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Posts Tagged ‘cos’

The Chain Rule

November 21, 2009 Leave a comment

The chain rule allows you to differentiate composite functions (functions of other functions) ie) f(g(x)) such as sin(3x2) or (5x3+2x+3)2. The rule is as follows
\frac{d}{dx}(f(g(x)) = \frac{dg}{dx}\frac{df}{dg}(g(x)) = g'(x)f'(g(x))
or to understand it more simply you differentiate the inner function and multiply it by the derivative of the outer function (leaving what’s inside alone).

Differentiating brackets raised to a power

The chain rule can be a great short cut to differentiating brackets raised to a power as it doesn’t require you to multiply them all out, it also enables you to differentiate brackets raised to an unknown power.
Consider \frac{d}{dx}((ax + b)^n)
This is the composite of the functions ax+b and tn. So we differentiate them both to get a and ntn-1 and then apply the formula to get
\frac{d}{dx}((ax + b)^n) = an(ax+b)^{n-1}
Notice how we multiplied the derivative of the inner function, a, by the derivative of the outer function ntn-1 but substituted ax+b back in for t.

To generalise we can replace the ax+b with f(x) and by applying the above get
\frac{d}{dx}((f(x))^n) = f'(x)n(f(x))^{n-1}

Differentiating Trigonometric functions

We can also use the chain rule when differentiating sin(f(x)) and cos(f(x)) since we know how to differentiate sin(x) and cos(x).
Using the chain rule we get
\frac{d}{dx}(sin(f(x)) = f'(x)cos(f(x))
and
\frac{d}{dx}(cos(f(x)) = -f'(x)sin(f(x))

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.

y = cos(x) - in radians

y = cos(x) - in radians

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

Categories: maths, trigonometry Tags: , ,

Sec, Cosec, Cot

January 10, 2009 2 comments

Sec, cosec and cot are all functions in trigonometry. They are simply equal to one over on of the other functions, ie cos, sin and tan.

so

Sec = 1/cos

Cosec = 1/sin

cot = 1/tan

You can remember which is paired with which using the third letter rule. This is that the third letter is the first letter of the corresponding function ie)

sec goes with cos
cosec goes with sin
cot goes with tan

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.

Tan = sin/cos

June 23, 2008 14 comments

this site is now at www.breakingwave.co.nr

This is often useful when solving trig equations so i thought i’d include it

basically:

sin = opp/hyp
and
cos=adj/hyp

so

sin/cos = (opp/hyp)/(adj/hyp)

so if we cancel the hyp’s we get

sin/cos = opp/adj

and since tan = opp/adj

tan = sin/cos

Categories: maths Tags: , , , ,

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))

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