## Friday, January 3, 2014

### 78-Basics of Trigonometry - 06

Blog-78
Dear Students,
Now we will study the next part of trigonometry.

Today we will study trigonometric ratios of three groups as shown bellow.
Group-01: 120° (90° + 30°)210° (180° + 30°)300° (270° + 30°).
Group-02: 135° (90° + 45°)225° (180° + 45°)315° (270° + 45°).
Group-03: 150° (90° + 60°)240° (180° + 60°)330° (270° + 60°).

1) Trigonometric ratios of Group-01: 120° (90° + 30°)210° (180° + 30°)300° (270° + 30°).

a) An angle q = 120°
Here angle XOP is of 120° (Anti-clock-wise-direction).

Here, inclination of ray OP is 120°, so angle AOP 60° and angle OPA is 30°

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

As point A is to the negative side of X-axis, x-coordinate of point A will be - r/2. In the same way point P is in the 2nd quadrant, so 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 = 120° with
x = r/2,
y = (√3 r)/ 2,
r = r.

 a) sin 120° = y/r     sin 120° = [(√3 r)/2]/r        sin 120° = √3/2 b) cos 120° = x/r     cos 120° = (- r/2)/r        cos 120° = - 1/2 c) tan 120° = y/x     tan 120° = [(√3 r)/2]/(- r/2)                    tan 120° = - √3 d) csc 120° = r/y     csc 120° =  r/[(√3 r)/2]          csc 120° = 2/√3 e) sec 120° = r/x     sec 120° = r/(- r/2)           sec 120° = - 2 f) cot 120° = x/y     cot 120° = (- r/2)/ [(√3 r)/2]        cot 120° = - 1/√3

b) An angle q =  210°
Here angle XOP is of 210° (Anti-clock-wise-direction).

Here, inclination of ray OP is 210°, so angle AOP 30° and angle OPA is 60°

We know that the side opposite to 30° is half the hypotenuse and side opposite to 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 (side opposite of 60°).

As point A is to the negative side of X-axis, x-coordinate of point A will be - √3 r/2. In the same way point P is in the 3rd quadrant, 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 = 210° with
x = (√3 r)/2,
y = - r/2,
r =  r.

 a) sin 210° = y/r     sin 210° = (- r/2)/r        sin 210° = -1/2 b) cos 210° = x/r     cos 210° = [(-√3 r)/2]/r        cos 210° = - √3/2 c) tan 210° = y/x     tan 210° = (- r/2)/[(-√3 r)/2]      tan 210° =  1/√3 d) csc 210° = r/y     csc 210° =  r/(- r/2)         csc 210° = -2 e) sec 210° = r/x     sec 210° = r/[(-√3 r)/2]           sec 210° = - 2/√3 f) cot 210° = x/y     cot 210° = [(-√3 r)/2]/ (- r/2)     cot 210° = √3

c) An angle q = 300°
Here angle XOP is of 300° (Anti-clock-wise-direction).

Here, inclination of ray OP is 300°, so angle AOP 60° and angle OPA is 30°

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

As point A is to the positive side of X-axis, x-coordinate of point A will be r/2. In the same way point P is in the 4th quadrant, so 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 = 300° with
x = r/2,
y = - (√3 r)/2,
r =  r.
 a) sin 300° = y/r     sin 300° = [(-√3 r)/2]/r        sin 300° = -√3/2 b) cos 300° = x/r     cos 300  = (r/2)/r        cos 300° = 1/2 c) tan 300° = y/x     tan 300° = [(-√3 r)/2]/(r/2)     tan 300° =  -√3 d) csc 300° = r/y     csc 300° =  r/[(-√3 r)/2]      csc 300° = -2/√3 e) sec 300° = r/x     sec 300° = r/(r/2)           sec 300° = 2 f) cot 300° = x/y     cot 300° = (r/2)/[(-√3 r)/2]     cot 300° = -1/√3

Group-02: 135° (90° + 45°)225° (180° + 45°)315° (270° + 45°) will be published in the next Blog.