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## Summary:

- 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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- #1

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## Summary:

- 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.

I

- #2

etotheipi

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##\cos{90^{\circ}} = 0##

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$$

\sin^2 \psi + \cos^2 \psi = 1$$

$$if~\psi = 90^{\circ} ~and~accepting ~\sin 90^{\circ} = 1$$

$$1 +\cos^2 90^{\circ} = 1$$

$$\cos^2 90^{\circ} = 0 $$

$$\cos 90^{\circ}=0

$$

- #4

etotheipi

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There are a few ways of determining the values of trigonometric functions. A useful tool is the so-called unit circle, which consists of a circle of radius 1 centred on the origin in the ##x##-##y## plane. The coordinates of any point on the circle at an angle ##t## to the positive ##x## axis are ##(\cos{t}, \sin{t})##, as shown:

If ##t = \frac{\pi}{2} = 90^{\circ}##, the point is at the top of the circle. The ##x##-coordinate is ##0##, and this is the cosine of the angle. It's also useful for determining the signs of trigonometric functions in the different quadrants (I've seen this called the "CAST Diagram" before).

Really you should memorise how ##\sin##, ##\cos## and ##\tan## behave at ##\theta = 0##, ##\theta = \frac{\pi}{6}##, ##\theta = \frac{\pi}{4}##, ##\theta = \frac{\pi}{3}## and ##\theta = \frac{\pi}{2}##. You can use the unit circle and trigonometric identities to deduce the values for lots more angles from them.

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symbolipoint

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Not Correct.But how is its cosine equal to 0.

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symbolipoint

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You can find the proof of Law Of Cosines in some Trigonometry books and some Calculus books. There is an old book by Anton which includes a proof.Summary::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.

IView attachment 264987

- #7

mathwonk

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My favorite proofs are propositions 12, 13 Book II, Euclid's Elements.

- #8

neilparker62

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cos 90 = sin(90-90) = sin(0) = 0.Summary::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.

Does that now beg the question "how is sin(0) = 0" ?

- #9

etotheipi

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But you have turned the question into something that is much easier to answer, if you know or can derive the series expansions. Namely...cos 90 = sin(90-90) = sin(0) = 0.

Does that now beg the question "how is sin(0) = 0" ?

##\sin{0} = 0 - \frac{1}{3!}0^3 + \frac{1}{5!}0^5 + \dots = 0##

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neilparker62

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Last edited:

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neilparker62

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My favorite proofs are propositions 12, 13 Book II, Euclid's Elements.

- #12

etotheipi

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Pythagoras on the left triangle: ##x^2 + h^2 = b^2##

Pythagoras on the right triangle: ##a^2 - 2ax + (x^2 + h^2) = c^2##

Now since ##b\cos{C} = x##, we finally have ##a^2 + b^2 - 2ab\cos{C} = c^2##.

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Oh you English people! I can see that you have not even used the scale for drawing that perpendicular ##h## and that ##h## is very pointy, it has sharp end points. It’s really an English’s writing (so tough to read).Another way, label the triangle like this:

View attachment 265361

Pythagoras on the left triangle: ##x^2 + h^2 = b^2##

Pythagoras on the right triangle: ##a^2 - 2ax + (x^2 + h^2) = c^2##

Now since ##b\cos{C} = x##, we finally have ##a^2 + b^2 - 2ab\cos{C} = c^2##.

- #14

etotheipi

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Yes my hand-writing is terrible, I never got my pen-license in primary schoolOh you English people! I can see that you have not even used the scale for drawing that perpendicular ##h## and that ##h## is very pointy, it has sharp end points. It’s really an English’s writing (so tough to read).

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Draw an unit circle and add an additional line on the positive y axis as the terminal arm. Let z denote the length of the terminal arm, we get: $$\cos(\theta)=\frac x z\implies \cos(90)=\frac 0 1 = 0$$

Last edited:

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etotheipi

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What do you mean by a terminal arm?Draw an unit circle and add an additional line on the positive y axis as the terminal arm. Let z denote the length of the terminal arm, we get: $$\cos(x)=\frac x z\implies \cos(90)=\frac 0 1 = 0$$

- #17

Mark44

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I'm pretty sure that he means the ray whose angle is measured relative to the reference direction, the positive x-axis.What do you mean by a terminal arm?

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$$\vec C = \vec A +\vec B$$

[itex] \begin{align*}

C^2 &= (\vec A+\vec B)\cdot (\vec A+\vec B) =A^2 +B^2 + 2AB\cos\theta \\

\cos\theta_{AB} &\equiv\frac{C^2-A^2-B^2}{2AB}

\end{align*} [/itex]

[itex] \begin{align*}

(\vec A+\vec B)\cdot (\vec A+\vec B) &=A^2 +B^2 + 2AB\cos\theta \\

(\vec A-\vec B)\cdot (\vec A-\vec B) &=A^2 +B^2 - 2AB\cos\theta \\

\cos\theta_{AB} &\equiv\frac{(\vec A+\vec B)\cdot (\vec A+\vec B) - (\vec A-\vec B)\cdot (\vec A-\vec B)}{4AB}

\end{align*} [/itex]

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