# HOW TO FIND VALUES OF TRIGONOMETRIC FUNCTIONS FROM THE TERMINAL SIDE

Example 1 :

(5/7, 2√6/7) is a point on the terminal side of an angle θ in standard position. Determine the trigonometric function values of angle θ.

Solution :

First let us mark the given point in a graph paper and draw a triangle.

 OA2  =  OB2 + AB2OA2  =  (5/7)2 + (2√6/7)2  =  25/49 + 24/49OA  =  1

Opposite side  =  AB  =  2√6/7

Adjacent side  =  OB  =  5/7

Hypotenuse side  =  OA  =  1

 sin θ  =  AB/OA  =  (2√6/7)/ 1  =  2√6/7 cos θ  =  OB/OA  =  (5/7)/ 1  =  5/7 cosec θ  =  OA/AB  =  1/(2√6/7)  =  7/2√6 sec θ  =  OB/OA  =  1/(5/7)=  7/5 tan θ  =  AB/OB  =  (2√6/7)/(5/7)   =  2√6/5 cot θ  =  OB/AB  =  (5/7)/(2√6/7)  =  5/2√6

Example 2 :

Find the values of other five trigonometric functions for the following:

(i) cos θ = -1/2, θ lies in the III quadrant.

(ii) cos θ = 2/3, θ lies in the I quadrant.

Solution :

(i) cos θ = -1/2, θ lies in the III quadrant.

cos θ  =  Adjacent side/hypotenuse side

cos θ  =  -1/2

Adjacent side  =  1, Hypotenuse side  =  2

(Hypotenuse side)2  =  (Opposite side)2 + (Adjacent side)2

Opposite side  =  √22 - 12

=  √(4 - 1)

=  3

Opposite side  =  3

Note : Since θ  lies in 3rd quadrant, all trigonometric ratios other than tan  and cot will have negative sign.

sin θ  =  Opposite side/hypotenuse side   -3/2

tan θ  =  Opposite side/Adjacent side  =  3/1  =  -3

cosec θ  =  Hypotenuse side/Opposite side  =  2/3

sec θ  =  Hypotenuse side/Adjacent side  =  2/1  =  2

cot θ  =  Adjacent side/Opposite side  -1/3

(ii) cos θ = 2/3, θ lies in the I quadrant.

Solution :

cos θ  =  Adjacent side/hypotenuse side

cos θ  =  2/3

Adjacent side  =  2, Hypotenuse side  =  3

(Hypotenuse side)2  =  (Opposite side)2 + (Adjacent side)2

Opposite side  =  √32 - 22

=  √(9 - 4)

=  √5

Opposite side  =  √5

Note : Since θ  lies in 1st quadrant, all trigonometric ratios will be positive.

sin θ  =  Opposite side/hypotenuse side   √5/3

tan θ  =  Opposite side/Adjacent side  =   √5/2

cosec θ  =  Hypotenuse side/Opposite side  =  3/√5

sec θ  =  Hypotenuse side/Adjacent side   3/2

cot θ  =  Adjacent side/Opposite side  =  2/√5

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