## Cosine Graph – y = cos x

The cosine graph is similar to the sine graph (it moves between 1 and -1 over a period of 180 degrees or 2π radians) but is shifted to the left by 90 degrees or π/4 radians. The graph of y=cos x is shown below.

Unlike the sine graph the cosine graph is an even function as it is symmetrical about the y axis. It has a maximum value of 1 and a minimum value of -1

By David Woodford

## Trigonometry Identities

There a number of “identities” in trigonometry that can be found from the basic ideas of sin, cos and tan as explained in my earlier post. These identities can help in solving equations involving trig functions, especially when there are 2 or more different functions as the often allow you to write the equation in terms of one function, eg sin, that you can then solve. One of the identities is: **sin ^{2} + cos^{2} = 1. ** To prove this consider a right angled triangle with side a,b and c as shown below

From this we can use Pythagoras theorem to say: a^{2}+b^{2}=c^{2 now we know sin t = b/c so b = csin t cos t = a/c so a = ccos t} substituting these values in the above equation we get (csint)^{2} +(ccost)^{2} = c^{2} canceling the c^{2} we get **sint ^{2} + cost^{2} = 1**

There are trig functions that are equal to 1 over sin, cos and tan called cosec = 1/sin, sec = 1/cos and cot = 1/tan. These can be remembered using the third letter rule as the third letter of each of these corresponds to the the function it is one over.

Using these a cos

^{2}+ sin

^{2}= 1 we can calculate other identities

**tan**We can obtain this by dividing through by cos

^{2}t + 1 = sec^{2}t^{2}as we know sin/cos = tan, cos/cos = 1 and 1/cos = sec. Other similar identities can be obtained for cosec and cot.

## Sine and Cos Graphs Differentiating sin and cos

This is the basics of the sine cos and tan graphs and how sine and cos relate to give you tan. It also shows how to differentiate sin and cos.

The output or range of both sine and cos is from -1 to 1 when given any angle. They can be shown on a graph where y = sin(x) and y = cos(x). In these graphs all the angles go along the x axis and you can see a wave type shape is formed

**Sine Graph**

**Cosine Graph**

As you can see both the sin and cos graphs move periodically between -1 and 1 as the angles change, this pattern continues indefinitely because once you pass 360 degrees or 2 pi radians you will return back to the beginning. If you try to perform sin^{-1} of a value out side the range -1 to 1 you will get an error.

**Differentiate Sin and Cos**

also notice that the gradient of the sin graph is the value of the cos graph for the same angle and that the gradient of the cos graph is the -value of the sin graph for that angle. This means that we can differentiate the sin and cos graphs:

if f(x) = sin(x) then f ‘ (x)=cos(x)

and

if f(x) = cos(x) then f ‘ (x) = -sin(x)

however if we use ax instead of x we must differentiate it by bringing the a out, when its just x this doesn’t matter as the differential of x is 1.

ie)

let y = sin(f(x))

now let u = f(x)

du/dx = f ‘ (x)

also

y=sin(u) as u = f(x)

dy/du = cos(u)

from the chain rule

dy/dx = du/dx * dy/du

therefore

**if y = sin(f(x))
dy/dx = f ‘ (x)cos(f(x))**

**and similarly for cos
if y = cos(f(x))
dy/dx = -f ‘ (x)sin(f(x))**

## Proof of Cosine Rule

Below is the proof by Pythagoras’s theorem of the cosine rule, a^{2}=b^{2}+c^{2}– 2bccosA.

This assumes you understand Pythagoras’s theorem (visit pythagoras’s theorm to view my lesson on it), how to use basic trigonometry(basic trigonometry lesson). If you want to learn how to use the cosine and sine rule, opposed to just learning the proof) visit by sine and cosine rule page.

The proof is done using the letters of the following triangle

and we are trying to prove the cosine rule:

**a ^{2}=b^{2}+c^{2}– 2bccosA**

**In triangle CBL**

a^{2} = (c-x)^{2} + h^{2}

a^{2} = c^{2} – 2cx + x^{2} + h^{2}

h^{2} = a^{2} -c ^{2}– x^{2} + 2cx *<<EQN1*

**in triangle CLA**

b^{2} = h^{2} + x^{2}

h^{2} = b^{2} – x^{2 }* <<EQN2*

*eqn1 – eqn2 ::* 0 = a^{2} – c^{2} – b^{2} +2cx

a^{2} = c ^{2}+ b^{2} – 2cx *<<EQN3*

** in CLA**

cosA = x/b

x = bcosA

**in eqn3**

**a ^{2} = c^{2} + b^{2} – 2bccosA**

So there is the proof for the cosine rule using pythagorases therom. If you found that usefull try looking at my other maths lessons

## Understand the Sine and Cosine Rules

This assumes you already have a knowledge of basic trigonometry(ir using sin, cos and tan in a right angled triangle, if you don’t click here to read my lesson on these) and aims to teach you how to use the sine and cosine rule.

In basic trigonometry you can only look at a right angled triangle which greatly limits its applications, however with these formula you can calculate sides and angles in any triangle provided you know enough information. They are proved by splitting one triangle in 1/2 so that the dividing line is perpendicular to one of the sides and therefore creating 2 right angled triangle in which the normal rules can be applied.

The following use symbols as defined in the above triangle. Note that side a is opposite angle A and b is opposite B etc

**Sine Rule**

a/sinA = b/sinB = c/sinC

This allows us to find both an angle and a side as we can invert all of the fractions and it remains true. This means if we know the side opposite the angle we want and any other side angle pair we can work out the angle we want, or we can work out a side if we know the angle opposite it and any other side angle pair.

EG)Lets say

a = 10cm

b = 5cm

B = 30^{o}

and we want to find angle A

we know a side angle pair, b and B, and we know the side opposite the angle we want so we can write the sine rule as

sinA / 10 = sinB/b >>note we don’t need to include the c parts as we dont know either c or C

sinA / 10 = sin30/5

sinA = 10sin30/5

sinA = 1

A = sin^{-1}1

A = 90^{o}

We can work out any angle or side in a similar way.

**Cosine rule**

This rule allows us to find an angle if we know all the sides or a side if we know the other 2 and a angle

c² = a² + b² – 2abcosC

To find an angle we can re-arrange it so

C = cos^{-1}((a^{2} + b^{2} – c^{2})/2ab)

Im sure you can put the numbers in yourself as ive show you how it can be written to find either an angle or side so ill leave you to it 🙂 enjoy

If you have any questions, improvements, or suggestions please leave a comment below or email me at woodford_4@hotmail.co.uk. Also visit my site at www.breakingwave.co.nr

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