Trigonometric  Ratios  of  45° 

Concept

 

In ∆ ABC, right-angled at B, if one angle is 45°, then the other angle is also 45°, i.e., ∠ A = ∠  C = 45°.


So,  BC = AB  ;Now, Suppose BC = AB = a.


Then by Pythagoras Theorem, $AC^2  = AB^2  + BC^2$  = $a^2  + a^2  = 2a^2$

and, therefore,          AC =$a√2$


Using the definitions of the trigonometric ratios, we have :
 

sin 45° = ${BC}/{AC}=a/{a√2}=1/√2$
 

cos 45° = ${AB}/{AC}=a/{a√2}=1/√2$

tan 45° =${BC}/{AB}=a/a=1$
 

Also,   cosec 45° =$1/{sin 45°} = √2$
 

sec 45° =$1/{cos 45°} = √2$
 

cot 45° =$1/{tan 45°} =1$ 

Trigonometric  Ratios  of  30°  and  60°
Concept
 
Consider an equilateral triangle ABC. Since each angle in an equilateral triangle is 60°, therefore,
∠ A = ∠ B = ∠ C = 60°.
Draw the perpendicular AD from A to the side BC 
Now          ∆ ABD ≅  ∆ ACD  Therefore,    BD =  DC
and   ∠ BAD =  ∠ CAD   (CPCT)
Now observe that:
∆ ABD is a right triangle, right-angled at D with ∠ BAD = 30° and ∠ ABD = 60°. 

 So, let the length of AB =BC=CA= a.

Then,    BD =$1/2BC$ =>$a/2$
and  $AD^2  = AB^2  – BD^2  =  (a)^2  – (a/2)^2  = 3/4a^2,$ 

Therefore, AD =  $√3/2a$
Now, we have : 

sin 30° = ${BD}/{AB} ={a}/{2a}=1/2$

cos 30° =  ${AD}/{AB}={a√3}/{2a} =√3/2 $    
 
tan 30° =  ${BD}/{AD}=a/{a√3} =1/√3$
 
cosec 30° = $1/{sin 30°} = 2$
 
sec 30° = $1/{cos 30°}$ = $2/{√3} $
 
cot 30°  =$1/{tan 30°}$= √3

Similarly,

sin 60° = ${AD}/{AB}$ =$ {a√3}/{2a} =√3/2$ 

cos 60°  =${BD}/{AB} ={a}/{2a}=1/2$

tan 60° =${AD}/{BD}={a√3}/a= √3$

cosec 60° =$2/√3$

sec 60°  = 2 

cot 60° =$1/√3$

Trigonometric Ratios of 0° and 90° 
Concept
 

 
Let us see what happens to the trigonometric ratios of angle A, if it is made smaller and smaller in the right triangle ABC,

 till it becomes zero. As ∠ A gets smaller and smaller, the length of the side BC decreases.The point C gets closer to point B,
 
 and finally when ∠ A becomes very close to 0°, AC becomes almost the same as AB.
 
When  ∠  A  is  very  close  to  0°,  BC  gets  very  close  to  0  and  so  the  value  of sinA=${BC}/{AC}$ is very close to 0.
 
Also, when ∠ A is very close to 0°, AC is nearly the same as AB and so the value of cos A =${AB }/{AC}$ is very close to 1. 

This helps us to see how we can define the values of sin A and cos A when A = 0°. 
We define : 
sin 0° = 0    
cos 0° =  1.

Using these, we have :

tan 0° =${sin 0°}/{cos 0°} =0$ ; cot 0°  =$1/{tan 0°} = $ which is not defined.

sec 0° =$1/{cos 0°} = 1 $ and cosec 0° =$1/{sin 0°} = $ which is not defined.

Now, let us see what happens to the trigonometric ratios of ∠ A, when it is made larger and larger in ∆ ABC till it becomes 90°.
 As ∠ A gets larger and larger, ∠ C gets smaller and smaller. Therefore, as in the case above, the length of the side AB goes 
 on decreasing. The point A gets closer to point B. Finally when ∠ A is very close to 90°, ∠  C  becomes  very  close  to  0° 
 and  the  side AC  almost  coincides  with  side  BC .

When ∠ C is very close to 0°, ∠ A is very close to 90°, side AC is nearly the same as side BC, and so sin A is very close to 1.
 Also when ∠ A is very close to 90°, ∠ C is very close to 0°, and the side AB is nearly zero, so cos A is very close to 0.

So, :                 
sin 90° =  1   
cos 90° =  0