Question

Using tables evaluate the following:
$\left( i \right){\text{ 4cot4}}{{\text{5}}^ \circ } - {\sec ^2}{60^ \circ } + \sin {30^ \circ }$
$\left( {ii} \right){\text{ co}}{{\text{s}}^2}0 + {\cos ^2}\dfrac{\pi }{6} + {\cos ^2}\dfrac{\pi }{3} + {\cos ^2}\dfrac{\pi }{2}$

Answer 1

Hint:
Here in this question we have to find the values of the given trigonometric function. So here the table has been attached to it. By using the table we can easily solve it. So we just need to add or subtract according to the question.

$\text{Angle} \implies$	$0^{\circ}$	${30}^{\circ}$	${45}^{\circ}$	${60}^{\circ}$	${90}^{\circ}$
$\sin \theta$	$0$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{1}{2}$	$1$
$\cos \theta$	$1$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{1}{2}$	$0$
$\tan \theta$	$0$	$\dfrac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$	$\text{Not Defined}$
$\cos ec \theta$	$\text{Not Defined}$	$2$	$\sqrt{2}$	$\dfrac{2}{\sqrt{3}}$	$1$
$\sec \theta$	$1$	$\dfrac{2}{\sqrt{3}}$	$\sqrt{2}$	$2$	$\text{Not Defined}$
\cot \theta	$\text{Not Defined}$	$\sqrt{3}$	$1$	$\dfrac{1}{\sqrt{3}}$	$0$

Complete step by step solution:
So here we have the question given as
$\left( i \right){\text{ 4cot4}}{{\text{5}}^ \circ } - {\sec ^2}{60^ \circ } + \sin {30^ \circ }$
Now by using the table the values of the above functions are
$\cot {45^ \circ } = 1,{\text{ sec6}}{{\text{0}}^ \circ } = 2,{\text{ sin3}}{{\text{0}}^ \circ } = \dfrac{1}{2}$
So now by using the above values and putting it into the question, we get
$ \Rightarrow 4 \times 1 - {\left( 2 \right)^2} + \dfrac{1}{2}$
Therefore, on solving the above we get
$ \Rightarrow 4 - 4 + \dfrac{1}{2}$
Now solving the addition and subtraction part of the equation we get
$ \Rightarrow \dfrac{1}{2}$
Therefore, on evaluating${\text{4cot4}}{{\text{5}}^ \circ } - {\sec ^2}{60^ \circ } + \sin {30^ \circ }$, we get$\dfrac{1}{2}$.
$\left( {ii} \right){\text{ co}}{{\text{s}}^2}0 + {\cos ^2}\dfrac{\pi }{6} + {\cos ^2}\dfrac{\pi }{3} + {\cos ^2}\dfrac{\pi }{2}$
Now by using the table the values of the above functions are
$\cos {0^ \circ } = 1,{\text{ }}\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2},{\text{ }}\cos {60^ \circ } = \dfrac{1}{2},{\text{ }}\cos {90^ \circ } = 1$
So now by using the above values and putting it into the question, we get
$ \Rightarrow {\left( 1 \right)^2} + {\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2} + {\left( {\dfrac{1}{2}} \right)^2} + {\left( 1 \right)^2}$
Now on solving the squaring, we get
$ \Rightarrow 1 + \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{2} + 1$
Now on adding the above equation, we get
$ \Rightarrow \dfrac{{2 + \sqrt 3 + 1 + 2}}{2}$
On solving the numerator part, we get
$ \Rightarrow \dfrac{{5 + \sqrt 3 }}{2}$

Therefore, on evaluating ${\text{co}}{{\text{s}}^2}0 + {\cos ^2}\dfrac{\pi }{6} + {\cos ^2}\dfrac{\pi }{3} + {\cos ^2}\dfrac{\pi }{2}$, we get $\dfrac{{5 + \sqrt 3 }}{2}$.

Additional information:
If we only concentrate on the sin/cos graph between$0{\text{ and 9}}{{\text{0}}^ \circ }$, the graph almost looks like a straight line. One can exploit this property to predict a very close approximate for any sine and cos value between$0{\text{ and 9}}{{\text{0}}^ \circ }$.
This procedure may appear to be somewhat befuddling from the start however when perceived, there's no snappier method to get all the mathematical proportions. Initially, for this technique, you have to know the fundamentals of Trigonometry.

Note:
As we have seen some tricks to learn the values or we can also memorize the values. Without the values, we cannot solve it. It will help everywhere once we learn it. If we don't have the foggiest idea about the essentials, at that point read on, because these rudiments help in recalling the proportions as well as go far in taking care of geometry issues-particularly in tallness and separations issues.