B.2 Trigonometry
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Angles — Radians vs Degrees
For mathematics, and especially in calculus, it is much better to measure angles in units called radians rather than degrees. By definition, an arc of length θ on a circle of radius one subtends an angle of θ radians at the centre of the circle.
The circle on the left has radius 1, and the arc swept out by an angle of θ radians has length θ. Because a circle of radius one has circumference 2π we have
2π radians =360∘π radians =180∘π2 radians =90∘π3 radians =60∘π4 radians =45∘π6 radians =30∘
More generally, consider a circle of radius r. Let L(θ) denote the length of the arc swept out by an angle of θ radians and let A(θ) denote the area of the sector (or wedge) swept out by the same angle. Since the angle sweeps out the fraction θ2π of a whole circle, we have
L(θ)=2πr⋅θ2π=θrandA(θ)=πr2⋅θ2π=θ2r2
Trig Function Definitions
The trigonometric functions are defined as ratios of the lengths of the sides of a right-angle triangle as shown in the left of the diagram below . These ratios depend only on the angle θ.
The trigonometric functions sine, cosine and tangent are defined as ratios of the lengths of the sides
sinθ=oppositehypotenusecosθ=adjacenthypotenusetanθ=oppositeadjacent=sinθcosθ.
These are frequently abbreviated as
sinθ=ohcosθ=ahtanθ=oa
which gives rise to the mnemonic
SOHCAHTOA
If we scale the triangle so that they hypotenuse has length 1 then we obtain the diagram on the right. In that case, sinθ is the height of the triangle, cosθ the length of its base and tanθ is the length of the line tangent to the circle of radius 1 as shown.
Since the angle 2π sweeps out a full circle, the angles θ and θ+2π are really the same.
Hence all the trigonometric functions are periodic with period 2π. That is
sin(θ+2π)=sin(θ)cos(θ+2π)=cos(θ)tan(θ+2π)=tan(θ)
The plots of these functions are shown below
The reciprocals (cosecant, secant and cotangent) of these functions also play important roles in trigonometry and calculus:
cscθ=1sinθ=hosecθ=1cosθ=hacotθ=1tanθ=cosθsinθ=ao
The plots of these functions are shown below
These reciprocal functions also have geometric interpretations:
Since these are all right-angled triangles we can use Pythagoras to obtain the following identities:
sin2θ+cos2θ=1tan2θ+1=sec2θ1+cot2θ=csc2θ
Of these it is only necessary to remember the first
sin2θ+cos2θ=1
The second can then be obtained by dividing this by cos2θ and the third by dividing by sin2θ.
Important Triangles
Computing sine and cosine is non-trivial for general angles — we need Taylor series (or similar tools) to do this. However there are some special angles (usually small integer fractions of π) for which we can use a little geometry to help. Consider the following two triangles.
The first results from cutting a square along its diagonal, while the second is obtained by cutting an equilateral triangle from one corner to the middle of the opposite side. These, together with the angles 0,π2 and π give the following table of values
θ | sinθ | cosθ | tanθ | cscθ | secθ | cotθ |
0 rad | 0 | 1 | 0 | DNE | 1 | DNE |
π2 rad | 1 | 0 | DNE | 1 | DNE | 0 |
π rad | 0 | -1 | 0 | DNE | -1 | DNE |
π4 rad | 1√2 | 1√2 | 1 | √2 | √2 | 1 |
π6 rad | 12 | √32 | 1√3 | 2 | 2√3 | √3 |
π3 rad | √32 | 12 | √3 | 2√3 | 2 | 1√3 |
Some More Simple Identities
Consider the figure below
The pair triangles on the left shows that there is a simple relationship between trigonometric functions evaluated at θ and at −θ:
sin(−θ)=−sin(θ)cos(−θ)=cos(θ)
That is — sine is an odd function, while cosine is even. Since the other trigonometric functions can be expressed in terms of sine and cosine we obtain
tan(−θ)=−tan(θ)csc(−θ)=−csc(θ)sec(−θ)=sec(θ)cot(−θ)=−cot(θ)
Now consider the triangle on the right — if we consider the angle π2−θ the side-lengths of the triangle remain unchanged, but the roles of “opposite” and “adjacent” are swapped. Hence we have
sin(π2−θ)=cosθcos(π2−θ)=sinθ
Again these imply that
tan(π2−θ)=cotθcsc(π2−θ)=secθsec(π2−θ)=cscθcot(π2−θ)=tanθ
We can go further. Consider the following diagram:
This implies that
sin(π−θ)=sin(θ)cos(π−θ)=−cos(θ)sin(π+θ)=−sin(θ)cos(π+θ)=−cos(θ)
From which we can get the rules for the other four trigonometric functions.
Identities — Adding Angles
We wish to explain the origins of the identity
sin(α+β)=sin(α)cos(β)+cos(α)sin(β).
