How do you find the other five trigonometric functions of x if $\cos x = \dfrac{3}{5}$.
Answer
Verified
444k+ 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
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
Find the value of x if the mode of the following data class 11 maths CBSE
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE
Trending doubts
10 examples of friction in our daily life
The correct order of melting point of 14th group elements class 11 chemistry CBSE
Difference Between Prokaryotic Cells and Eukaryotic Cells
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
State and prove Bernoullis theorem class 11 physics CBSE
What organs are located on the left side of your body class 11 biology CBSE