130-Basics of Trigonometry - 16 Important key points

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129-Basics of Trigonometry - 15 Important key points

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128-Basics of Trigonometry - 14 Important key points

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111-Basics of Trigonometry - 13 Important key points

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Example-7

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110-Basics of Trigonometry - 12 Important key points

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109-Basics of Trigonometry - 11 Important key points

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Example-3

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108-Basics of Trigonometry - 10 Important key points

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Example-1

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81-Basics of Trigonometry - 09 Important key points

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Basics of trigonometry

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1) Fundamental Identities:

      a) sin² Θ + cos² Θ = 1

      b) sec² Θ - tan² Θ = 1

      c) csc² Θ - cot² Θ = 1

Proof:

      a) sin² Θ + cos² Θ = 1 

Here, 

       1)  x² + y²  = r²  -----------------(1)

            Dividing equation 1 by r² , we get,  

       2)  (x² / r² + y² / r² = r² / r²

       3)  (x / r) ² + (y / r) ² = 1 -------(2)

       4) We know that, 

          y/r = sin Θ

          x/r  =  cos Θ, ----- from equation 2, we have,

      5)  (cos Θ) ² + (sin Θ) ² = 1 ---(3)         

      6)  (sin Θ) ² + (cos Θ) ² = 1

      b) sec² Θ - tan² Θ = 1

Here, 

       1)  x² + y²  = r²  -------------------------(1)

            Dividing equation 1 by x²,we get,  

       2)  (x² / x² + y² / x² = r² / x²

       3)  (x / x) ² + (y / x) ² = (r / x) ² -------(2)

       4) We know that, 

           y/x =  tan Θ

           r/x =  sec Θ, ----- from equation 2, we have,

      5)  (1) ² + (tan Θ) ² =  (sec Θ) ² ---(3)         

      6)  (sec Θ) ² - (tan Θ) ² = 1

      c) csc² Θ - cot² Θ = 1

Here, 

       1)  x² + y²  = r²  -------------------------(1)

            Dividing equation 1 by y²,we get,  

       2)  (x² / y² + y² / y² = r² / y²

       3)  (x / y) ² + (y / y) ² = (r / y) ² -------(2)

       4) We know that, 

           x/y =  cot Θ

           r/y =  csc Θ, ----- from equation 2, we have,

      5)  (cot Θ) ² + (1) ² =  (csc Θ) ² ---(3)         

      6)  (csc Θ) ² - (cot Θ) ² = 1

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Anil Satpute

80-Basics of Trigonometry - 08 Important key points

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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/2, r/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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Anil Satpute

79-Basics of Trigonometry - 07 Important key points

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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-02: 135° (90° + 45°), 225° (180° + 45°), 315° (270° + 45°).

2) Trigonometric ratios of Group-02: 135° (90° + 45°), 225° (180° + 45°), 315° (270° + 45°).

a) An angle q = 135°

Here angle XOP is 135° (Anti-clock-wise-direction).

Here, an inclination of ray OP is 135°, so angle AOP 45° and angle OPA is 45°

We know that the side opposite 45° is 1/√2 times the hypotenuse.

So if hypotenuse OP = r, then  AP = OA = r/√2 (side opposite of 45°).

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 r/√2. So, the coordinates of point P will be (-r/√2, r/√2).

So, all the trigonometric ratios of q = 135° with

 x = - r/√2,

 y = r/√2,

 r = r.

a) sin 135° = y/r     sin 135° = (r/√2)/r        sin 135° = 1/√2	b) cos 135° = x/r     cos 135° = (- r/√2)/r        cos 135° = - 1/√2	c) tan 135° = y/x     tan 135° = (r/√2)/(- r/√2)                      tan 135° = - 1
d) csc 135° = r/y     csc 135° =  r/(r/√2)          csc 135° = √2	e) sec 135° = r/x     sec 135° = r/(- r/√2)           sec 135° = - √2	f) cot 135° = x/y     cot 135° = (- r/√2)/(r/√2)        cot 135° = - 1

b) An angle q = 225°

Here angle XOP is 225° (Anti-clock-wise-direction).

Here, an inclination of ray OP is 225°, so angle AOP 45° and angle OPA is 45°

We know that the side opposite 45° is 1/√2 times the hypotenuse.

So if hypotenuse OP = r, then  AP = OA = r/√2 (side opposite of 45°).

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 3rd quadrant, the y-coordinate of point P will be - r/√2. So, the coordinates of point P will be (-r/√2, -r/√2)

So, all the trigonometric ratios of q = 225° with

 x = - r/√2,

 y = - r/√2,

 r =  r.

a) sin 225° = y/r     sin 225° = (- r/√2)/r    sin 225° = -1/√2	b) cos 225° = x/r     cos 225° = (-r/√2)/r        cos 225° = -1/√2	c) tan 225° = y/x     tan 225° = (- r/√2)/(-r/√2)                  tan 225° =  1
d) csc 225° = r/y     csc 225° =  r/(- r/√2)     csc 225° = -√2	e) sec 225° = r/x     sec 225° = r/(- r/√2)           sec 225° = -√2	f) cot 225° = x/y     cot 225° = (- r/√2)/(- r/√2)     cot 225° = 1

c) An angle q = 315°

Here angle XOP is 315° (Anti-clock-wise-direction).

Here, an inclination of ray OP is 315°, so angle AOP 45° and angle OPA is 45°

We know that the side opposite 45° is 1/√2 times the hypotenuse.

So if hypotenuse OP = r, then  AP = OA = r/√2 (side opposite of 45°).

As point A is to the positive side of the X-axis, the x-coordinate of point A will be 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 (r/√2, -r/√2)

So, all the trigonometric ratios of q = 315° with

 x = r/√2,

 y = - r/√2,

 r =  r.

a) sin 315° = y/r     sin 315° = (-r/√2)/r        sin 315° = -1/√2	b) cos 315° = x/r     cos 315= (r/√2) r        cos 315° = 1/√2	c) tan 315° = y/x     tan 315° = (-r/√2)/(r/√2)      tan 315° =  -1
d) csc 315° = r/y     csc 315° =  r/(-r/√2)     csc 315° = -√2	e) sec 315° = r/x     sec 315° = r/(r/√2)           sec 315° = √2	f) cot 315° = x/y     cot 315° = (r/√2)/(-r/√2)     cot 315° = -1

Group-03: 150° (90° + 60°), 240° (180° + 60°), 330° (270° + 60°) will be published in the next Blog.

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Anil Satpute