# Trignometry

• Sep 27th 2009, 10:31 PM
rohith14
Trignometry
Angle is introduced and there by we can make a triangle.How trignometric ratios are intoduced?I am unable to understand myself why sinA=opp/hyp(rt anld trile).It is difficult to me to understand trignometric ratio though i saw hds of problems up to now.(sorry if i have asked you a wrong qestion)
• Sep 28th 2009, 12:45 AM
gtmarnisel
Quote:

Originally Posted by rohith14
Angle is introduced and there by we can make a triangle.How trignometric ratios are intoduced?I am unable to understand myself why sinA=opp/hyp(rt anld trile).It is difficult to me to understand trignometric ratio though i saw hds of problems up to now.(sorry if i have asked you a wrong qestion)

uhmm rohith i think this would help SOH CAH TOA
Sine; Opp/Hyp
Tangent; Opp/Adj try to memorise this expression its quite rythmic [SOH CAH TOA]
• Sep 28th 2009, 02:06 AM
Hello rohith14
Quote:

Originally Posted by rohith14
Angle is introduced and there by we can make a triangle.How trignometric ratios are intoduced?I am unable to understand myself why sinA=opp/hyp(rt anld trile).It is difficult to me to understand trignometric ratio though i saw hds of problems up to now.(sorry if i have asked you a wrong qestion)

Think about triangles that have angles of $\displaystyle 30^o, 60^o$ and $\displaystyle 90^o$. Do you agree that, no matter how big or small they are, they all have the same shape? (In fact, in mathematics, we say they are similar.)

Well, whenever you draw a triangle this shape, you'll always find that the length of the shortest side is exactly one-half the length of the longest side. (Try it and see!) So there's something special about that fraction $\displaystyle \tfrac12$, as far as this triangle is concerned: it is the ratio of the smallest side to the longest.

Now we call (as you probably know) the longest side of a right-angled triangle the hypotenuse; and you'll see that the smallest side in our $\displaystyle 30-60-90$ triangle is opposite to the $\displaystyle 30^o$ angle. So, as far as this angle of $\displaystyle 30^o$ is concerned, we call this side the opposite (that seems fair, doesn't it?).

And what we have just discovered is that, for an angle of $\displaystyle 30^o$ in a right-angled triangle, the ratio of the opposite side to the hypotenuse is exactly $\displaystyle \tfrac12$. We call this ratio the sine of the angle, and we can write this as a formula:

$\displaystyle \text{Sine of angle} = \frac{\text{opposite}}{\text{hypotenuse}}$

Well, if we now draw a right-angled triangle a different shape, we shall have a different angle and a different ratio, but the formula that I've just given you is the one that will tell you (if you know the size of the angle) what that ratio is.

Trigonometry is a very big subject, but that should be enough to get you started. We can define cosine and tangent in a similar way, using the name of the third side of the right-angled triangle, which is the adjacent.

These definitions are:

$\displaystyle \text{Cosine of angle} = \frac{\text{adjacent}}{\text{hypotenuse}}$

$\displaystyle \text{Tangent of angle} = \frac{\text{opposite}}{\text{adjacent}}$

• Sep 28th 2009, 06:35 AM
pacman
this pics may help from Steven G. Johnson . . . .

g
• Sep 29th 2009, 02:12 AM
rohith14
Trignometry
In the same way could you explain me about sin0^o,sin180^o.
• Sep 29th 2009, 02:43 AM
Hello rohith14
Quote:

Originally Posted by rohith14
In the same way could you explain me about sin0^o,sin180^o.

$\displaystyle \sin0^o$ is simply what you get if the angle in the right-angled triangle closes right up to zero. The length of the opposite side is then reduced to zero, and the fraction $\displaystyle \frac{\text{opposite}}{\text{hypotenuse}}$ then has the value zero.

So $\displaystyle \sin0^o = 0$

For angles greater than $\displaystyle 90^o$, you need an extended definition for sine, cosine and tangent. You can study the diagram that pacman has posted if you wish, or, for angles up to $\displaystyle 180^o$, you can use the formulae:

$\displaystyle \sin\theta = \sin(180^o-\theta), \cos\theta = -\cos(180^o-\theta), \tan\theta = -\tan(180^o-\theta)$

This will mean, for instance, that $\displaystyle \sin180^o = \sin(180^o-180^o)=\sin0^o=0$

• Sep 29th 2009, 03:09 AM
rohith14
Trignometry
Does opp,adj,hyp exist if angle is reduced to zero?
• Sep 29th 2009, 04:03 AM
Hello rohith14
Quote:

Originally Posted by rohith14
Does opp,adj,hyp exist if angle is reduced to zero?

No, in fact they don't - I was giving you the simple answer, assuming that you just wanted to handle right-angled triangles, where the smaller angles of the triangle lay between $\displaystyle 0^o$ and $\displaystyle 90^o$.

For a more mathematically complete reply, you need to re-define sine, cosine and tangent, using a diagram similar to pacman's, which uses co-ordinate geometry. It goes something like this:

Consider a circle centre $\displaystyle (0,0)$, radius $\displaystyle 1$ unit. A is the point $\displaystyle (1,0)$ where the circle cuts the positive $\displaystyle x$-axis. A point P then starts at A and moves anticlockwise around the circle. The angle $\displaystyle \theta$ is the angle through which the radius OP has turned.

Then we can define $\displaystyle \sin\theta$ and $\displaystyle \cos\theta$ as the $\displaystyle y$- and $\displaystyle x$-coordinates of P, respectively. In other words, P is the point whose co-ordinates are $\displaystyle (\cos\theta, \sin\theta)$.

This, then, provides a definition of sine and cosine for all possible angles - including negative ones, and angles more than $\displaystyle 360^o$.

The definition of the tangent of an angle $\displaystyle \theta$ is then simply:

$\displaystyle \tan\theta = \frac{\sin\theta}{\cos\theta}$

(You need to refine this a little to take account of what happens when $\displaystyle \cos\theta = 0$, but that's another story!)