A very geometric demonstration uses the figure below and an observation about areas.
- The left-most figure shows two right-angled triangles with angles α and β and both with hypotenuse length 1.
- The next figure simply rearranges the triangles — translating and rotating the lower triangle so that it lies adjacent to the top of the upper triangle.
- Now scale the lower triangle by a factor of q so that edges opposite the angles α and β are flush. This means that qcosβ=cosα. ie
q=cosαcosβ
Now compute the areas of these (blue and red) trianglesAred=12q2sinβcosβAblue=12sinαcosα
So twice the total area is
2Atotal=sinαcosα+q2sinβcosβ - But we can also compute the total area using the rightmost triangle:
2Atotal=qsin(α+β)
Since the total area must be the same no matter how we compute it we have
qsin(α+β)=sinαcosα+q2sinβcosβsin(α+β)=1qsinαcosα+qsinβcosβ=cosβcosαsinαcosα+cosαcosβsinβcosβ=sinαcosβ+cosαsinβ
as required.
We can obtain the angle addition formula for cosine by substituting α↦π/2−α and β↦−β into our sine formula:
sin(α+β)=sin(α)cos(β)+cos(α)sin(β)becomessin(π/2−α−β)⏟cos(α+β)=sin(π/2−α)⏟cos(α)cos(−β)+cos(π/2−α)⏟sin(α)sin(−β)cos(α+β)=cos(α)cos(β)−sin(α)sin(β)
where we have used sin(π/2−θ)=cos(θ) and cos(π/2−θ)=sin(θ).
It is then a small step to the formulas for the difference of angles. From the relation
sin(α+β)=sin(α)cos(β)+cos(α)sin(β)
we can substitute β↦−β and so obtain
sin(α−β)=sin(α)cos(−β)+cos(α)sin(−β)=sin(α)cos(β)−cos(α)sin(β)
The formula for cosine can be obtained in a similar manner. To summarise
sin(α±β)=sin(α)cos(β)±cos(α)sin(β)cos(α±β)=cos(α)cos(β)∓sin(α)sin(β)
The formulas for tangent are a bit more work, but
tan(α+β)=sin(α+β)cos(α+β)=sin(α)cos(β)+cos(α)sin(β)cos(α)cos(β)−sin(α)sin(β)=sin(α)cos(β)+cos(α)sin(β)cos(α)cos(β)−sin(α)sin(β)⋅sec(α)sec(β)sec(α)sec(β)=sin(α)sec(α)+sin(β)sec(β)1−sin(α)sec(α)sin(β)sec(β)=tan(α)+tan(β)1−tan(α)tan(β)
and similarly we get
tan(α−β)=tan(α)−tan(β)1+tan(α)tan(β)
Identities — Double-angle Formulas
If we set β=α in the angle-addition formulas we get
sin(2α)=2sin(α)cos(α)cos(2α)=cos2(α)−sin2(α)=2cos2(α)−1since sin2θ=1−cos2θ=1−2sin2(α)since cos2θ=1−sin2θtan(2α)=2tan(α)1−tan2(α)=2cot(α)−tan(α)divide top and bottom by tan(α)
Identities — Extras
Sums to Products
Consider the identities
sin(α+β)=sin(α)cos(β)+cos(α)sin(β)sin(α−β)=sin(α)cos(β)−cos(α)sin(β)
If we add them together some terms on the right-hand side cancel:
sin(α+β)+sin(α−β)=2sin(α)cos(β).
If we now set u=α+β and v=α−β (i.e. α=u+v2,β=u−v2) then
sin(u)+sin(v)=2sin(u+v2)cos(u−v2)
This transforms a sum into a product. Similarly:
sin(u)−sin(v)=2sin(u−v2)cos(u+v2)cos(u)+cos(v)=2cos(u+v2)cos(u−v2)cos(u)−cos(v)=−2sin(u+v2)sin(u−v2)
Products to sums
Again consider the identities
sin(α+β)=sin(α)cos(β)+cos(α)sin(β)sin(α−β)=sin(α)cos(β)−cos(α)sin(β)
and add them together:
sin(α+β)+sin(α−β)=2sin(α)cos(β).
Then rearrange:
sin(α)cos(β)=sin(α+β)+sin(α−β)2
In a similar way, start with the identities
cos(α+β)=cos(α)cos(β)−sin(α)sin(β)cos(α−β)=cos(α)cos(β)+sin(α)sin(β)
If we add these together we get
2cos(α)cos(β)=cos(α+β)+cos(α−β)
while taking their difference gives
2sin(α)sin(β)=cos(α−β)−cos(α+β)Hence
sin(α)sin(β)=cos(α−β)−cos(α+β)2cos(α)cos(β)=cos(α−β)+cos(α+β)2