Loading [MathJax]/jax/output/HTML-CSS/jax.js
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

B.2 Trigonometry

( \newcommand{\kernel}{\mathrm{null}\,}\)

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 =360π radians =180π2 radians =90π3 radians =60π4 radians =45π6 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)=2πrθ2π=θrandA(θ)=πr2θ2π=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

sinθ
cosθ
tanθ

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

cscθ
secθ
cotθ

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 12 12 1 2 2 1
π6 rad 12 32 13 2 23 3
π3 rad 32 12 3 23 2 13

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 π2θ the side-lengths of the triangle remain unchanged, but the roles of “opposite” and “adjacent” are swapped. Hence we have

sin(π2θ)=cosθcos(π2θ)=sinθ

Again these imply that

tan(π2θ)=cotθcsc(π2θ)=secθsec(π2θ)=cscθcot(π2θ)=tanθ

We can go further. Consider the following diagram:

This implies that

sin(πθ)=sin(θ)cos(πθ)=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) triangles

    Ared=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 απ/2α and ββ into our sine formula:

sin(α+β)=sin(α)cos(β)+cos(α)sin(β)becomessin(π/2αβ)cos(α+β)=sin(π/2α)cos(α)cos(β)+cos(π/2α)sin(α)sin(β)cos(α+β)=cos(α)cos(β)sin(α)sin(β)

where we have used sin(π/2θ)=cos(θ) and cos(π/2θ)=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 ββ and so obtain

sin(αβ)=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)1sin(α)sec(α)sin(β)sec(β)=tan(α)+tan(β)1tan(α)tan(β)

and similarly we get

tan(αβ)=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=1cos2θ=12sin2(α)since cos2θ=1sin2θtan(2α)=2tan(α)1tan2(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(αβ)=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(αβ)=2sin(α)cos(β).

If we now set u=α+β and v=αβ (i.e. α=u+v2,β=uv2) then

sin(u)+sin(v)=2sin(u+v2)cos(uv2)

This transforms a sum into a product. Similarly:

sin(u)sin(v)=2sin(uv2)cos(u+v2)cos(u)+cos(v)=2cos(u+v2)cos(uv2)cos(u)cos(v)=2sin(u+v2)sin(uv2)

Products to sums

Again consider the identities

sin(α+β)=sin(α)cos(β)+cos(α)sin(β)sin(αβ)=sin(α)cos(β)cos(α)sin(β)

and add them together:

sin(α+β)+sin(αβ)=2sin(α)cos(β).

Then rearrange:

sin(α)cos(β)=sin(α+β)+sin(αβ)2

In a similar way, start with the identities

cos(α+β)=cos(α)cos(β)sin(α)sin(β)cos(αβ)=cos(α)cos(β)+sin(α)sin(β)

If we add these together we get

2cos(α)cos(β)=cos(α+β)+cos(αβ)

while taking their difference gives

2sin(α)sin(β)=cos(αβ)cos(α+β)

Hence

sin(α)sin(β)=cos(αβ)cos(α+β)2cos(α)cos(β)=cos(αβ)+cos(α+β)2


This page titled B.2 Trigonometry is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform.

Support Center

How can we help?