# Ken Ward's Mathematics Pages

## Trigonometry Basic Formulae

Trigonometry Contents

## Area of Triangle (Proof)

[1.1]
In the triangle below, the height is h. The area is:
[1.2]
(Half the base times the height, of course)

h=b·sinC  [1.3]
So substituting in 1.2, we have:

## Sine Rule (Proof)

Re-using the above triangle, in triangle AXC,
h/b=sin A
h=b·sin A  [2.1]

In triangle XBC,
h/a=sin B
h=a·sin B [2.2]

Equating Equations 2.1 and 2.2, we have
h=b·sin A=a·sin B
So,
b/sin B=a/sin A

Using a perpendicular from A to BC, we can show that
b/sin B=c/sin C
Hence we have the Sine Rule:
[2.3]

## Cosine Rule (Proof)

The Cosine Rule is:

[3.1]

To prove it we use the triangle below:

h is the height (CX) and x is the distance AX, and, because AB=c, then XB=c-x

In triangle AXC, by Pythagoras' Theorem:
b2=h2+x2
h2=b2−x2   [3.2]

In triangle XBC, by Pythagoras' Theorem:

a2=(c-x)2+h2 [3.3]
a2=c2+x2-2cx+h2  [3.4]

Substituting the value for h2 in Equation 3.2 in Equation 3.4:

a2=c2+x2−2cx+b2−x2   [3.5]

The x2 cancels and by slight rearranging:
a2=b2+ c2−2cx [3.6]

In triangle AXC, we note:
x=b·cosA   [3.7]

Using this value in Equation 3.6, we get the Cosine Rule:

## Compound Angle Proof

The compound angle formula is: [4.1]

We construct a triangle, ABC, with CX being perpendicular to AB, and of length h. The line CS divides the angle C into two angles α and β.

We recall that the area of triangle ABC is:
[4.2]

Also, the area of triangle ACX is:
[4.3]

And the area of triangle XBC is:
[4.4]

We also note that
h=b·cos α=a·cos β  [4.5]

And for the set up, we note that the area of ABC is equal to the sum of the areas of triangles AXC and XBC:
[4.6]
There are choices in substituting for h from Equation 4.5, and choosing the appropriate one to give 1/2ab throughout gives:
[4.7]
Choosing an inappropriate value leads to more algebra, but the same result! A similar method is used to prove Pythagoras' Theorem

On division throughout by 1/2·a·b gives us the compound angle formula:

Should we wish to prove the formula for a compound angle clearly greater than 90° (π/2), we can use the following triangle, with the same argument as before.

## Trigonometric Ratios and Pythagoras

The following triangle is a right-angled triangle, with angle ABC a right angle.

We call AB, b (for base); BC p (for perpendicular) and AC as h (for hypotenuse).

By definition,
sinθ=p/h   [5.1]
cosθ=b/h   [5.2]
tanθ=p/b   [5.3]

h2=b2+p2 (Pythagoras' Theorem) [5.4]

## Sine and Cosine

Dividing [5.4] throughout by h2
1=b2/h2+p2/h2 [5.5]

And using the relations in Equations 5.1 and 5.2 we get:
[5.6]

From which we can express:
sin2θ=1−cos2θ  [5.7]
cos2θ=1−sin2θ  [5.8]

We can note some more relationships, by definition:
cosecθ=h/p=1/sinθ [5.9]
secθ=h/b=1/cosθ   [5.10]
cotθ=b/p=1/tanθ    [5.11]

## Tangent and Cotangent

If we divide Equation 5.6 by cos2θ, we get:
[5.12]

And by rearranging Equation 5.12:
tan2θ=sec2−1  [5.13]

## Cotangent and Cosecant

If we divide Equation 5.6 by sin2θ, we get:
[5.14]

## Tangent, Sine and Cosine

sin2θ=1−cos2θ  [5.7, repeated]
Multiplying by  sec2θ/sec2θ
sin2θ=(sec2θ-1)/sec2θ

As tan2θ=sec2−1, and sec2=1+tan2θ
We find:
[5.15]

cos2θ=1−sin2θ  [5.8, repeated]
Multiplying by  sec2θ/sec2θ
cos2θ=(sec2θ−tan2θ)/sec2θ

Noting:
sec2θ=1+tan2θ [5.12 repeated]

We find:
[5.16]

Because
tanθ = sinθ/cosθ, and consequently tan2θ=sin2θ/cos2θ
And using the relationship between the squares of cosine and sine, cos2θ+sin2θ=1

We find:
[5.17]
And
[5.18]

## Negative Angles

We will just state:
sin(−θ)= −sinθ  [6.1]
cos(−θ)= cosθ  [6.2]
tan(−θ)= −tanθ  [6.3]

Trigonometry Contents

Ken Ward's Mathematics Pages

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