Archive

Posts Tagged ‘Trigonometry’

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.

Advertisements

Auxiliary Angle Method for Solving Trigonometry Equations

April 19, 2009 2 comments

This is a method of solving equations in the form asinx+bcosx = c where a and b are constants and c is another expression.

It involves rewriting letting asinx + bcosx = rsin(x+y) (or you could use cos(x+y)) where y is acute and then finding values for r and y, then with only one trig function to deal with the equation can be solved more easily.


For example

consider 2sinx + 3cosx = 3

Let 2sinx + 3cosx = rsin(x+y)

Now expand the sin(x+y) to get

2sinx + 3cosx = rsinx cosy + rcosx siny

Since y is constant and therefore cosy and sin y are constant we can compare the coefficients to get

2 = rcosy —–(1)
3 = rsiny ——(2)

We can solve these to find values for r and y.
To find y consider (2)/(1) to get

3/2 = tany
since sin/cos = tan and the r’s cancel
so y = 56.3 °

To find r consider (1)2+(2)2 to get
22+32 = r2
since sin2+cos2 = 1
so r =√13

So we can write

2sinx + 3 cosx = √13 cos(x+56.3) = 3

so x = cos<sup>-1</sup>(3/√13) -56.3

so x = cos-1(3/√13) -56.3

since cos-1(3/√13) = 33.7 for solutions between 0° and 90°

x = ±33.7 -56.3 + 180n where n is an integer


In General

asinx + bcosx = √(a2+b2) sin(x+tan-1(b/a))

If you have any questions, comments or corrections please leave them as a comment below

By David Woodford

Compound Angles – sin(A+B) = cosAsinB+sinAcosB

February 7, 2009 4 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 sin(A+B):

sin(A+B) = sinAcosB+sinBcosA

Compound angle of A+B showing how they relate

Compound angle of A+B showing how they relate

From the definition of sin=opp/hyp we find

sin(A+B) = RT/OR

But sinceRT comprises of RS+ST

Compund Angle derivation

By David Woodford

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.

Hyperbolic Functions

December 10, 2008 Leave a comment

This post can new be viewed from www.breakingwave.co.nr :):)

Hyperbolic functions are similar to sin,cos tan etc in trigonometry and share many similar rules. Usually hyperbolic functions are written like the trigonometric ones but with a h on the end, eg sinh and cosh.

The hyperbolic functions can be all written in terms of e, sinh and cosh are as follows

sinh(x) = (ex – e-x)/2
cosh(x) = (ex + e-x)/2

And tanh can be defined as sinh/cosh so

tanh = (ex – e-x) / (ex + e-x)

though this is often written as
tanh = (e2x – 1) / (e2x + 1)
by timesing the top and bottom by ex

the other other hyperbolic functions sinh as sech, coth etc can be found in the same way as they would be in trigonometry, by using 1 over the other functions, ie sech = 1/cosh

Most of the identities in trigonometry have a similar identity with hyperbolic functions, however in most of these whenever there is a sin2 it changes to a -sinh2
so
cosh2 – sinh2=1
which you can work out by placing the equations with e’s in the place of sinh and cosh


IF you have any questions pleas ask below

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: , , , ,