All Basic Trigonometric functions

Basic trigonometric functions

Let θ be an angle in standard position with (x,y) a point on the terminal side of angle θ and $\dpi{120} r=\sqrt{x^2+y^2}\neq 0$

Sinθ = y/r Cosec = r/y

Cosθ = x/r Secθ = r/x

Tanθ = y/x Cotθ = x/y

Since r can’t be 0 so sine and cosine are defined for every real value of . However if x=0, then tan and sec are undefined. Similarly if y=0, then cosec and cot are undefined.

Example: Given point (-3,4) on terminal side of an angle in standard position. Determine the exact value of all six trigonometric functions.

Solution: From given point (x, y)= (-3,4), we have x=-3 and y= 4

$\dpi{120} r=\sqrt{x^2+y^2}$

$\dpi{120} r=\sqrt{(-3)^2+(4)^2}=\sqrt{25}=5$

So, we have the functions as follows:

Sinθ = 4/5 Cosecθ = 5/4

Cosθ = – 3/5 Secθ = – 5/3

Tanθ = – 4/3 Cot = – 3/4

ASTC rule:

X and Y axis divide whole space in 4 parts called quadrants. To know the sign of trigonometric ratios in these different quadrants, we use ASTC rule. We can use any mnemonic to memorize this rule like “ All Students Take Calculus” or “Add Sugar To Coffee”

All trigonometric ratios positive in first quadrant.

Only Sin and its reciprocal Cosec are positive in 2^nd quadrant, rest all are negative.

Only Tan and its reciprocal cot are positive in 3^rd quadrant , rest all are negative.

Only Cos and its reciprocal Sec are positive in 4^th quadrant, rest all are negative.

Quadrants	Range in degrees	Range in radians.
I	0 -90°	$\dpi{120} 0- \frac{\pi }{2}$
II	90° -180°	$\dpi{120} \frac{\pi }{2}-\pi$
III	180° -270°	$\dpi{120} \pi -\frac{3\pi }{2}$
IV	270° -360°	$\dpi{120} \frac{3\pi }{2}-2\pi$

Example: Find the values of the other five trigonometric functions of θ with given function value .

$\dpi{120} {\color{Red} tan\theta =- \frac{15}{8}}$ and constraint sinθ >0.

Solution: Note that tan is negative and sin is positive which is possible only in 2^nd quadrant.

Given that $\dpi{120} tan\theta =- \frac{15}{8}$

$\dpi{120} \frac{y}{x}=-\frac{15}{8}$

In 2^nd quadrant x is negative and y is positive.

So x=-8 and y= 15

Therefore $\dpi{120} r=\sqrt{(-8)^2+(15)^2}=\sqrt{289}= 17$

Sinθ = 15/17 Cosecθ = 17/15

Cosθ = -8/17 Secθ = – 17/8

Tanθ = -15/8 Cot = – 8/15

Since angle was in second quadrant, so we got only sin and cosec positive and rest all negative.

Example: If cosθ=-1/2 and $\dpi{120} {\color{Red} \pi <\theta <\frac{3\pi }{2}}$ then find the value of $\dpi{120} {\color{Red} 4tan^{2}\theta -3csc^{2}\theta }$

Solution: Since θ lies in third quadrant, therefore sin is negative and tan is positive.

Using basic Pythagorean identities given in previous section, we have,

$\dpi{120} sin\theta =\pm \sqrt{1-cos^{2}\theta }$

$\dpi{120} sin\theta =-\sqrt{1-\left ( \frac{-1}{2} \right )^{2} }$

$\dpi{120} sin\theta =-\sqrt{1-\frac{1}{4}} =-\sqrt{\frac{3}{4}} =-\frac{\sqrt{3}}{2}$

$\dpi{120} csc\theta =\frac{-2}{\sqrt{3}}$

Using basic trigonometric identities we have,

$\dpi{120} tan\theta =\frac{sin\theta }{cos\theta }= -\frac{\sqrt{3}}{2}*\frac{-2}{1}=\sqrt{3}$

$\dpi{120} 4tan^{2}\theta -3csc^{2}\theta = 4\left ( \sqrt{3} \right )^{2}-3\left ( \frac{-2}{\sqrt{3}} \right )^{2}$

=4(3)-3(4/3) = 12-4 = 8

Practice problems:

Given point (5,-12) on terminal side of an angle in standard position. Determine the exact value of all six trigonometric functions.
Find the values of the other five trigonometric functions of θ with given function value cosθ = -12/13 and constraint sinθ < 0.
If cosθ =-3/5 and $\dpi{120} \pi <\theta <\frac{3\pi }{2}$ then find the value of $\dpi{120} \frac{csc\theta +cot\theta }{sec\theta -tan\theta }$

Answers:

sin =-12/13 ,cos =5/13 ,tan =-12/5 ,cosec =-13/12 ,sec =13/5 , cot =-5/12
sin =-5/13 ,tan =5/12 ,cosec =-13/5 ,sec =-13/12 , cot =12/5
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