Question

How do you calculate\[\arccos \left( { - \dfrac{{\sqrt 3 }}{2}} \right)\]?

Answer 1

Hint: We use the concept that arc means inverse of the function. We are given the value of sine of an angle. Use the trigonometric identity \[{\sin ^2}x + {\cos ^2}x = 1\]to find the value of cosine of the same angle. Divide sine of the angle by cosine of the angle to calculate tangent of the angle. Use a quadrant diagram to write the value of angle for cosine.
* Arc cosine of a function is defined as the inverse cosine of the function. When \[\cos y = x\], then we can write \[\arccos x = {\cos ^{ - 1}}x = y\].
* We know the values of all trigonometric angles are positive in the first quadrant.
Values of only \[\sin \theta \]are positive in the second quadrant.
Values of only \[\tan \theta \]are positive in the third quadrant.
Values of only\[\cos \theta \] are positive in the fourth quadrant.

Complete step-by-step answer:
Let us assume the angle as ‘x’
We are given that \[\arccos \left( { - \dfrac{{\sqrt 3 }}{2}} \right)\]
Since we know that \[\arccos x = {\cos ^{ - 1}}x = y\]
Then we can write \[\arccos \left( { - \dfrac{{\sqrt 3 }}{2}} \right) = {\cos ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right) = y\]
\[ \Rightarrow {\cos ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right) = y\]
Take cosine on both sides of the equation
\[ \Rightarrow \cos \left[ {{{\cos }^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)} \right] = \cos y\]
Cancel cosine and cosine inverse on left side of the equation
\[ \Rightarrow - \dfrac{{\sqrt 3 }}{2} = \cos y\] … (1)
We know that the value of \[\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}\]
Also, we know cosine is an even function, so, \[\cos ( - x) = - \cos x\] and that cosine function is negative in the second and third quadrant.
We can write \[\cos \left( {\pi + \dfrac{\pi }{6}} \right) = \dfrac{{ - \sqrt 3 }}{2}\]and \[\cos \left( {\pi - \dfrac{\pi }{6}} \right) = \dfrac{{ - \sqrt 3 }}{2}\]
i.e. \[\cos \left( {\dfrac{{6\pi + \pi }}{6}} \right) = \dfrac{{ - \sqrt 3 }}{2}\]and \[\cos \left( {\dfrac{{6\pi - \pi }}{6}} \right) = \dfrac{{ - \sqrt 3 }}{2}\]
i.e. \[\cos \left( {\dfrac{{7\pi }}{6}} \right) = \dfrac{{ - \sqrt 3 }}{2}\]and \[\cos \left( {\dfrac{{5\pi }}{6}} \right) = \dfrac{{ - \sqrt 3 }}{2}\]
So equation (1) becomes
\[ \Rightarrow \cos \left( {\dfrac{{7\pi }}{6}} \right) = \cos y\]and \[\cos \left( {\dfrac{{5\pi }}{6}} \right) = \cos y\]
Taking inverse sine function on both sides
\[ \Rightarrow {\cos ^{ - 1}}\left[ {\cos \left( {\dfrac{{7\pi }}{6}} \right)} \right] = {\cos ^{ - 1}}\left[ {\cos y} \right]\]and \[{\cos ^{ - 1}}\left[ {\cos \left( {\dfrac{{5\pi }}{6}} \right)} \right] = {\cos ^{ - 1}}\left[ {\cos y} \right]\]
Cancel inverse of the function by the function
\[ \Rightarrow y = \dfrac{{5\pi }}{6}\]and \[y = \dfrac{{7\pi }}{6}\]

\[\therefore \] The values of \[\arccos \left( { - \dfrac{{\sqrt 3 }}{2}} \right)\]are \[\dfrac{{5\pi }}{6}\]and \[\dfrac{{7\pi }}{6}\]

Note:
Many students make mistake of calculating the angle inside cosine as positive which is wrong, the value of cosine is given negative so we will have to look at the quadrant diagram and figure out what to add or subtract the angle from. Keep in mind we can use table for trigonometric terms if we don’t remember the values at some common angles\[{0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ },{90^ \circ }\]

ANGLEFUNCTION	\[{0^ \circ }\]	\[{30^ \circ }\]	\[{45^ \circ }\]	\[{60^ \circ }\]	\[{90^ \circ }\]
Sin	0	\[\dfrac{1}{2}\]	\[\dfrac{1}{{\sqrt 2 }}\]	\[\dfrac{{\sqrt 3 }}{2}\]	1
Cos	1	\[\dfrac{{\sqrt 3 }}{2}\]	\[\dfrac{1}{{\sqrt 2 }}\]	\[\dfrac{1}{2}\]	0
Tan	0	\[\dfrac{1}{{\sqrt 3 }}\]	1	\[\sqrt 3 \]	Not defined