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Question: How do you evaluate sine, cosine and tangent of \(\dfrac{{10\pi }}{3}\) without using a calculator?...

How do you evaluate sine, cosine and tangent of 10π3\dfrac{{10\pi }}{3} without using a calculator?

Explanation

Solution

We will first write the angle in the form of nπ+θn\pi + \theta and then use the identities of sine, cosine and tangent as required to find the required values.

Complete step-by-step answer:
We have the angle 10π3\dfrac{{10\pi }}{3} given to us and we need to find the sine, cosine and tangent of it without using a calculator.
We can write the given angle 10π3\dfrac{{10\pi }}{3} as 12π−2π3\dfrac{{12\pi - 2\pi }}{3}.
Now, we will use the fact that: a+bc=ac+bc\dfrac{{a + b}}{c} = \dfrac{a}{c} + \dfrac{b}{c}.
So, we will obtain: 10π3=12π−2π3=12π3−2π3\dfrac{{10\pi }}{3} = \dfrac{{12\pi - 2\pi }}{3} = \dfrac{{12\pi }}{3} - \dfrac{{2\pi }}{3}
On simplifying it, we will then get:- 10π3=4π−2π3\dfrac{{10\pi }}{3} = 4\pi - \dfrac{{2\pi }}{3}
Now, we will find the sine, cosine and tangent of this angle.
Let us first find the sine of this angle.
⇒sin⁡(10π3)=sin⁡(4π−2π3)\Rightarrow \sin \left( {\dfrac{{10\pi }}{3}} \right) = \sin \left( {4\pi - \dfrac{{2\pi }}{3}} \right)
Now, we will use the fact that: sin⁡(4π−θ)=−sin⁡(θ)\sin \left( {4\pi - \theta } \right) = - \sin \left( \theta \right)
⇒sin⁡(10π3)=−sin⁡(2π3)\Rightarrow \sin \left( {\dfrac{{10\pi }}{3}} \right) = - \sin \left( {\dfrac{{2\pi }}{3}} \right)
Now, we can further modify it like following:-
⇒sin⁡(10π3)=−sin⁡(π−π3)\Rightarrow \sin \left( {\dfrac{{10\pi }}{3}} \right) = - \sin \left( {\pi - \dfrac{\pi }{3}} \right)
Now, we will use the fact that: sin⁡(π−θ)=sin⁡(θ)\sin \left( {\pi - \theta } \right) = \sin \left( \theta \right)
⇒sin⁡(10π3)=−sin⁡(π3)\Rightarrow \sin \left( {\dfrac{{10\pi }}{3}} \right) = - \sin \left( {\dfrac{\pi }{3}} \right)
Since, we know that sin⁡(π3)=32\sin \left( {\dfrac{\pi }{3}} \right) = \dfrac{{\sqrt 3 }}{2}
⇒sin⁡(10π3)=−32\Rightarrow \sin \left( {\dfrac{{10\pi }}{3}} \right) = - \dfrac{{\sqrt 3 }}{2}
Let us now find the cosine of this angle.
⇒cos⁡(10π3)=cos⁡(4π−2π3)\Rightarrow \cos \left( {\dfrac{{10\pi }}{3}} \right) = \cos \left( {4\pi - \dfrac{{2\pi }}{3}} \right)
Now, we will use the fact that: cos⁡(4π−θ)=cos⁡(θ)\cos \left( {4\pi - \theta } \right) = \cos \left( \theta \right)
⇒cos⁡(10π3)=cos⁡(2π3)\Rightarrow \cos \left( {\dfrac{{10\pi }}{3}} \right) = \cos \left( {\dfrac{{2\pi }}{3}} \right)
Now, we can further modify it like following:-
⇒cos⁡(10π3)=cos⁡(π−π3)\Rightarrow \cos \left( {\dfrac{{10\pi }}{3}} \right) = \cos \left( {\pi - \dfrac{\pi }{3}} \right)
Now, we will use the fact that: cos⁡(π−θ)=−cos⁡(θ)\cos \left( {\pi - \theta } \right) = - \cos \left( \theta \right)
⇒cos⁡(10π3)=−cos⁡(π3)\Rightarrow \cos \left( {\dfrac{{10\pi }}{3}} \right) = - \cos \left( {\dfrac{\pi }{3}} \right)
Since, we know that cos⁡(π3)=12\cos \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{2}
⇒sin⁡(10π3)=−12\Rightarrow \sin \left( {\dfrac{{10\pi }}{3}} \right) = - \dfrac{1}{2}
Let us now find the tangent of this angle.
⇒tan⁡(10π3)=tan⁡(4π−2π3)\Rightarrow \tan \left( {\dfrac{{10\pi }}{3}} \right) = \tan \left( {4\pi - \dfrac{{2\pi }}{3}} \right)
Now, we will use the fact that: tan⁡(4π−θ)=−tan⁡(θ)\tan \left( {4\pi - \theta } \right) = - \tan \left( \theta \right)
⇒tan⁡(10π3)=−tan⁡(2π3)\Rightarrow \tan \left( {\dfrac{{10\pi }}{3}} \right) = - \tan \left( {\dfrac{{2\pi }}{3}} \right)
Now, we can further modify it like following:-
⇒tan⁡(10π3)=−tan⁡(π−π3)\Rightarrow \tan \left( {\dfrac{{10\pi }}{3}} \right) = - \tan \left( {\pi - \dfrac{\pi }{3}} \right)
Now, we will use the fact that: tan⁡(π−θ)=−tan⁡(θ)\tan \left( {\pi - \theta } \right) = - \tan \left( \theta \right)
⇒tan⁡(10π3)=tan⁡(π3)\Rightarrow \tan \left( {\dfrac{{10\pi }}{3}} \right) = \tan \left( {\dfrac{\pi }{3}} \right)
Since, we know that tan⁡(π3)=3\tan \left( {\dfrac{\pi }{3}} \right) = \sqrt 3
⇒tan⁡(10π3)=3\Rightarrow \tan \left( {\dfrac{{10\pi }}{3}} \right) = \sqrt 3

Thus, we have the required answer.

Note:
The students must note that sine, cosine and tangent of any angle is positive or negative depending upon ADD SUGAR TO COFFEE, here the letters are bold A means All, S means sine, T means tangent and C means coffee. It suggests that all trigonometric ratios are positive in the first quadrant, sine and cosecant are positive in the second quadrant, tangent and cotangent are positive in the third quadrant and finally cosine and secant are positive in the fourth quadrant.
The students must commit to memory the following formulas:-
sin⁡(4π−θ)=−sin⁡(θ)\sin \left( {4\pi - \theta } \right) = - \sin \left( \theta \right)
sin⁡(π−θ)=sin⁡(θ)\sin \left( {\pi - \theta } \right) = \sin \left( \theta \right)
cos⁡(4π−θ)=cos⁡(θ)\cos \left( {4\pi - \theta } \right) = \cos \left( \theta \right)
cos⁡(π−θ)=−cos⁡(θ)\cos \left( {\pi - \theta } \right) = - \cos \left( \theta \right)
tan⁡(4π−θ)=−tan⁡(θ)\tan \left( {4\pi - \theta } \right) = - \tan \left( \theta \right)
tan⁡(π−θ)=−tan⁡(θ)\tan \left( {\pi - \theta } \right) = - \tan \left( \theta \right)