Relations linking the trigonometric functions

Complimentary angle

You will have noted in the preceding work that certain values of the functions are the same; for example, sin 30º = cos 60º = 1/2 or tan 45º = cot 45º = 1 or sin 40º = cos 50º = 0.6428, etc. The corresponding angles 30º and 60º or 45º and 45º or 40º and 50º always add up to 90º, that is they complement each other. This suggests that there exist basic relations between the values of the trigonometric functions at complementary angles.

In the right-angled triangle ABC, = 90º - is the complementary angle of . Thus

sin(90º-)=b/c	cos(90º-)=a/c	tan(90º-)=b/a	cot(90º-)=a/b
cos=b/c	sin=a/c	cot=b/a	tan= a/b
sin(90º-)=cos	cos(90º-)=sin	tan (90º-)=cotan	cot(90º-)=tan

cosin means complimentary of sin. It was written originally co-sin and later on cos. The same explanation applies to cot, which originally was co-tan. For this reason, sometimes people talk of co-functions.

The formulae above yield the Rule:

Every trigonometric function of an angle equals
the co-function of its complimentary angle 90º -.

Multiplication and division of side ratios

a/b·b/a = 1		(a/c)/(b/c) = a/b		(b/c)/(a/c) = b/a
tan·cot=1		sin/cos=tan		cos/sin=cot

Pythagoras' equation

If you divide Pythagoras' equation a² + b² = c² by one of its three quadratic terms, it becomes an equation of ratios and must therefore lead to relations between the trigonometric functions:

Division by c²:

(a/c)² + (b/c)² = 1

sin² + cos² =1

For (sin)² you write sin², which, however, must not be mistaken for
(sin²). You read it "sin square ". The same applies to the other trigonometric functions.

Division by b²:		(a/b)²+1=(c/b)²		tan²+1=1/cos²		cos=1/
Division by a²:		1+ (b/a)²=(c/a)²		1+1/tan²=1/sin²		sin=tan/

If you produce for every trigonometric function a special right-angled triangle, in which the side in the denominator every time is unity, that is for cos and sin the hypotenuse, for tan the opposite side, for cot the adjacent side, and compute the third side by means of Pythagoras' Rule, then the reduction of all the functions to a selected one can be read off the relevant triangle below:

For example, if you want to express all other trigonometric functions in terms of tan , you use the third triangle above:

sin=tan/		cos= 1/
tan=tan		cot=1/tan

For example, the fourth triangle yields

sin = 1/

You can also prove this by means of the earlier formulae: Thus

1/(1 + cot²)^1/2 = 1/(1 + cos²/sin²)^1/2 = 1/([sin²+ cos²]/sin²)^1/2 =
sin/(sin²+ cos²) ^1/2=sin .

Task: Given sin=2/3, find cos, tan and cotwithout using the table:

1. By formula:

tan = sin/cos= (3/5)/(4/5) = 3/4,

cot = 1/tan = 4/3;
2. By means of a triangle: Since sin = 3/5, by Pythagoras' Rule, you can compute b using a = 3 and c = 5. You find from a² + b² = c² that 9 + b² = 25, whence b = 4 and cos = 4/5, tan = 3/4, cot = 4/5.

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