Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

How do you evaluate the sine, cosine, and tangent of the angle $53$ degrees without using a calculator?

seo-qna
Last updated date: 23rd Jun 2024
Total views: 375k
Views today: 7.75k
Answer
VerifiedVerified
375k+ views
Hint: In this question we have been asked to calculate the value of three different trigonometric ratios for the same given angle. From the basic concept we know that these trigonometric ratios are related as $\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$ and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ . We will simplify and calculate the values using these values.

Complete answer:
Now considering from the question we have been asked to calculate the value of three different trigonometric ratios the sine, cosine, and tangent of the angle $53$ degrees.
From the basic concepts we know that these trigonometric ratios are related as $\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$ and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$.
We know that from the concept the expansion of sine angle is given as$\sin x=\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}}{\left( 2n+1 \right)!}{{x}^{2n+1}}=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}.........\forall x}$.
As we have been asked for the angle ${{53}^{\circ }}$ we will convert it into the radians and then substitute it in the above expansion.
From the basics of trigonometry we know that ${{1}^{\circ }}$ is equal to $\dfrac{\pi }{{{180}^{\circ }}}$ radians. So we can say that ${{53}^{\circ }}$ will be equal to $0.925$ radians.
By substituting $x=0.9$ we will get an approximate value of the sine ratio.
Now we will have $\sin {{53}^{\circ }}=\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}}{\left( 2n+1 \right)!}{{\left( 0.9 \right)}^{2n+1}}=\left( 0.9 \right)-\dfrac{{{\left( 0.9 \right)}^{3}}}{3!}+\dfrac{{{\left( 0.9 \right)}^{5}}}{5!}.........}$ .
By calculating its value we will have $\sin {{53}^{\circ }}=0.7986$ .
The cosine ratio will be given by $\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$. After substituting the value of theta we will have
$\begin{align}
  & \cos {{53}^{\circ }}=\sqrt{1-{{\left( 0.7 \right)}^{2}}} \\
 & \Rightarrow \cos {{53}^{\circ }}=\sqrt{1-0.49} \\
 & \Rightarrow \cos {{53}^{\circ }}=0.6018 \\
\end{align}$
Now we need to analyse the value of tangent ratio for ${{53}^{\circ }}$ angle. We will use the relation $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ .
By substituting the respective values we will have
$\begin{align}
  & \tan {{53}^{\circ }}=\dfrac{\sin {{53}^{\circ }}}{\cos {{53}^{\circ }}} \\
 & \Rightarrow \tan {{53}^{\circ }}=\dfrac{0.7986}{0.6018} \\
 & \Rightarrow \tan {{53}^{\circ }}=1.3270 \\
\end{align}$
These all are approximate values.

Note:
While answering we should be sure with our calculations and the transformations and basic arithmetic simplifications we make. Similar to the sine expansion we have expansions for all the trigonometric expansions like for cosine we have $\cos x=\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}}{\left( 2n \right)!}{{x}^{2n}}=1-\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}-........\forall x}$ we can also use this expansion while solving the above question.