Trevor Pythagoras Maths » cos

CAST Diagarams for finding values of sin, cos and tan between 0 and 360 or 2 pi

admin — Sat, 05 Jun 2010 21:04:43 +0000

When we work out the inverse of sin cos and tan of a positive number we always find a value between and or 90^o however between 0 and or 360^o there are more values for inverse trig functions, you can see where these are by looking at a graph (below). The CAST diagram is a method of working out these other values so that we can find all the solutions of sin(x)=a,cos(y)=b,tan(z)=c for x,y,z with a,b,c constants.

There are two solutions to sin(x)=0.65 we get solution (1) when we take the inverse but we need a way of finding (2) from (1)

The four sections of the CAST are cos,all,sin and tan starting by labelling the bottom right and working round in an anti-clockwise direction.

Use the following method to work out the values for the inverse between 0 and 2pi or 360

find the value of the inverse of the positive between 0 and 90 or pi/2 using a normal method (eg a calculator).
So if we want to find sin^-1(0.6) we calculate 0.6435 rads or 36.87 deg and if we want sin^-1(-0.8) we calculate sin^-1(0.8) to get 53.13 deg or 0.927 rads
draw in the four lines on the CAST diagram (shown in green) that represent the angle. Do this by measuring from the horizontal the angle calculated in 1
If the value we are finding the inverse for is negative (eg we are finding sin^-1(-0.6)) consider the quadrants that don’t include the name of the function. (so if we are finding inverse sin only consider Tan and Cos, if we are finding inverse cos only consider sin and tan and if we are finding inverse tan only consider sin and cos)
If the value we are finding the inverse for is positive (eg we are finding sin^-1(0.6)) only consider the quadrants with the name of the function and all. (so if we are finding inverse sin we only consider sin and all, if we are finding inverse cos only consider only consider cos and all and if we are finding inverse ta only consider tan and all)
Calculate the angle from the zero line anti-clockwise to the lines in the quadrant’s we are considering. The angle labels I’ve put on the axes should make this easier. These values are your solutions.

All four lines representing the angle x drawn but we will only be interested in two of them

You can check that this works by putting the values back into you calculator and if you want to check that you’ve got all of the solutions check against the graph of the function.

Example

Find all the solutions of cos(x) = 0.7 between 0 and 360^o or rads.

We can use a calculator to find the value between 0 and 90 deg or pi/2 rads
so write y = cos^-1(0.7) = 45.57 deg (we’ll work in degrees for the example to avoid repeating all the calculations)

We no draw this on the cast diagram and choose the two lines we need (in red), in this case the lines in the cos and “all” quadrants.

We then calculate the anti-clockwise angles to these lines from the horizontal:
so we get 45.57 deg for the in the “all” quadrant and 360-45.57=314.43 for the cos quadrant
so our set of solutions for x between 0 and 360 of cos(x)=0.7 are 45.57 and 314.43

Also see

Sec, Cosec, Cot (0)
Tan = sin/cos (6)
Sin, Cos and Tan (15)
Integrate cos and sin squared using double angles (2)
The Chain Rule (0)
Trigonometry Identities (11)

Integrate cos and sin squared using double angles

admin — Wed, 27 Jan 2010 17:08:40 +0000

The integration of cos²x and sin²x comes up quite a lot and an easy trick for finding them is to use cos(2x). You do this using the following identities:

and

These are derived from the formula for double cos as follows

by subsitituion of

and similarly for sin²x but using the substitution for cos²x instead of sin²x.

To use these identities we simply substitute them into the integral and find the integral as normal since we know

so we get

and

Also see

The Chain Rule

admin — Sat, 21 Nov 2009 15:17:27 +0000

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

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
This is the composite of the functions ax+b and tⁿ. So we differentiate them both to get a and nt^n-1 and then apply the formula to get

Notice how we multiplied the derivative of the inner function, a, by the derivative of the outer function nt^n-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

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

and

Also see

Cosine Graph – y = cos x

admin — Wed, 26 Aug 2009 11:08:28 +0000

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

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

Also see

Sec, Cosec, Cot

admin — Sat, 10 Jan 2009 16:56:41 +0000

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.

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

Also see

Trigonometry Identities

admin — Tue, 23 Dec 2008 15:09:09 +0000

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

From this we can use Pythagoras theorem to say: a²+b²=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 = 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.

Also see

Proof of Cosine Rule (4)
Understand the Sine and Cosine Rules (11)
Sin, Cos and Tan (15)
Proof of Pythagorases Theroem (0)
Sine and Cos Graphs Differentiating sin and cos (7)
Area of a Triangle (3)

Tan = sin/cos

admin — Mon, 23 Jun 2008 20:29:02 +0000

This is often useful when solving trigonometric

We know that

sin = opp/hyp
and
cos=adj/hyp

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

If we cancel the hyp’s we get

sin/cos = opp/adj

and since tan = opp/adj

tan = sin/cos

Also see

Sine and Cos Graphs Differentiating sin and cos

admin — Mon, 23 Jun 2008 20:23:30 +0000

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))

Also see

Sin, Cos and Tan

admin — Sun, 03 Feb 2008 21:57:47 +0000

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

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.