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Question: How do you evaluate the sine, cosine, and tangent of the angle \(53\) degrees without using a calcul...

How do you evaluate the sine, cosine, and tangent of the angle 5353 degrees without using a calculator?

Explanation

Solution

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 cosθ=1sin2θ\cos \theta =\sqrt{1-{{\sin }^{2}}\theta } and tanθ=sinθcosθ\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 5353 degrees.
From the basic concepts we know that these trigonometric ratios are related as cosθ=1sin2θ\cos \theta =\sqrt{1-{{\sin }^{2}}\theta } and tanθ=sinθcosθ\tan \theta =\dfrac{\sin \theta }{\cos \theta }.
We know that from the concept the expansion of sine angle is given assinx=n=0(1)n(2n+1)!x2n+1=xx33!+x55!.........x\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{{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{{1}^{\circ }} is equal to π180\dfrac{\pi }{{{180}^{\circ }}} radians. So we can say that 53{{53}^{\circ }} will be equal to 0.9250.925 radians.
By substituting x=0.9x=0.9 we will get an approximate value of the sine ratio.
Now we will have sin53=n=0(1)n(2n+1)!(0.9)2n+1=(0.9)(0.9)33!+(0.9)55!.........\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 sin53=0.7986\sin {{53}^{\circ }}=0.7986 .
The cosine ratio will be given by cosθ=1sin2θ\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }. After substituting the value of theta we will have
cos53=1(0.7)2 cos53=10.49 cos53=0.6018 \begin{aligned} & \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{aligned}
Now we need to analyse the value of tangent ratio for 53{{53}^{\circ }} angle. We will use the relation tanθ=sinθcosθ\tan \theta =\dfrac{\sin \theta }{\cos \theta } .
By substituting the respective values we will have
tan53=sin53cos53 tan53=0.79860.6018 tan53=1.3270 \begin{aligned} & \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{aligned}
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 cosx=n=0(1)n(2n)!x2n=1x22!+x44!........x\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.