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Trigonometry & Coordinate

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CAT 2025 Lesson : Trigonometry & Coordinate - Quadrants & Graph

2.3 Trigonometric ratios across Quadrants

Where the angles are more than

90\degree

, the trigonometric ratios can be found by applying the following.

Note that a full circle of

360\degree

can be divided across

4

quadrants as shown below.


The first letters of the words “All Silver Tea Cups”, i.e. A, S, T and C signify the trigonometric functions that would be positive across the 4 quadrants in that order, i.e. All, Sin, Tan and Cos. ('All' means all the trigonometric functions, i.e., Sin, Tan, Cos, etc.) As $\cosec \theta = \dfrac{1}{\sin \theta}$ , $\sec \theta = \dfrac{1}{\cos \theta}$ and $\cot \theta \dfrac{1}{\tan \theta}$ , these functions will also have the same sign (positive/negative) as their respective reciprocal functions. The table below summarises this.

The first letters of the words “All Silver Tea Cups”, i.e. A, S, T and C signify the trigonometric functions that would be positive across the 4 quadrants in that order, i.e. All, Sin, Tan and Cos. ('All' means all the trigonometric functions, i.e., Sin, Tan, Cos, etc.)

As

\cosec \theta = \dfrac{1}{\sin \theta}

\sec \theta = \dfrac{1}{\cos \theta}

and

\cot \theta \dfrac{1}{\tan \theta}

, these functions will also have the same sign (positive/negative) as their respective reciprocal functions.

The table below summarises this.

Quadrant	Range	Positive Ratios	Negative Ratios
Quadrant 1	0 $\degree$ to 90 $\degree$	All (Sin, Cosec, Tan, Cot, Cos, Sec)	-
Quadrant 2	90 $\degree$ to 180 $\degree$	Sin, Cosec	Cos, Sec, Tan, Cot
Quadrant 3	180 $\degree$ to 270 $\degree$	Tan, Cot	Sin, Cosec, Cos, Sec
Quadrant 4	270 $\degree$ to 360 $\degree$	Cos, Sec	Sin, Cosec, Tan, Cot

The following steps, shown with an example of Sin 240

\degree

, is to be followed simplify trigonometric ratios where the angle is greater than 90

\degree

Steps	Example: Sin 240 $\degree$
1) Find the multiples of 90 between which the angle is. You can use any one of these to get the answer.	180 $\degree$ and 270 $\degree$
2) Write the angle with respect to the multiple of 90 $\degree$	Sin (180 $\degree$ + 60 $\degree$ ) (270 $\degree$ - 30 $\degree$ )
3) Value will be positive/negative depending on the angle (i.e. 240 $\degree$ ) and the function (i.e., Sin) for which the value needs to be found (in this case negative as Sin is negative in Quadrant 3). 4) If an odd multiple of 90 $\degree$ is used, then switch the functions (as stated above). Retain the function if it is an even multiple of 90. (180 is an even multiple, while 270 is an odd multiple of 90)	- Sin 60 $\degree$ - Cos 30 $\degree$
5) Write down the final answer	$\dfrac{-\sqrt{3}}{2}$ $\dfrac{-\sqrt{3}}{2}$

Note:
1) The answer always be the same in both the cases.
2) For odd multiples of 900, the switches are as follows.
(a) Sin for Cos and Cos for Sin
(b) Tan for Cot and Cot for Tan
(c) Sec for Cosec and Cosec for Sec

Example 6

What are the values of the following?

(i) tan 240

\degree

(ii) cos 510

\degree

(iii) sec 720

\degree

(iv) tan 1500

\degree

Solution

(i)

\tan 240\degree = \tan (180\degree + 60\degree) = + \tan 60\degree = +\sqrt{3}

(ii)

\cos 510\degree = \cos(510\degree - 360\degree) = \cos 150\degree

= \cos (180\degree - 30\degree) = - \cos 30\degree = - \dfrac{\sqrt{3}}{2}

(iii)

\sec 720\degree = \sec (720\degree - 2 \times 360\degree) = \sec 0\degree = 1

(iv)

\tan 1500\degree \tan (1500\degree - 4 \times 360\degree) = \tan 60\degree = \sqrt{3}

Answer: (i)

+\sqrt{3}

; (ii)

\dfrac{-\sqrt{3}}{2}

; (iii)

1

; (iv)

\sqrt{3}

Example 7

Where 0 rad < θ < π rad, how many values of θ satisfy if

2 \sin^{2}

θ -

(2 + \sqrt{3}) \sinθ + \sqrt{3} = 0

Solution

2 \sin^{2} θ - (2 + \sqrt{3}) \sin

θ +

\sqrt{3} = 0

Let

x

\sin

θ
⇒

2x^{2} - (2 + \sqrt{3})x + \sqrt{3} = 0

⇒

2x^{2} - 2x - \sqrt{3x} + \sqrt{3} = 0

⇒

2x(x - 1) - \sqrt{3}(x - 1) = 0

⇒

(x - 1)(2x - \sqrt{3}) = 0

⇒

x = 1, \dfrac{\sqrt{3}}{2}

0 rad < θ < π rad, which means θ is an angle between 0

\degree

to 180

\degree

.

∴ The angle can be in the first or second quadrants, wherein sin also has a positive value.

In the first quadrant,

\sin60\degree = \dfrac{\sqrt{3}}{2}

and

\sin90\degree = 1

and .

In the second quadrant,

\sin120\degree = \sin(180\degree - 60\degree) = \sin60\degree = \dfrac{\sqrt{3}}{2}

∴ The values that satisfy the above equation are

60\degree, 90\degree

and

120\degree

Answer:

3

2.4 Range and Graph

The graph for a few functions is given for your reference and basic understanding,

Functions	Range
Sin $x$ and Cos $x$	[-1, 1]
Tan $x$ and Cot $x$	(-∞, +∞)
Sec $x$ and Cosec $x$	(-∞, -1] $\cup$ [1, +∞)


Graph for $f(x) = \sin x$
Graph for $f(x) = \cos x$
Graph for $f(x) = \tan x$

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