Archive

Posts Tagged ‘sine’

Sine Graph

The sine function is a periodic function meaning that it repeats itself every so many (in the case of sine 2pi radians or 360o). It has a range of -1 to 1 and has a domain for -∞ to ∞. Starting at the origin it increase to 1 at 90<sup>o</sup> or pi/2 radians and then decrease to -1 at 270<sup>0</sup> or 3pi/2 radians and then returns to 0 and 360<sup>o</sup> or 2pi radians.

On the graph below the angle, in radians, is along the x axis and the value of the sine function for that angle is on the y axis.

Graph of y=sin(x)

Graph of y=sin(x)

Categories: maths, trigonometry Tags: , ,

Differentiate Inverse Cos – Proof

January 8, 2009 7 comments

Now available from trevorpythag.blogspot.com

How to differentiate cos-1x

y=cos-1x
Bring the cos across
cosy = x
Differentiate both sides, remember when differentiating y time by dy/dx
-sin(y) dy/dx = 1
dy/dx = -1/siny

However we want to get the differential in terms of x, to do this we can use the identity
sin2t+cos2t = 1
so
sint = √(1 – cos2t)

putting this into our expression for dy/dx we get

dy/dx = -1/√(1-cos2y)
but cosy = x so

dy/dx =- 1/√(1-x2)

Differentiate Inverse Sine

Now avalible from trevorpythag.blogspot.com

This tutorial explain how to differentiate inverse sine, this applies when using radians.

begin with

y = sin-1 x
bring sin-1 across to become sin
sin y = x
differentiate
cos y dy/dx = 1
note that the derivative of sint wrtt is cos t as explained in an earlier tutorial and by the chain rule when we differentiated sin y it became cosy time dy/dx as we are differnetiatiny a y and the derivative of y is dy/dx

then make dy/dx the subject

dy/dx = 1/cosy

We know the identity
sin2t + cos2t = 1
so we can wrtie
cos t =√(1 – sin2t)

we can now put this into the expression for dy/dx to get
dy/dx = 1/√(1 – sin2y)
but we know from the second line that sin y = x so

dy/dx = 1/√(1 – x2)

Categories: maths Tags: , , ,

Trigonometry Identities

December 23, 2008 11 comments

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: sin2 + cos2 = 1. To prove this consider a right angled triangle with side a,b and c as shown below

Right Angled Triangle

From this we can use Pythagoras theorem to say: a2+b2=c2 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 = c2 canceling the c2 we get sint2 + cost2 = 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 cos2 + sin2 = 1 we can calculate other identities tan2t + 1 = sec2t We can obtain this by dividing through by cos2 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

June 23, 2008 7 comments

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
graph of y=sin(x)

Cosine 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))

Understand the Sine and Cosine Rules

February 13, 2008 11 comments

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.

Sine/Cosine rule triangle

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 = 30o
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-11
A = 90o

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((a2 + b2 – c2)/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

Trigonometry: Sin, Cos and Tan

February 3, 2008 15 comments

This is the basics of using sine, co-sine and tangent for a right angled triangle. To do this you’ll probably need a scientific calculator

Trigonometry Triangle

To perform calculations we are going to use the triangle above.
The three main relationships are:

Tan(x) = o/a
Sin(x) = o/h
Cos(x) = a/h

so if h = 5 and x = 30
a = Cos(30)h = 4.330

We can also use a inverse of the functions
ie) x = tan-1(o/a)
x = sin-1(o/h)
x = cos-1(a/h)

so if o = 5 and a = 10
x = tan-1(5/10) = 26.565

Using this information we can work out any side or angle in a right angled triangle as long as we have to other pieces of information (like a side and a angle or 2 sides). This is used a lot in resolving forces in physics and allows us to derive some other more complex equations.

Soon ill be adding a maths section to my site Breakingwave