## Compound tan – tan(A+B)

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

we can now substitue in

sin(A+B) = sinAcosB + sinBcosA

and

cos(A+B) = cosAcosB – sinAsinB

to get

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

or

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

finally by substituting the identity we find our result

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

## Tan Graph – y=tan(x)

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

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.

## 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)

se**c **goes with **c**os

co**s**ec goes with **s**in

co**t** goes with **t**an

## 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**

## Trigonometry: Sin, Cos and Tan

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.

Soon ill be adding a maths section to my site Breakingwave

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