Introduction to Trig Ratios

We have six trigonometric ratios with respect to ∠ BAC = θ, and they are as follows:
•    Sine θ

•    Cosine θ

•    Tangent θ

•    Cosecant θ

•    Secant θ

•    Cotangent θ


Let the Hypotenuse in Δ ABC = h
Adjacent in Δ ABC = b
Opposite in Δ ABC = a

Now, we define the above mentioned trigonometric ratios:
•    Sine θ or Sin θ = Opposite / Hypotenuse = a / h

•    Cosine θ or Cos θ = Adjacent / Hypotenuse = b / h

•    Tangent θ or Tan θ = Opposite / Adjacent = a / b

•    Cosecant θ or Cosec θ = Hypotenuse / Opposite = h / a

•    Secant θ or Sec θ = Hypotenuse / Adjacent = h / b

•    Cotangent θ or Cot θ = Adjacent / Opposite = b / a

From the above discussion, it is clear that the last three trigonometric ratios are opposite of the first three trigonometric ratios respectively.

That is,
•    Cosecant θ or Cosec θ = 1 / Sine θ

•    Secant θ or Sec θ = 1 / Cosine θ

•    Cotangent θ or Cot θ = 1 / Tangent θ

SOH CAH TOA

There is one short method for remembering all six trigonometric ratios.

SOH

‘S’ stands for Sine
‘O’ stands for Opposite
‘H’ stands for Hypotenuse

Sine = Opposite / Hypotenuse

CAH

‘C’ stands for Cosine
‘A’ stands for Adjacent
‘H’ stands for Hypotenuse

Cosine = Adjacent / Hypotenuse
 

TOA

‘T’ stands for Tangent
‘O’ stands for Opposite
‘A’ stands for Adjacent

Tangent = Opposite / Adjacent

As discussed above, that Cosecant, Secant, and Cotangent are opposites of Sine, Cosine and Tangent respectively.

Let’s solve few problems based on the above discussion:

In Δ ABC, right angled at A, if AB = 12, AC = 5 and BC = 13, find all the six trigonometric ratios of angle B.

With reference to above Δ ABC we have,
Opposite = AC = 5
Adjacent = AB = 12
Hypotenuse = BC = 13

Using the definitions of trigonometric ratios, we have
Sine B = Opposite / Hypotenuse = AC / BC = 5 / 13
Cosine B = Adjacent / Hypotenuse = AB / BC = 12 / 13
Tangent B = Opposite / Adjacent = AC / AB = 5 / 12
Cosecant B = Hypotenuse / Opposite = BC / AC = 13 / 5
Secant B = Hypotenuse / Adjacent = BC / AB = 13 / 12
Cotangent B = Adjacent / Opposite = AB / AC = 12 / 5

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Reference Links:

•     http://www.purplemath.com/modules/basirati.htm
•     http://en.wikipedia.org/wiki/Sine
•     http://en.wikipedia.org/wiki/Trigonometric_functions#Sine.2C_cosine.2C_and_tangent
•     http://en.wikipedia.org/wiki/Tangent
•     http://en.wikipedia.org/wiki/Trigonometry