How do you find the value of $ \cos {300^ \circ } $ ?
Answer
523.2k+ views
Hint: In the given question, we are required to find the value of $ \cos {300^ \circ } $ . We will use the trigonometric formulae and identities such as $ \cos \left( {{{360}^ \circ } - \theta } \right) = \cos \theta $ to find the value of the trigonometric function at the particular angle. We should be clear with the signs of all trigonometric functions in the four quadrants.
Complete step-by-step answer:
So, we have to find the value of the trigonometric function $ \cos {300^ \circ } $ .
Now, we know that the trigonometric function cosine is positive in the fourth quadrant. Also, the value of the cosine function gets repeated after a regular interval of $ 2\pi $ radians.
We know that the angle $ {300^ \circ } $ lies in the fourth quadrant. So, the cosine of the angle will be a positive value.
So, we have, $ \cos {300^ \circ } $
$ \Rightarrow \cos {300^ \circ } = \cos \left( {{{360}^ \circ } - {{60}^ \circ }} \right) $
Now, we know that the value of $ \cos \left( {{{360}^ \circ } - \theta } \right) $ is equal to $ \cos \theta $ . So, we get,
$ \Rightarrow \cos {300^ \circ } = \cos \left( {{{60}^ \circ }} \right) $
Now, we know that cosine and sine are complementary ratios of each other. So, we have, $ \sin \left( {{{90}^ \circ } - \theta } \right) = \cos \theta $ . Hence, we get,
$ \Rightarrow \cos {300^ \circ } = \sin \left( {{{90}^ \circ } - {{60}^ \circ }} \right) $
Simplifying the expression,
$ \Rightarrow \cos {300^ \circ } = \sin \left( {{{30}^ \circ }} \right) $
Now, we also know that the value of $ \sin \left( {{{30}^ \circ }} \right) $ is $ \left( {\dfrac{1}{2}} \right) $ .
Hence, we get,
$ \Rightarrow \cos {300^ \circ } = \dfrac{1}{2} $
Therefore, the value of $ \cos {300^ \circ } $ is $ \left( {\dfrac{1}{2}} \right) $ .
So, the correct answer is “$ \left( {\dfrac{1}{2}} \right) $ ”.
Note: We can also solve the given problem using the periodicity of the cosine and sine functions. Periodic Function is a function that repeats its value after a certain interval. For a real number $ T > 0 $ , $ f\left( {x + T} \right) = f\left( x \right) $ for all x. If T is the smallest positive real number such that $ f\left( {x + T} \right) = f\left( x \right) $ for all x, then T is called the fundamental period. The fundamental period of cosine is $ 2\pi $ radians. So, $ \cos \left( \theta \right) = \cos \left( {\theta - 2\pi } \right) $ . Hence, we get, $ \cos {300^ \circ } = \cos \left( {{{300}^ \circ } - {{360}^ \circ }} \right) = \cos \left( { - {{60}^ \circ }} \right) = \dfrac{1}{2} $ as we know that cosine is also positive in fourth quadrant of Cartesian plane.
Complete step-by-step answer:
So, we have to find the value of the trigonometric function $ \cos {300^ \circ } $ .
Now, we know that the trigonometric function cosine is positive in the fourth quadrant. Also, the value of the cosine function gets repeated after a regular interval of $ 2\pi $ radians.
We know that the angle $ {300^ \circ } $ lies in the fourth quadrant. So, the cosine of the angle will be a positive value.
So, we have, $ \cos {300^ \circ } $
$ \Rightarrow \cos {300^ \circ } = \cos \left( {{{360}^ \circ } - {{60}^ \circ }} \right) $
Now, we know that the value of $ \cos \left( {{{360}^ \circ } - \theta } \right) $ is equal to $ \cos \theta $ . So, we get,
$ \Rightarrow \cos {300^ \circ } = \cos \left( {{{60}^ \circ }} \right) $
Now, we know that cosine and sine are complementary ratios of each other. So, we have, $ \sin \left( {{{90}^ \circ } - \theta } \right) = \cos \theta $ . Hence, we get,
$ \Rightarrow \cos {300^ \circ } = \sin \left( {{{90}^ \circ } - {{60}^ \circ }} \right) $
Simplifying the expression,
$ \Rightarrow \cos {300^ \circ } = \sin \left( {{{30}^ \circ }} \right) $
Now, we also know that the value of $ \sin \left( {{{30}^ \circ }} \right) $ is $ \left( {\dfrac{1}{2}} \right) $ .
Hence, we get,
$ \Rightarrow \cos {300^ \circ } = \dfrac{1}{2} $
Therefore, the value of $ \cos {300^ \circ } $ is $ \left( {\dfrac{1}{2}} \right) $ .
So, the correct answer is “$ \left( {\dfrac{1}{2}} \right) $ ”.
Note: We can also solve the given problem using the periodicity of the cosine and sine functions. Periodic Function is a function that repeats its value after a certain interval. For a real number $ T > 0 $ , $ f\left( {x + T} \right) = f\left( x \right) $ for all x. If T is the smallest positive real number such that $ f\left( {x + T} \right) = f\left( x \right) $ for all x, then T is called the fundamental period. The fundamental period of cosine is $ 2\pi $ radians. So, $ \cos \left( \theta \right) = \cos \left( {\theta - 2\pi } \right) $ . Hence, we get, $ \cos {300^ \circ } = \cos \left( {{{300}^ \circ } - {{360}^ \circ }} \right) = \cos \left( { - {{60}^ \circ }} \right) = \dfrac{1}{2} $ as we know that cosine is also positive in fourth quadrant of Cartesian plane.
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