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Compound tan – tan(A+B)

September 24, 2009 Leave a comment

We can use expressions for cos(A+B) and sin(A+B) to help us find tan(A+B).

Using the identity tanx = sinx / cosx we can write

tan(A+B) = \frac{sin(A+B)}{cos(A+B)}

we can now substitue in
sin(A+B) = sinAcosB + sinBcosA
and
cos(A+B) = cosAcosB – sinAsinB
to get

tan(A+B) = \frac{sinAcosB + sinBcosA}{cosAcosB - sinAsinB}

We can now divide both the top and bottom by cosAcosB to get

tan(A+B) = \cfrac{\cfrac{sinAcosB + sinBcosA}{cosAcosB}}{\cfrac{cosAcosB - sinAsinB}{cosAcosB}}
or
tan(A+B) = \cfrac{\cfrac{sinAcosB}{cosAcosB} + \cfrac{sinBcosA}{cosAcosB}}{\cfrac{cosAcosB}{cosAcosB} - \cfrac{sinAsinB}{cosAcosB}}

We can now simplify this by cancelling any cosA and cosB to get

tan(A+B) = \cfrac{\cfrac{sinA}{cosA} + \cfrac{sinB}{cosB}}{1 - \cfrac{sinA sinB}{cosA cosB}}

finally by substituting the identity tan(x) = \frac{sinx}{cosx} we find our result

tan(A+B) = \cfrac{ tanA + tanB}{1 - tanAtanB}

And it can be shown that this result can be extended to

tan(A \pm B) = \cfrac{tanA \pm tanB}{1 \mp tanAtanB}

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Maclaurin Series with example sin(x)

September 23, 2009 Leave a comment

The Maclaurin series is a way of approximating a function f(x) using a power series of x. It only works on functions you can differentiate but you can gain any level of accuracy by stopping the series at different points. It is equal to the taylor series for a function about 0. One example of a use of the Maclaurin series is to calculate a value for sin(x).

The Maclaurin series for a function f(x) is as follows:

f(x) = f(0) + xf'(0) + \frac{x^2 f''(x)}{2!} + \frac{x^3 f'''(x)}{3!} + ...
or
f(x) = \sum_{i=0}^{\infty} \frac{x^i f^i (0)}{i!}

Example sin(x)

An expression for sin(x) can be calculated using this method. Firstly we must decide what level of accuracy to go to, for the example we will go to the 3rdterm. The next step is to differentiate sinx. Firstly let f(x) = sinx

then

f(0) = 0

f'(x) = cos(x) \quad \quad \quad \Rightarrow \quad \quad \quad f'(0) = 1

f''(x) = -sin(x) \quad \quad \quad \Rightarrow \quad \quad \quad f''(0) = 0

f'''(x) = -cos(x) \quad \quad \quad \Rightarrow \quad \quad \quad f'''(0) = -1

f''''(x) = sin(x) \quad \quad \quad \Rightarrow \quad \quad \quad f''''(0) = 0

f'''''(x) = cos(x) \quad \quad \quad \Rightarrow \quad \quad \quad f'''''(0) = 1

We can now combine these into the series to get

f(x) = sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + ...

f(x) = sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} + ...

which can be used to calculate the value of sin(x)  — though only for the radian measure of angle.

eg)
sin(\frac{\pi}{3}) = \frac{\pi}{3} - \frac{\pi ^3}{6 \cdot 3^3} + \frac{\pi ^5}{120 \cdot 3^5} + ... \simeq   0.8663

Which is approximately the value you would get if you type sin(pi/3) into a calculator.

Using Polar Co-ordinates and Converting to and from Cartesian

September 19, 2009 Leave a comment

Polar co-ordinates are a different co-ordinate scheme to the standard Cartesian co-ordinates.  Again any point in a 2D plane can be located using only two numbers.

It works by taking an “initial line” (shown in red — the equivalent of the positive x axis in Cartesian co-ordinates) with the origin at one end. Then any point can be found by drawing a line from the point to the origin and quoting the length of this line (the radius) and the angle the line makes with the origin. These numbers are usually written in brackets in the same way as Cartesian co-ordinates with the radius first followed by the angle. For example look at the point (3,π/3) below.

The point (3,pi/3) in Polar Co-ordinates

The point (3,pi/3) in Polar Co-ordinates

Note// The angles are usually measured in radians

Cartesian Equivalent of Polar Co-ordinates

Since both Cartesian and polar co-ordinates are a way of describing a point position in a 2D plane it is possible to convert between then. When doing this the initial line is taken as the x-axis.

To find the Cartesian co-ordinates we must use trigonometry by drawing a vertical line down from the point to the x-axis to form a right angled triangle. The length of the vertical line then gives the y- coordinate and its distance from the origin gives the x-coordinate.

To find the x and y values for the point (r,Θ) we must therefore use the equations

x = rcosΘ

and

y = rsinΘ

If we want to go in reverse to find the polar co-ordinates of the point (x,y) in a Cartesian system we must solve these equations simultaneously.

We can eliminate Θ by squaring both of the equations to obtain

x2 = r2cos2Θ
and
y2 = r2sin2Θ

and then adding these equations to get

x2 + y2 = r2(cos2Θ + sin2Θ)

by substituting the trigonometric identity sin2+cos2 = 1 to get

x2 + y2 = r2

we have removed Θ and can therefore calculate r using
r =√(x2 + y2)

To find Θ we can divide the two equations given at the start such that the r’s cancel to get
y/x = sin Θ/cos Θ
which using the identity tan Θ=sin Θ/cos Θ gives

Θ = tan-1(y/x)

Tan Graph – y=tan(x)

September 1, 2009 Leave a comment

The graph of y=tanx is different from the other cos and sin graphs as it has a range from -∞ to ∞ and a period of 180° or π radians. The graph of y=tan(x) in radians is shown below

Graph of y=tan(x) in radians

Graph of y=tan(x) in radians

As can be seen from the graph the curve passes through the origin. It has  vertical asymtopes (lines it tends toward but never touches — in this case where the graph goes to infinity) at x =π/2,3π/2,5π/2 and x=-π/2,-3π/2 etc radians or at x=90,270,450 and x=-90,-270 etc degrees.

The graph has a stationary (flat) point whenver it crosses the x-axis.

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: , ,

Compound Angles: Cos(A+B) = CosACosB – SinASinB

July 20, 2009 2 comments

Compound angles are angles made by adding two other angles together. When using trigonometry unfortunately you cant just “times out” the trig function but have to use an identity. This post will consider how we get the identity for cos(A+B):


Compound angle of A+B showing how they relate

Compound angle of A+B showing how they relate

From the definition of cos we find

cos(A+B) = OT/OR

but
OT = OP – PT
and PT = SQ so
OT = OP – SQ

so

cos(A+B) = ( OP – SQ ) / OR
so
cos(A+B) = OP/OR – SQ/OR

if we now times the both the top and bottom of the first term by OQ and do the same for the second term but with RQ we can get

compcos1

but OP/OQ = cosB,
OQ/OR = cosA,
SQ/RQ = sinB,
RQ/OR = sinA

so we get, when these are substituted in and re arranged

cos(A+B) = cosAcosB-sinAsinB

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: , ,