Signs of trigonometric functions

Signs of trigonometric functions are different in different quadrant. Lets acquire the knowledge of them. But before studying this make sure you know the basics of trigonometry.If not you can look into my previous posts.{Introduction of trigonometry and trigonometric table}

Let us understand this with the following image.

we will try to understand this quadrant wise.

First quadrant: ( 0º to 90º )

Have a look at the first quadrant .Let ΔAFO be a right-angled triangle,right angled at F.Here ∠AOF= θ

the coordinates of the points are;

A (a,b) ,O(0,0) & F(a,0).

In first quadrant both a & b are positive.

So here perpendicular will be = +b.

base=+a and

hypotenuse(h) =\sqrt { { a }^{ 2 }+{ b }^{ 2 } } {Hypotenuse will be always +ve in each quadrant}

Thus the signs trigonometric functions will be:

sinθ=\frac { perpendicular }{ hypotenuse }= \frac { +b }{ +h } =+Ve

cosecθ=\frac { hypotenuse }{ perpendicular } = \frac { +h }{ +b } =+Ve

cosθ=\frac { base }{ hypotenuse } = \frac { +a }{ +h } =+Ve

secθ=\frac { hypotenuse }{ base } = \frac { +h }{ +a } =+Ve

tanθ=\frac { perpendicular }{ base } = \frac { +b }{ +a } =+Ve

cotθ=\frac { base }{ perpendicular } = \frac { +a }{ +b } =+Ve

 Therefore all trigonometric functions in first quadrant are positive. 


Second quadrant: (90º to 180º)

Let ΔBEO be a right-angled triangle,right angled at E.

Here ∠BOE=θ the coordinates of the points are;

B (-a,b) ,O(0,0) & E(-a,0).

In second quadrant a is negative & b is positive.

So, perpendicular = +b , base= -a (negative sign) and

hypotenuse(h) =\sqrt { { a }^{ 2 }+{ b }^{ 2 } }

sinθ=\frac { perpendicular }{ hypotenuse }= \frac { +b }{ +h } =+Ve

cosecθ=\frac { hypotenuse }{ perpendicular } = \frac { +h }{ +b } =+Ve

cosθ=\frac { base }{ hypotenuse } = \frac { -a }{ +h } =-Ve

secθ=\frac { hypotenuse }{ base } = \frac { +h }{ -a } =-Ve

tanθ=\frac { perpendicular }{ base } = \frac { +b }{ -a } =-Ve

cotθ=\frac { base }{ perpendicular } = \frac { -a }{ +b } =-Ve

 In 2nd quadrant only sin and cosec are positive. 


3rd quadrant: (180º to 270º)

Let ΔCEO be a right-angled triangle,right angled at E.Here ∠COE= θ

the coordinates of the points are;

C (-a,-b) ,O(0,0) & E(-a,0). In third quadrant both a & b are negative.

So perpendicular = -b(negative), base= -a(negative) and

hypotenuse(h) =\sqrt { { a }^{ 2 }+{ b }^{ 2 } }

Thus

sinθ=\frac { perpendicular }{ hypotenuse }= \frac { -b }{ +h } =-Ve

cosecθ=\frac { hypotenuse }{ perpendicular } = \frac { +h }{ -b } =-Ve

cosθ=\frac { base }{ hypotenuse } = \frac { -a }{ +h } =-Ve

secθ=\frac { hypotenuse }{ base } = \frac { +h }{ -a } =-Ve

tanθ=\frac { perpendicular }{ base } = \frac { -b }{ -a } =+Ve

cotθ=\frac { base }{ perpendicular } = \frac { -a }{ -b } =+Ve

 In 3rd quadrant only tan and cot are positive. 


Forth quadrant: (270º to 360º)

Let ΔFDO be a right-angled triangle,right angled at F.Here∠ FOD=θ

the coordinates of the points are;

F (a,0) ,O(0,0) & D(a,-b).

In fourth quadrant a is positive & b is negative.

So perpendicular = -b. base= +a and

hypotenuse =

Thus

sinθ=\frac { perpendicular }{ hypotenuse }= \frac { -b }{ +h } =-Ve

cosecθ=\frac { hypotenuse }{ perpendicular } = \frac { +h }{ -b } =+Ve

cosθ=\frac { base }{ hypotenuse } = \frac { +a }{ +h } =+Ve

secθ=\frac { hypotenuse }{ base } = \frac { +h }{ +a } =+Ve

tanθ=\frac { perpendicular }{ base } = \frac { -b }{ +a } =-Ve

cotθ=\frac { base }{ perpendicular } = \frac { +a }{ -b } =-Ve

 In fourth quadrant only cos and sec are positive. 


In general we can say signs of trigonometric functions in different quadrants are:

 

1st

2nd

3rd

4th

sin x

+

+

cos x

+

+

tan x

+

+

cosec x

+

+

sec x

+

+

cot x

+

+

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