nav.gif (1245 bytes)mathematics

Ken Ward's Mathematics Pages

Trigonometry Basic Formulae

Trigonometry Contents
See also:

Page Contents

  1. Area of Triangle:  formulaHalfABsinC.gif
  2. Sine Rule (Law of Sines): sineRule.gif
  3. Cosine Rule (Law of Cosines): cosineRuleFormula.gif
  4. Compound Angle Proof: compoundAngleFormula.gif
  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)

formulaHalfABsinC.gif [1.1]
In the triangle below, the height is h. The area is:
triangleArea1.gif  [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
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:
sineRule.gif [2.3]

Cosine Rule (Proof)

The Cosine Rule is:

cosineRuleFormula.gif [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:
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:compoundAngleFormula.gif [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.compoundAngleDiag1.gif

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

Also, the area of triangle ACX is:
compoundAngle1.gif [4.3]

And the area of triangle XBC is:
compoundAngle2.gif [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:
compoundAngle4.gif [4.6]
There are choices in substituting for h from Equation 4.5, and choosing the appropriate one to give 1/2ab throughout gives:
compoundAngle3.gif [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:
sinsquaredPlusCosSquared.gif [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:
secSquaredIsTanSquaredAndOne.gif [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:
cosecSquaredEqualsOnePlusCotSquared.gif [5.14]

Tangent, Sine and Cosine

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

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

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

sec2θ=1+tan2θ [5.12 repeated]

We find:
cosAndTan.gif [5.16]

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

We find:
tanAndSin.gif [5.17]
tanAndcos.gif [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

Faster Arithmetic - by Ken Ward

Ken's book is packed with examples and explanations that enable you to discover more than 150 techniques to speed up your arithmetic and increase your understanding of numbers. Paperback and Kindle:
Buy now at
Faster Arithmetic for Adults