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A question

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Collin Li:

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To prove this with algebra and trigonometry, start from the RHS, realise it is an addition of two angles (or subtraction), so your next step should be: of it. Note I did not use , because then the expanded version of the compound angle formula for would result in (undefined) coming up.

Since this is boring, and I demonstrated it above, I will explain the geometric meaning of this proof, by proving it geometrically:

Let , and draw a right-angled triangle of angle theta. The opposite is length 1 and the adjacent is length so that .

Now, let the remaining angle be (by definition, using the fact that the interior angles of a triangle add up to ).

From triangle (reading the opposite and adjacent relative to ):








(as required)

I'm not sure whether the first proof () can be done as easily.

AppleXY:
WOW. That's awesome man. Thanks and yeah :P

Glockmeister:

--- Quote from: coblin on March 20, 2008, 01:52:38 am ---
--- Quote from: Glockmeister on March 20, 2008, 01:36:40 am ---

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Let and

Since inverse trigonometric functions represent angles, realise that this is the sum of two angles:







Using triangles, or the trigonometric identity:

, and







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Thanks for that man, knew it had to have something to do with the inverse.

Mao:

--- Quote from: coblin on March 20, 2008, 08:28:10 am ---I'm not sure whether the first proof () can be done as easily.

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it turns out a lot simpler :P

by the same token, if we assume a right angled triangle with hypotenuse of 1, angle of and opposite of :





let be the complementary angle







adding these two together:

Ahmad:
This might be obvious, but another method:

Let



i.e. constant function





Works the same with the other problems.

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