See also:

- Area of Triangle:
- Sine Rule (Law of Sines):
- Cosine Rule (Law of Cosines):
- Compound Angle Proof:
- Trigonometric Ratios and Pythagoras
- Negative Angles

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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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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[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:

b

h

In triangle XBC, by Pythagoras' Theorem:

a

a

Substituting the value for h

a

The x2 cancels and by slight rearranging:

a

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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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:

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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.

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]

h

1=b

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

[5.6]

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From which we can express:

sin

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]

[5.12]

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And by rearranging Equation 5.12:

tan

[5.14]

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Multiplying by sec

sin

As tan

We find:

[5.15]

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cos

Multiplying by sec

cos

Noting:

sec

We find:

[5.16]

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Because

tanθ = sinθ/cosθ, and consequently tan

And using the relationship between the squares of cosine and sine, cos

We find:

[5.17]

And

[5.18]

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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