Question

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

Answer 1

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.