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TRIGONOMETRIC RATIOS OF SOME SPECIFIC ANGLES

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Introduction to Trig Ratios

The major functions of trigonometric ratios are sine, cosine, tangent, cosecant, secant and cotangent.

trig_r_spec_1These functions are used to find the relationship between the angle and side of a right angled triangle.


Some specific angles are:

• 0° and 90°
• 45°
• 30° and 60°



Trigonometric Ratios of 0° and 90°trig_r_spec_4

In ΔABC, right – angled at B, and ∠BAC = θ

So, from ΔABC, we have

Sin θ = BC / AC
Cos θ = AB / AC
Tan θ = BC / AB



Case I: ∠A is becoming small

If ∠A is made smaller and smaller in the ΔABC, till it becomes zero. As ∠A gets smaller and smaller, the length of the 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

Sin A = BC / AC is very close to 0.

Also, when A is very close to 0°, AC is same as AB and so the value of

Cos A = AB/AC is very close to 1.

From the above discussion, we have

Sin 0° = 0
Cosec 0° = 1 / Sin 0° = 1/0 = not defined
Cosec 0° = ∞
Cos 0° = 1
Sec 0° = 1 / Cos 0° = 1/1 = 1
Sec 0° = 1


Using Sin and Cos values, we can find Tan 0°

Tan 0° = Sin 0° / Cos 0° = 0
Tan 0° = 0

Also, Cot 0° = 1 / Tan 0° = 1/0 = not defined

Cot 0° = ∞



Case II: ∠A is becoming large

Now, let’s see when ∠A is made larger and larger in ΔABC till it becomes 90°. As ∠A gets larger and larger, ∠C gets smaller and smaller. So, 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.


So, Sin A is very close to 1.


From above discussion we get, 

Sin 90° = 1
Cosec 90° = 1
Cos 90°= 0
Sec 90° = ∞
Tan 90°= ∞
Cot 90° = 0



trig_r_spec_2Trigonometric Ratios of 45°

In ΔABC, right angled at B, if one angle is 45, then the other angle by angle sum property of triangle will also be 45.
∠A = ∠C = 45°

So, BC = AB (Isosceles triangle property)


Let, AB = BC = ‘a’


Then by Pythagoras theorem, AC2 = AB2 + BC2

AC2 = a2 + a2 = 2a2

AC = a√2.


Using formulas for trigonometric ratios:

Sin 45° = Side opposite to angle 45°    = a/a√2 = 1/√2
                         Hypotenuse


Cos 45° = Side adjacent to angle 45°    = a/a√2 = 1/√2

                          Hypotenuse


Tan 45° = Side opposite to angle 45°     = a/a = 1

              Side adjacent to angle 45°
 

Also, Cosec 45° = √2, Sec 45° = √2, Cot 45° = 1



Trigonometric Ratios of 30° and 60°

Let ΔABC, be an equilateral triangle. So, ∠A = ∠B = ∠C = 60°

trig_r_spec_3Draw perpendicular AD from A to the side BC.


Now, ΔABD ≅ ΔACD (by ASA)


Therefore, BD = DC

∠BAD = ∠CAD (by CPCT)


Consider, ΔABD

A = 30, B = 60, D = 90


Let AB = x


So, BD = x/2


And we will find the length of AD by Pythagoras theorem.

AB2 = AD2 + BD2

AB2 – BD2 = AD2

x2 – x2/4 = AD2

AD2 = 3x2/4

AD = x√3/2


Using formulas for trigonometric ratios:

Sin 30° = Side opposite to angle 30°    =x/2 / x = 1/2
                        Hypotenuse


Cos 30° = Side adjacent to angle 30°    = x√3/2 / x = √3/2

                        Hypotenuse


Tan 30° = Side opposite to angle 30°     = x/2 / x√3/2 = 1/√3

              Side adjacent to angle 30°
 

Also, Cosec 30° = 2, Sec 30° = 2/√3, Cot 30° = √3


Similarly,

Sin 60° = √3/2

Cosec 60° = 2/√3

Cos 60° = ½

Sec 60° 2

Tan 60° = √3

Cot 60° = 1/√3



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Reference Links:

 http://www.purplemath.com/modules/basirati.htm

    

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