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

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

Latest Math Problems

Email subscription

Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.