Question

How do I find the value of \[\cos 45{}^\circ \]?

Answer 1

Hint:
Using the Pythagoras theorem in a right angled triangle and using the property of an isosceles triangle when the two angles are equal then the opposite sides are also equal. Then using the pythagorean theorem and after simplifying we can calculate the Hypotenuse in terms of the other side which is \[\sqrt{2}\] times of the perpendicular or base.

Complete step by step solution:
Let consider a right angled triangle ABC, \[m\angle C=90{}^\circ \]
And the opposite sides to corresponding angle be \[a,b,c\]
Also given that, \[m\angle A=45{}^\circ \]

Now using angle sum property of a triangle
\[\Rightarrow m\angle A+m\angle B+m\angle C=180{}^\circ \]
\[\Rightarrow \angle B=45{}^\circ \]
Since the two angles are equal in a triangle that means the triangle is isosceles
\[\Rightarrow a=b--(1)\]
Now using Pythagoras theorem
\[{{H}^{2}}={{P}^{2}}+{{B}^{2}}\]
\[\Rightarrow {{c}^{2}}={{a}^{2}}+{{b}^{2}}--(2)\]
From equation \[(1)\] and \[(2)\]
\[\Rightarrow {{c}^{2}}=2{{a}^{2}}\]
\[\Rightarrow a=\dfrac{c}{\sqrt{2}}--(3)\]
Now using trigonometric ratios
\[\cos \theta =\dfrac{Base}{Hypotenuse}\]
\[\Rightarrow \cos B=\dfrac{a}{c}\]
\[\Rightarrow \cos 45{}^\circ =\dfrac{c}{\sqrt{2}}.\dfrac{1}{c}\] (From equation \[3\])
\[\Rightarrow \dfrac{1}{\sqrt{2}}\]

Hence, \[\cos 45{}^\circ =\dfrac{1}{\sqrt{2}}\]

Note:
In this type of questions we use the Pythagoras theorem and the properties if trigonometric ratio to find their values usually for the angle of \[(2n+1){}^{\pi }/{}_{4}\] where \[n\in Z\] the length of hypotenuse is \[\sqrt{2}\] times of the perpendicular or base.