How do you find the other five trigonometric functions of x if $\cos x = \dfrac{3}{5}$.
Answer
629.4k+ views
Hint:n order to determine exact values of all six trigonometric function of the angle whose cosine is given to us in the above question, first assume the a right angled triangle and express all the trigonometric ratios with respect to and in terms of the sides of that triangle and then find all the trigonometric ratios knowing the lengths of hypotenuse, altitude and base.
Complete step by step solution:
We are given a point $\cos x = \dfrac{3}{5}$. Now we assume a right angled triangle ABC and let $\angle BAC = x$
Now, $\angle ACB = {90^ \circ }$.
So, $\cos x = \dfrac{{Base}}{{{\text{Hypotenuse}}}} = \dfrac{3}{5}$.
So, we know the ratio of Base and Hypotenuse. Let ${\text{Base = 3x}}$ and ${\text{Hypotenuse = 5x}}$.
Now, calculating opposite side of the triangle using Pythagoras theorem,
${\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Altitude} \right)^2}$
$ \Rightarrow {\left( {5x} \right)^2} = {\left( {3x} \right)^2} + {\left( {Altitude} \right)^2}$
$ \Rightarrow {\left( {Altitude} \right)^2} = 25{x^2} - 9{x^2}$
$ \Rightarrow {\left( {Altitude} \right)^2} = 16{x^2}$
$ \Rightarrow Altitude = 4x$
Therefore Calculating all the trigonometric ratios as:
\[\sin x = \dfrac{{{\text{Altitude}}}}{{{\text{Hypotenuse}}}} = \dfrac{4}{5}\]
\[\tan x = \dfrac{{{\text{Altitude}}}}{{{\text{Base}}}} = \dfrac{4}{3}\]
\[\cot x = \dfrac{{{\text{Base}}}}{{{\text{Altitude}}}} = \dfrac{3}{4}\]
\[\cos ecx = \dfrac{{{\text{Hypotenuse}}}}{{{\text{Altitude}}}} = \dfrac{5}{4}\]
\[\sec x = \dfrac{{{\text{Hypotenuse}}}}{{{\text{Base}}}} = \dfrac{5}{3}\]
So, these are the values of trigonometric ratios except cosine which was given beforehand in the question itself.
Note: Trigonometry is one of the significant branches throughout the entire existence of mathematics and has wide ranging applications in various fields of mathematics such as Geometry, Algebra and Calculus. One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer. Trigonometric ratios are the ratios of the sides of a triangle and thus the trigonometric ratios can be found by expressing the ratios in the terms of the sides of a triangle.
Complete step by step solution:
We are given a point $\cos x = \dfrac{3}{5}$. Now we assume a right angled triangle ABC and let $\angle BAC = x$
Now, $\angle ACB = {90^ \circ }$.
So, $\cos x = \dfrac{{Base}}{{{\text{Hypotenuse}}}} = \dfrac{3}{5}$.
So, we know the ratio of Base and Hypotenuse. Let ${\text{Base = 3x}}$ and ${\text{Hypotenuse = 5x}}$.
Now, calculating opposite side of the triangle using Pythagoras theorem,
${\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Altitude} \right)^2}$
$ \Rightarrow {\left( {5x} \right)^2} = {\left( {3x} \right)^2} + {\left( {Altitude} \right)^2}$
$ \Rightarrow {\left( {Altitude} \right)^2} = 25{x^2} - 9{x^2}$
$ \Rightarrow {\left( {Altitude} \right)^2} = 16{x^2}$
$ \Rightarrow Altitude = 4x$
Therefore Calculating all the trigonometric ratios as:
\[\sin x = \dfrac{{{\text{Altitude}}}}{{{\text{Hypotenuse}}}} = \dfrac{4}{5}\]
\[\tan x = \dfrac{{{\text{Altitude}}}}{{{\text{Base}}}} = \dfrac{4}{3}\]
\[\cot x = \dfrac{{{\text{Base}}}}{{{\text{Altitude}}}} = \dfrac{3}{4}\]
\[\cos ecx = \dfrac{{{\text{Hypotenuse}}}}{{{\text{Altitude}}}} = \dfrac{5}{4}\]
\[\sec x = \dfrac{{{\text{Hypotenuse}}}}{{{\text{Base}}}} = \dfrac{5}{3}\]
So, these are the values of trigonometric ratios except cosine which was given beforehand in the question itself.
Note: Trigonometry is one of the significant branches throughout the entire existence of mathematics and has wide ranging applications in various fields of mathematics such as Geometry, Algebra and Calculus. One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer. Trigonometric ratios are the ratios of the sides of a triangle and thus the trigonometric ratios can be found by expressing the ratios in the terms of the sides of a triangle.
Recently Updated Pages
Which will be the least stable resonating structure class 11 chemistry CBSE

How many 5 digit telephone numbers can be construc-class-11-maths-CBSE

How do you find the angle of the resultant vector class 11 physics CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

What is the need and importance of classification class 11 biology CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Potato is a stem and sweet potato is a root Justify class 11 biology CBSE

Draw a well labelled diagram of Euglena class 11 biology CBSE

What is the maximum height of a projectile class 11 physics CBSE

