Tuesday, January 21, 2014

80-Basics of Trigonometry - 08 Important key points

Click here for the previous basics of trigonometry.

Basics of trigonometry
Now we will study the next part of trigonometry. 

Today we will study the trigonometric ratios of three groups as shown below.
Group-03: 150° (90° + 60°)240° (180° + 60°)330° (270° + 60°).

3) Trigonometric ratios of Group-03: 150° (90° + 60°)240° (180° + 60°)330° (270° + 60°).

a) An angle q = 150°
Here angle XOP is 150° (Anti-clock-wise-direction).

Here, an inclination of ray OP is 150°, so angle AOP 30° and angle OPA is 30°

We know that the side opposite 30° is 1/2 times the hypotenuse.
So if hypotenuse OP = r, then  AP = r/2 (side opposite of 30°).

As point A is to the negative side of the X-axis, the x-coordinate of point A will be - 3r/2. In the same way, point P is in the 2nd quadrant so the y-coordinate of point P will be r/2. So, the coordinates of point P will be (-3r/2r/2).

So, all the trigonometric ratios of q = 150° with
 x = 3r/2,
 y = r/2,
 r = r.
a) sin 150° = y/r
    sin 150° = (r/2)/r   
    sin 150° = 1/2
b) cos 150° = x/r
    cos 150° = (-√3 r/2)/r  
    cos 150° = - √3/2
c) tan 150° = y/x
    tan 150° = (r/2)/(-√3 r/2)         tan 150° = - 1/√3
d) csc 150° = r/y
    csc 150° =  r/(r/2)    
     csc 150° = 2
e) sec 150° = r/x
    sec 150° = r/(-√3 r/2)         sec 150° = - 2/√3
f) cot 150° = x/y
    cot 150° = (-√3 r/2)/(r/2)   
    cot 150° = - √3

b) An angle q = 240°
Here angle XOP is 240° (Anti-clock-wise-direction).

Here, an inclination of ray OP is 240°, so angle AOP 60° and angle OPA is 30°

We know that the side opposite 30° is 1/2 times the hypotenuse.


So if hypotenuse OP = r, then  AO = r/2 (side opposite to 30°)
& AP = √3 r/2 (side opposite of 60°).

As point A is to the negative side of the X-axis, the x-coordinate of point A will be - r/2. In the same way, point P is in the 2nd quadrant so the y-coordinate of point P will be √3 r/2. So, the coordinates of point P will be (-r/2, √3 r/2).

So, all the trigonometric ratios of q = 150° with
 x = - r/2,
 y = - √3 r/2,
 r = r.
a) sin 240° = y/r
    sin 240° = [(-√3 r)/2]/r   
    sin 240° = -√3/2
b) cos 240° = x/r
    cos 240°  = (-r/2)/r       cos 240° = -1/2
c) tan 240° = y/x
    tan 240° = [(-√3 r)/2]/(-r/2)
     tan 240° =  √3
d) csc 240° = r/y
    csc 240° =  r/[(-√3 r)/2]  
    csc 240° = -2/√3
e) sec 240° = r/x
    sec 240° = r/(-r/2)        sec 240° = -2
f) cot 240° = x/y
    cot 240° = (-r/2)/[(-√3 r)/2]
    cot 240° = 1/√3
c) An angle q = 330°
Here angle XOP is 330° (Anti-clock-wise-direction).

Here, an inclination of ray OP is 330°, so angle AOP 30° and angle OPA is 60°

We know that the side opposite 30° is half the hypotenuse and the side opposite 60° is √3/2 times the hypotenuse.
So if hypotenuse OP = r, then  AP = r/2 (side opposite of 30°)
and OA = (√3 r)/2 (the side opposite of 60°).

As point A is to the positive side of the X-axis, the x-coordinate of point A will be (√3 r)/2. In the same way, point P is in the 4th quadrant so the y-coordinate of point P will be -r/2. So, the coordinates of point P will be (√3 r/2, -r/2)

So, all the trigonometric ratios of q = 330° with
 x = (√3 r)/2,
 y = - r/2,
 r =  r.
a) sin 330° = y/r
    sin 330° = (-r/2)/r   
    sin 330° = -1/2
b) cos 330° = x/r
    cos 330= (√3 r)/2/ r        cos 330° = √3/2
c) tan 330° = y/x
    tan 330° = (-r/2)/(√3 r)/2
     tan 330° =  -1/√3
d) csc 330° = r/y
    csc 330° =  r/(-r/2) 
   csc 330° = -2
e) sec 330° = r/x
    sec 330° = r/(√3 r)/2        sec 330° = 2/√3
f) cot 330° = x/y
    cot 330° = (√3 r)/2/(-r/2)
    cot 330° = -√3

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