Ken Ward's Mathematics Pages

Trigonometry Basic Formulae

Trigonometry Contents
See also:

• Compound Angles
• Half Angles

Page Contents

1. Area of Triangle:  
2. Sine Rule (Law of Sines): 
3. Cosine Rule (Law of Cosines): 
4. Compound Angle Proof:
5. Trigonometric Ratios and Pythagoras
1. Sine and Cosine 
2. Tangent and Cotangent
3. Cotangent and Cosecant
4. Tangent, Sine and Cosine
6. Negative Angles

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:

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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]
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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:

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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]

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

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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]
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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]
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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]
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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]
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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]

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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]
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Trigonometry Contents

Ken Ward's Mathematics Pages

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Faster Arithmetic - by Ken Ward