Or we can substitute the value of \[\pi = {180^\circ }\] and solve this question too, get the final output. The angles are either measured in radians or degrees. The trigonometric ratios of a triangle are also called the trigonometric functions.

Complete step by step answer: Given that,
\[{\sin ^2}\dfrac{{2\pi }}{3} + {\cos ^2}\dfrac{{5\pi }}{6} - {\tan ^2}\dfrac{{3\pi }}{4}= {\sin ^2}(\pi - \dfrac{\pi }{3}) + {\cos ^2}(\pi - \dfrac{\pi }{6}) - {\tan ^2}(\pi - \dfrac{\pi }{4})\]
We know that,
\[\sin (\pi - x) = \sin x\]
\[\Rightarrow \cos (\pi - x) = - \cos x\]
\[\Rightarrow \tan (\pi - x) = - \tan x\]
Using this above trigonometry rules, we will get,
\[{\sin ^2}\dfrac{\pi }{3} + ( - {\cos ^2}\dfrac{\pi }{6}) - ( - {\tan ^2}\dfrac{\pi }{4})\]
\[\Rightarrow {(\sin \dfrac{\pi }{3})^2} + {( - \cos \dfrac{\pi }{6})^2} - {( - \tan \dfrac{\pi }{4})^2}\]

We also know that, the trigonometry values of:
\[\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}\]
\[\Rightarrow \cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}\]
\[\Rightarrow \tan \dfrac{\pi }{4} = 1\]
Substituting these values in the above expression, we will get,
\[{(\sin \dfrac{\pi }{3})^2} + {( - \cos \dfrac{\pi }{6})^2} - {( - \tan \dfrac{\pi }{4})^2} = {(\dfrac{{\sqrt 3 }}{2})^2} + {( - \dfrac{{\sqrt 3 }}{2})^2} - {( - 1)^2}\]
Removing the brackets, we get,
\[ \Rightarrow {(\sin \dfrac{\pi }{3})^2} + {( - \cos \dfrac{\pi }{6})^2} - {( - \tan \dfrac{\pi }{4})^2}= \dfrac{3}{4} + \dfrac{3}{4} - 1\]
On evaluating this, we will get,
\[ \Rightarrow{(\sin \dfrac{\pi }{3})^2} + {( - \cos \dfrac{\pi }{6})^2} - {( - \tan \dfrac{\pi }{4})^2} = \dfrac{6}{4} - 1\]
\[ \Rightarrow{(\sin \dfrac{\pi }{3})^2} + {( - \cos \dfrac{\pi }{6})^2} - {( - \tan \dfrac{\pi }{4})^2} = \dfrac{{6 - 4}}{4}\]
\[ \Rightarrow{(\sin \dfrac{\pi }{3})^2} + {( - \cos \dfrac{\pi }{6})^2} - {( - \tan \dfrac{\pi }{4})^2}= \dfrac{2}{4}\]
\[ \therefore {(\sin \dfrac{\pi }{3})^2} + {( - \cos \dfrac{\pi }{6})^2} - {( - \tan \dfrac{\pi }{4})^2} = \dfrac{1}{2}\]


Another Method:
\[{\sin ^2}\dfrac{{2\pi }}{3} + {\cos ^2}\dfrac{{5\pi }}{6} - {\tan ^2}\dfrac{{3\pi }}{4}\]
We know that, \[\pi = {180^\circ }\] and so substituting this value of \[\pi \] , we will get,
\[{\sin ^2}(\dfrac{{2 \times 180}}{3}) + {\cos ^2}(\dfrac{{5 \times 180}}{6}) - {\tan ^2}(\dfrac{{3 \times 180}}{4})\]
\[\Rightarrow {\sin ^2}{120^\circ } + {\cos ^2}{150^\circ } - {\tan ^2}{135^\circ }\]
\[\Rightarrow {\sin ^2}{(180 - 60)^\circ } + {\cos ^2}{(180 - 30)^\circ } - {\tan ^2}{(180 - 45)^\circ }\]
\[\Rightarrow {\sin ^2}{60^\circ } + ( - {\cos ^2}{30^\circ }) - ( - {\tan ^2}{45^\circ })\]
\[\Rightarrow {(\sin {60^\circ })^2} + {( - \cos {30^\circ })^2} - {( - \tan {45^\circ })^2}\]
Substituting the values of the trigonometry ratios, we will get,
\[{(\dfrac{{\sqrt 3 }}{2})^2} + {( - \dfrac{{\sqrt 3 }}{2})^2} - {( - 1)^2}=\dfrac{1}{2}\]

Hence, for the given trigonometric expression, the value of \[{\sin ^2}\dfrac{{2\pi }}{3} + {\cos ^2}\dfrac{{5\pi }}{6} - {\tan ^2}\dfrac{{3\pi }}{4} = \dfrac{1}{2}\].


Note: Here, we will study the relationship between the sides and angles of a right-angled triangle. Trigonometry is one of those divisions in mathematics that helps in finding the angles and missing sides of a triangle with the help of trigonometric ratios. The trigonometric ratios such as sine, cosine and tangent of these angles are easy to memorize.To find these angels we have to draw a right-angled triangle, in which one of the acute angles will be the corresponding trigonometry angle.