
How do I find the value of \[\cos 330{}^\circ \] ?
Answer
544.8k+ views
Hint:Since we have remembered the value of the trigonometric ratios only in the first quadrant and also know that these functions are periodic that means repeat its value after a particular interval so to make the question simple just split the given angle in terms of \[360{}^\circ \] that is \[330{}^\circ =360{}^\circ -30{}^\circ \] and we also know that \[\cos x\] is a positive fourth quadrant. The final value will be \[\cos 30{}^\circ \].
Complete step by step solution:
Since the given angle is in fourth quadrant, but we know its value only in first quadrant so we will split such that:
\[330{}^\circ =360{}^\circ -30{}^\circ \]
\[\Rightarrow \cos 330{}^\circ =\cos (360{}^\circ -30{}^\circ )\]
And we know that \[\cos x\] function is periodic in nature that means it repeats its value after an interval of \[360{}^\circ \]
\[\begin{align}
& \Rightarrow \cos (360{}^\circ -\theta )=\cos \theta \\
& \cos (360{}^\circ +\theta )=\cos \theta \\
\end{align}\]
Therefore,
\[\Rightarrow \cos 330{}^\circ =\cos (360{}^\circ -30{}^\circ )=\cos 30{}^\circ \]
Since we know that \[\cos 30{}^\circ ={}^{\sqrt{3}}/{}_{2}\].
Hence the value of \[\cos 330{}^\circ \] is \[{}^{\sqrt{3}}/{}_{2}\].
Note: During solving this type of question just remember the periodicity of the basic trigonometric functions and also note that \[\cos x\] is positive in the fourth quadrant. We can also split in terms of \[270{}^\circ \] but note that in this case the output value changes to \[\sin \theta \] but the cos function is positive.
Complete step by step solution:
Since the given angle is in fourth quadrant, but we know its value only in first quadrant so we will split such that:
\[330{}^\circ =360{}^\circ -30{}^\circ \]
\[\Rightarrow \cos 330{}^\circ =\cos (360{}^\circ -30{}^\circ )\]
And we know that \[\cos x\] function is periodic in nature that means it repeats its value after an interval of \[360{}^\circ \]
\[\begin{align}
& \Rightarrow \cos (360{}^\circ -\theta )=\cos \theta \\
& \cos (360{}^\circ +\theta )=\cos \theta \\
\end{align}\]
Therefore,
\[\Rightarrow \cos 330{}^\circ =\cos (360{}^\circ -30{}^\circ )=\cos 30{}^\circ \]
Since we know that \[\cos 30{}^\circ ={}^{\sqrt{3}}/{}_{2}\].
Hence the value of \[\cos 330{}^\circ \] is \[{}^{\sqrt{3}}/{}_{2}\].
Note: During solving this type of question just remember the periodicity of the basic trigonometric functions and also note that \[\cos x\] is positive in the fourth quadrant. We can also split in terms of \[270{}^\circ \] but note that in this case the output value changes to \[\sin \theta \] but the cos function is positive.
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