Question

Find the value of $\cos \left( -{{1710}^{\circ }} \right)$.

Answer 1

In this question, we need to find the value of $\cos \left( -{{1710}^{\circ }} \right)$. For this, we will first use the property of $\cos \left( -\theta \right)=\cos \theta$ to ease our calculation. After that, we will change the given angle from degrees to radian using the formula ${{1}^{\circ }}=\dfrac{\pi }{180}\text{radians}$. After that, we will try to find the required value using formula:
(I) cosine repeats its function after an interval of $2\pi$.
(II) $\cos {{90}^{\circ }}=0$ from the trigonometric ratio table.

Complete step-by-step solution:
Here we are given the function as $\cos \left( -{{1710}^{\circ }} \right)$. We need to find its value. For this, let's first simplify the angle of cosine function. We know that, $\cos \left( -\theta \right)=\cos \theta$ so our expression becomes $\cos \left( -{{1710}^{\circ }} \right)=\cos \left( {{1710}^{\circ }} \right)$.
We are given the angle in degrees, to easily calculate the values let us first change the angle from degrees to radian. Formula for changing degrees to radian is given by,
${{1}^{\circ }}=\dfrac{\pi }{180}\text{radians}$.
Therefore, ${{1710}^{\circ }}=\dfrac{1710\pi }{180}\text{radians}\Rightarrow {{1710}^{\circ }}=\dfrac{171\pi }{18}\text{radians}$.
Dividing the numerator and the denominator by 9 we get, ${{1710}^{\circ }}=\dfrac{19\pi }{2}\text{radians}$.
Hence our expression becomes $\cos \left( -{{1710}^{\circ }} \right)=\cos \left( \dfrac{19\pi }{2} \right)$.
Now we know that $\dfrac{19\pi }{2}$ can be written as $10\pi -\dfrac{\pi }{2}$ (Because $10\pi -\dfrac{\pi }{2}=\dfrac{20\pi -\pi }{2}=\dfrac{19\pi }{2}$)
So our expression becomes $\cos \left( -{{1710}^{\circ }} \right)=\cos \left( 10\pi -\dfrac{\pi }{2} \right)$.
As we know that cosine function repeats itself after every $2\pi$ which means $\cos \theta =\cos \left( 2\pi +\theta \right)=\cos \left( 4\pi +\theta \right)=\cdots \cdots \cdots =\cos \left( 2n+\theta \right)$ where n is any integer. So, we can apply this here as $10\pi =5\times 2\pi$.
Hence we can say $\cos \left( -{{1710}^{\circ }} \right)=\cos \left( -\dfrac{\pi }{2} \right)$.
Now let us apply $\cos \left( -\theta \right)=\cos \theta$ again we get, $\cos \left( -{{1710}^{\circ }} \right)=\cos \dfrac{\pi }{2}$.
From the trigonometric ratio table we know that $\cos \dfrac{\pi }{2}={{0}^{\circ }}$ therefore we get $\cos \left( -{{1710}^{\circ }} \right)=0$.
Hence the required value of $\cos \left( -{{1710}^{\circ }} \right)$ is 0.

Note: Students should keep in mind all the trigonometric properties for solving this type of sums. They should note that, not every trigonometric function absorbs the negative sign as cosine does. Try to convert degrees to radian to calculate the angle easily. Keep in mind the trigonometric ratio table for easily finding values of such function.