Sec, Cosec, Cot

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

Differentiate Inverse Sine (arcsin) - proof

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)

by David Woodford

Differentiating Trigonometry

Sine and Cos Graphs Differentiating sin and cos

June 23, 2008

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))

Trigonometry Identities

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=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

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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.

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Using these a cos² + sin² = 1 we can calculate other identities

tan²t + 1 = sec^2t

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

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) = (e^x - e^-x)/2
cosh(x) = (e^x + e^-x)/2

And tanh can be defined as sinh/cosh so

tanh = (e^x - e^-x) / (e^x + e^-x)

though this is often written as
tanh = (e^2x - 1) / (e^2x + 1)
by timesing the top and bottom by e^x

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

Tan=Sin/Cos

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

Sine and Cosine Rules

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))