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# How do you evaluate sine, cosine, tangent of $\dfrac{{10\pi }}{3}$ without using a calculator?

Last updated date: 29th Feb 2024
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Hint: The sine of the point is the proportion of the length of the side inverse the point partitioned by the length of the hypotenuse. The cosine of the point is the proportion of the length of the side near the point partitioned by the length of the hypotenuse. The digression of the point is the proportion of the length of the side inverse the point isolated by the length of side adjoining the point.

You consider that,
$\dfrac{{10\pi }}{3} = 2\pi + \pi + \dfrac{\pi }{3}$, so that
$\Rightarrow$sin$(\dfrac{{10\pi }}{3})$$= sin(2\pi + \pi + \dfrac{\pi }{3}) = sin(\pi + \dfrac{\pi }{3}) \Rightarrow sin\pi cos(\dfrac{\pi }{3})$$ +$cossin$(\dfrac{\pi }{3})$
$\Rightarrow -$sin$\dfrac{\pi }{3}$$= - \dfrac{{\sqrt 3 }}{2} \Rightarrow Cos(\dfrac{{10\pi }}{3}) = cos(2\pi + \pi + \dfrac{\pi }{3}) = cos(\pi + \dfrac{\pi }{3}) \Rightarrow cos\pi cos(\dfrac{\pi }{3})$$ -$sin$\pi$sin$(\dfrac{\pi }{3})$$=$$ -$cos$(\dfrac{\pi }{3})$
$\Rightarrow - \dfrac{1}{2}$
$\Rightarrow$Tan$(\dfrac{{10\pi }}{3}) = \dfrac{{\sin (\dfrac{{10\pi }}{3})}}{{\cos (\dfrac{{10\pi }}{3})}} = \dfrac{{ - \dfrac{{\sqrt 3 }}{2}}}{{ - \dfrac{1}{2}}} = \sqrt 3$

Note: In arithmetic, geometrical capacities called round capacities, point capacities or goniometric capacities are genuine capacities which relate the point of a correct point triangle to proportions of two side lengths.