Question

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

Answer 1

Hint: Use the property of the cosine function which is $\cos \left( -x \right)=\cos x$ to simplify the given expression. Write the angle ${{1710}^{\circ }}$ in the range of ${{0}^{\circ }}-{{90}^{\circ }}$ by dividing it by ${{180}^{\circ }}$. Use the property of cosine function which is $\cos \left( \left( 2n+1 \right)\pi \pm x \right)=-\cos x$ and $\cos \left( 2n\pi \pm x \right)=\cos x$ to find the value of given angle.

Complete step-by-step answer:
We have to find the value of $\cos \left( -{{1710}^{\circ }} \right)$.
We will use properties of cosine function to evaluate the value of a given angle.
We know that $\cos \left( -x \right)=\cos x$.
Substituting $x={{1710}^{\circ }}$ in the above equation, we have $\cos \left( -{{1710}^{\circ }} \right)=\cos \left( {{1710}^{\circ }} \right)$.
We will now write ${{1710}^{\circ }}$ in the range of ${{0}^{\circ }}-{{90}^{\circ }}$ by dividing it by ${{180}^{\circ }}$.
Thus, we have ${{1710}^{\circ }}=9\times {{180}^{\circ }}+{{90}^{\circ }}$.
So, we have $\cos \left( {{1710}^{\circ }} \right)=\cos \left( 9\times {{180}^{\circ }}+{{90}^{\circ }} \right)$.
We know that $\cos \left( \left( 2n+1 \right)\pi \pm x \right)=-\cos x$.
Thus, we have $\cos \left( -{{1710}^{\circ }} \right)=\cos \left( {{1710}^{\circ }} \right)=\cos \left( 9\times {{180}^{\circ }}+{{90}^{\circ }} \right)=-\cos \left( {{90}^{\circ }} \right)$.
We know that $\cos \left( {{90}^{\circ }} \right)=0$.
Thus, we have $\cos \left( -{{1710}^{\circ }} \right)=\cos \left( {{1710}^{\circ }} \right)=\cos \left( 9\times {{180}^{\circ }}+{{90}^{\circ }} \right)=-\cos \left( {{90}^{\circ }} \right)=0$.
Hence, the value of $\cos \left( -{{1710}^{\circ }} \right)$ is 0.
Trigonometric functions are real functions which relate any angle of a right angled triangle to the ratios of any two of its sides. The widely used trigonometric functions are sine, cosine and tangent. However, we can also use their reciprocals, i.e., cosecant, secant and cotangent. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius 1). We also write these trigonometric functions as infinite series or as solutions to differential equations. Thus, allowing us to expand the domain of these functions from the real line to the complex plane. One should be careful while using the trigonometric identities and rearranging the terms to convert from one trigonometric function to the other one.

Note: We can also solve this question by dividing ${{1710}^{\circ }}$ by ${{90}^{\circ }}$ and observing that ${{1710}^{\circ }}$ is a multiple of ${{90}^{\circ }}$. Then use the fact that $\cos \left( \left( \dfrac{2n+1}{2} \right)\pi \right)=0$ for all values of ‘n’. If we don’t use the fact that $\cos \left( -x \right)=\cos x$, we will have to write $-{{1710}^{\circ }}$ as a multiple of ${{90}^{\circ }}$.