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# Trigonometry (Law of Cosines)

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In case of right angled triangles, C^2 = A^2 +B^2 - 2AB cos(Ψ) is shortened to C^2 = A^2 +B^2 because the cosine of the angle "Ψ" which is 90° is equal to 0. But how is its cosine equal to 0.

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This illustration of sine and cosine curves might help: https://www.mathsisfun.com/algebra/trig-sin-cos-tan-graphs.html

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1 minute ago, Strange said:

This illustration of sine and cosine curves might help: https://www.mathsisfun.com/algebra/trig-sin-cos-tan-graphs.html

I know that but they don't answer my question.

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One way of defining cosine is as proportional to the horizontal projection of what you call B in your figure. The segment B would be the one that sweeps the angle. That's why it's zero.

The sine of 90º is 1, on the contrary, because the vertical projection of B is just B.

1 hour ago, King E said:

I know that but they don't answer my question.

The post does answer your question, although it may be difficult to see because it's moving. Take a look at the horizontal projection of the triangle when it goes through 90º.

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2 hours ago, King E said:

In case of right angled triangles, C^2 = A^2 +B^2 - 2AB cos(Ψ) is shortened to C^2 = A^2 +B^2 because the cosine of the angle "Ψ" which is 90° is equal to 0. But how is its cosine equal to 0.

Did you really learn about the cosine rule before you learned a definition of the cosine?

What is your definition of a cosine?

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2 hours ago, King E said:

I know that but they don't answer my question.

Can you explain why not?  Or ask your question more clearly

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Posted (edited)
27 minutes ago, Strange said:

50 minutes ago, studiot said:

27 minutes ago, Strange said:

Can you explain why not?  Or ask your question more clearly

Yes because you find faults in the question whenever your answer doesn't help.

Edited by King E

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18 minutes ago, King E said:

Yes because you find faults in the question whenever your answer doesn't help.

Uh, OK.

Did this answer help more:

1 hour ago, joigus said:

The sine of 90º is 1, on the contrary, because the vertical projection of B is just B.

(I thought it was spot on.)

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1 minute ago, Strange said:

Uh, OK.

Did this answer help more:

(I thought it was spot on.)

Yes it helped and I got my answer.

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Posted (edited)
7 hours ago, King E said:

Yes it helped and I got my answer.

Well I'm glad you got your answer.

However I'm disappointed with your response to my earlier question, which I asked for a particular reason, I had intended to develop for your benefit.

In my first encounter with cosines I was given this definition, which referred to right angled triangles only.
Pythagoras had already been proven.

$\cos \psi = \frac{{adjacent}}{{hypotenuse}}$

Using this definition can you see that there is a problem with this, in respect of your question ?

Since c has to be the hypotenuse, which of the two sides corresponds to the adjacent side ?  a or b ?

Clearly there are two possible answers here and the solution to this problem is geometrically interesting.

It is very important to get this straight in your head because both Pythagoras and the Cosine rule are deeply embedded in fundamental Physics.

Edited by studiot

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Another approach cos(x)=sin(90-x), so cos(90)=sin(0).  In a right triangle, it is obvious sin(0)=0.

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