A Question of Definitions of Trig Functions

You may feel this question is really basic or consequently really tough to explain.

as we know sin=Opposite/Hypotenuse

so if you have given some value of Opposite side and some value of Hypotenuse then you can easily calculate the sine.

But What I really wanted to know is How sine ends up to be one of the side in right angled triangle and Why only the particular side is sin and not other.

I maybe not explaining it well.If so please let me know , I can try once again to explain , otherwise.

re: A Question of Definitions of Trig Functions

Quote:

Originally Posted by

**thapchi** But What I really wanted to know is How sine ends up to be one of the side in right angled triangle and Why only the particular side is sin and not other.

Yes, I am not sure I understand the question. First, as you said, the sine is a ratio, not a side, unless the hypotenuse has length 1. Second, it's the ratio of the opposite side (and not the adjacent side) to the hypotenuse by definition.

Are you asking why every right triangle with the same acute angle has the same ratio of the opposite side to the hypotenuse? This is because all such triangles are similar.

re: A Question of Definitions of Trig Functions

Quote:

Originally Posted by

**thapchi** You may feel this question is really basic or consequently really tough to explain.

as we know sin=Opposite/Hypotenuse

so if you have given some value of Opposite side and some value of Hypotenuse then you can easily calculate the sine.

But What I really wanted to know is How sine ends up to be one of the side in right angled triangle and Why only the particular side is sin and not other.

I maybe not explaining it well.If so please let me know , I can try once again to explain , otherwise.

Well, the sine function actually does show up in every side-ratio of a right triangle.

As you already noted:

$\displaystyle \sin\theta=\frac{\mathrm{opposite}}{\mathrm{hypote nuse}}$

Another ratio is:

$\displaystyle \cos\theta=\frac{\mathrm{adjacent}}{\mathrm{hypote nuse}}$

which is equal to:

$\displaystyle \sin(\theta+90^{\circ})=\frac{ \mathrm{adjacent} }{ \mathrm{hypotenuse} }$

since the cosine is in essence nothing more than a $\displaystyle 90^{\circ}$ phase-shifted sine.

Does that help?

Re: A Question of Definitions of Trig Functions

Welcome to MHF, thapchi! :)

According to wikipedia:

"The word comes from the Latin *sinus* for gulf or bay,[1] since, given a unit circle, it is the side of the triangle on which the angle opens."

Re: A Question of Definitions of Trig Functions

First of all , I am sorry for naming it stupidly , But I was not able to find a suitable title.

Thank you Moderator sir , to fix the title of the thread and making it more related to content.

And, thank you ILikeSerena :)

I kind of understood what corsica said, Well, I knew that but.

What I wanted to know is if sin is a ratio , then how come it is also one of the triangles side in the unit circle and/or why it is one of the triangles side when hypotenuse is 1.

Third question would be How do you decide which side of triangle is sin and which is cos

I mean , they are just in 90 degree phase shift that means you must be able to switch them, or not. If not kindly tell why?

I am sorry , I am making it more complex to understand and maybe now its sounding a bit stupid , but above are the question I feel I should know answer of.

Re: A Question of Definitions of Trig Functions

A right triangle has three sides so there are 3(2)= 6 ways of forming ratios. "Sine" is simply the name given to one of those ratios, cosine is the name given to another (the other four are "tangent", "cotangent", "secant" and "cosecant". Yes, it is true that $\displaystyle sin(\theta)= cos(\pi/2- \theta)$. You are free to use whichever of those ratios you wish in solving a problem.

Re: A Question of Definitions of Trig Functions

Definition says: sine = opposing / hypotenuse

If the hypotenuse = 1 (as it is in the unit circle), then:

sine = opposing / hypotenuse = opposing / 1 = opposing.

You can only switch sine and cosine if you shift the angle (like corsica showed).

But perhaps I do not understand your question...?

Re: A Question of Definitions of Trig Functions

Quote:

Originally Posted by

**thapchi** What I wanted to know is if sin is a ratio , then how come it is also one of the triangles side in the unit circle and/or why it is one of the triangles side when hypotenuse is 1.

It takes three variables to fix the shape of a triangle (this a very fundamental property of any triangle). If you know three side lengths and/or angles, then there's only one possible shape the triangle can have.

Going back to your example: you know two side lenghts, so that already gives you two variables. You also know that one angle is 90 degrees, because it's a right triangle (this is your third variable). **So because you know three variables, the shape of the triangle is fully determined.** That's why it's possible to have an equation like $\displaystyle \sin\theta= \mathrm{opposite} / \mathrm{hypotenuse}$

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Re: A Question of Definitions of Trig Functions

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Originally Posted by

**thapchi** Third question would be How do you decide which side of triangle is sin and which is cos.

Be careful when you say sine. It's not just a 'sine', but it's always the **sine of a specific angle in the triangle** that you should be talking about.

Take this right triangle:

Attachment 23156

Let's take a look at the side ratio a/c. There are two possible equations for this ratio, depending on which angle of the triangle you're going to tuse:

$\displaystyle \sin A=\frac{a}{c}$

$\displaystyle \cos B=\frac{a}{c}$

**The combination of two sides and angle that you're taking determines whether you use sine or cosine.**

Re: A Question of Definitions of Trig Functions

Thank you corsica and ILikeSerena very much for telling me opp/1 concept I never thought about it that way also Thank you everyone else :)

I am doing trigonometory from 3-4 years now , but I am still not cleared with basic concepts how things work , and I sometimes find trigonometry a pain as you have to memorize tons of formulas :(

Re: A Question of Definitions of Trig Functions

Quote:

Originally Posted by

**corsica** Be careful when you say sine. It's not just a 'sine', but it's always the

**sine of a specific angle in the triangle** that you should be talking about.

Take this right triangle:

Attachment 23156
Let's take a look at the side ratio a/c. There are two possible equations for this ratio, depending on which angle of the triangle you're going to tuse:

$\displaystyle \sin A=\frac{a}{c}$

$\displaystyle \cos B=\frac{a}{c}$

**The combination of two sides and angle that you're taking determines whether you use sine or cosine.**

So does that mean we can say that ,

$\displaystyle \sin B=\frac{b}{c}$

$\displaystyle \cos A=\frac{b}{c}$

Re: A Question of Definitions of Trig Functions

Quote:

Originally Posted by

**thapchi** So does that mean we can say that ,

$\displaystyle \sin B=\frac{b}{c}$

$\displaystyle \cos A=\frac{b}{c}$

That's exactly right! :)

Re: A Question of Definitions of Trig Functions

Thanks alot again :)

Well another question most probably last one this time.

Does all this trigonometry only limited on Right angled triangle ? can you find angle of any triangle?

Re: A Question of Definitions of Trig Functions

Quote:

Originally Posted by

**thapchi** Thanks alot again :)

Well another question most probably last one this time.

**Does all this trigonometry only limited on Right angled triangle ? can you find angle of any triangle?**

research the Law of Sines and the Law of Cosines.

Re: A Question of Definitions of Trig Functions

Quote:

Originally Posted by

**thapchi** Thanks alot again :)

Well another question most probably last one this time.

Does all this trigonometry only limited on Right angled triangle ? can you find angle of any triangle?

Recall that three parameters are enough to fix the shape of a triangle. So if you know three side lengths and/or angles, then you can always find all other side lengths and angles (this is true for any triangle imaginable).

As Skeeter already pointed out, you use the Law of Sines and Law of Cosines when you're dealing with a non-right triangle.