$\cos{\theta} \, > \, 0$

The sign of value of cos function in first quadrant is always positive.

$\Delta AOB$ is a right triangle in the first quadrant of the two dimensional space. The angle of the right triangle is taken as theta. The sign of cos function in first quadrant is evaluated by the ratio of lengths of adjacent side to hypotenuse. Therefore, letâ€™s find the sign of cos function in first quadrant geometrically.

In first quadrant, the $x$-axis and $y$-axis both represent positive values. So, the lengths of the adjacent side and opposite side are denoted by $x$ and $y$ respectively. Mathematically, they are written as $x > 0$ and $y > 0$.

$\cos{\theta} \,=\, \dfrac{OB}{OA}$

$\implies \cos{\theta} \,=\, \dfrac{x}{\sqrt{x^2+y^2}}$

The length of adjacent side is $x$ and it is positive. The length of opposite side is $y$ and it is also positive in this case. So, the value in the numerator is positive and the square root of sum of squares of $x$ and $y$ in denominator is also positive.

$x > 0$ and $\sqrt{x^2+y^2} > 0$. Therefore, the sign of ratio between them must be positive mathematically and it proves that the sign of the value of cos function in first quadrant is positive.

$\, \therefore \,\,\, \cos{\theta} \, > \, 0$

Latest Math Topics

Jul 24, 2022

Jul 15, 2022

Latest Math Problems

Jul 29, 2022

Jul 17, 2022

Jun 02, 2022

Apr 06, 2022

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved