Unexplained properties beside Law of Sine and Cosine.

Larrousse

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According to Pythagoreans theorem these are an unexplained properties of all triangles without using the Law of Sine and Cosine.


\(\displaystyle (c\times\sin A)^2+(a\times\cos C)^2=a^2\)

\(\displaystyle (c\times\sin B)^2+(b\times\cos C)^2=b^2\)

\(\displaystyle (a\times\cos A)^2+(b\times\cos B)^2=c^2\)

\(\displaystyle (b\times\cos C)+(c\times\cos B)=a\)

\(\displaystyle (a\times\cos C)+(c\times\cos A)=b\)

\(\displaystyle (a\times\cos B)+(b \times\cos A)=c\)

\(\displaystyle (c\times\cos B)+(b \times\cos C)=a\)

\(\displaystyle (c\times\cos A)+(a\times\cos C)=b\)

and to find three of the altitudes given three sides:

\(\displaystyle h_a\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}\)

\(\displaystyle h_b\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2b}\)

\(\displaystyle h_c\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2c}\)

I want to know if it is new ?
 
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According to Pythagoreans theorem these are an unexplained properties of all triangles without using the Law of Sine and Cosine.

\(\displaystyle (c\times\sin A)^2+(a\times\cos C)^2=a^2\)

\(\displaystyle (c\times\sin B)^2+(b\times\cos C)^2=b^2\)

\(\displaystyle (\sin B)^2+(\cos B)^2=c^2\)

\(\displaystyle (\sin A)^2+(\cos A)^2=c^2\)

\(\displaystyle (b\times\cos C)+(c\times\cos B)=a\)

\(\displaystyle (a\times\cos C)+(c\times\cos A)=b\)

\(\displaystyle (a\times\cos B)+(b \times\cos A)=c\)

\(\displaystyle (c\times\cos B)+(b \times\cos C)=a\)

\(\displaystyle (c\times\cos A)+(a\times\cos C)=b\)

and to find three of the altitudes given three sides:

\(\displaystyle h_a\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}\)

\(\displaystyle h_b\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2b}\)

\(\displaystyle h_c\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2c}\)

What is your point?

Please state in words what you are trying to ask or state. What is unexplained? What property are you referring to? Is there a question here?

Your third and fourth lines are false, unless c = 1. I presume you left something out.
 
According to Pythagoreans theorem these are an unexplained properties of all triangles without using the Law of Sine and Cosine.


\(\displaystyle (c\times\sin A)^2+(a\times\cos C)^2=a^2\)

\(\displaystyle (c\times\sin B)^2+(b\times\cos C)^2=b^2\)



For side c so the base is not only 1the equations are:

\(\displaystyle (c\times\sin B)^2+(c\times\cos B)^2=c^2\)

\(\displaystyle (c\times\sin A)^2+(c\times\cos A)^2=c^2\)

\(\displaystyle (a\times\cos B)+(b\times\cos A)=c\)

\(\displaystyle (b\times\cos C)+(c\times\cos B)=a\)

\(\displaystyle (a\times\cos C)+(c\times\cos A)=b\)

\(\displaystyle (a\times\cos B)+(b \times\cos A)=c\)

\(\displaystyle (c\times\cos B)+(b \times\cos C)=a\)

\(\displaystyle (c\times\cos A)+(a\times\cos C)=b\)

and to find three of the altitudes given three sides:

\(\displaystyle h_a\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}\)

\(\displaystyle h_b\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2b}\)

\(\displaystyle h_c\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2c}\)

I want to know if it is new ? and are there more explanation for the 3th line?
 
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According to Pythagoreans theorem these are an unexplained properties of all triangles without using the Law of Sine and Cosine.


\(\displaystyle (c\times\sin A)^2+(a\times\cos C)^2=a^2\)

\(\displaystyle (c\times\sin B)^2+(b\times\cos C)^2=b^2\)

\(\displaystyle (a\times\cos B)^2+(b\times\cos A)^2=c^2\)

\(\displaystyle (b\times\cos C)+(c\times\cos B)=a\)

\(\displaystyle (a\times\cos C)+(c\times\cos A)=b\)

\(\displaystyle (a\times\cos B)+(b \times\cos A)=c\)

\(\displaystyle (c\times\cos B)+(b \times\cos C)=a\)

\(\displaystyle (c\times\cos A)+(a\times\cos C)=b\)

and to find three of the altitudes given three sides:

\(\displaystyle h_a\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}\)

\(\displaystyle h_b\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2b}\)

\(\displaystyle h_c\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2c}\)

I want to know if it is new ? and are there more explanation for the 3th line?
By that, do you mean unproven?
How do you know these are unexplained?
 
By that, do you mean unproven?
How do you know these are unexplained?

I have three altitudes and all three sides and I'm able to find all three angles with the help of SohCahToa, but I unable to find a general equation with the help of pythagorean identities for side of \(\displaystyle c^2\) without using one of the altitudes or height.
 
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I have three altitudes and all three sides and I'm able to find all three angles with the help of SohCahToa, but I unable to find a general equation with the help of pythagorean identities for side of \(\displaystyle c^2\) without using one of the altitudes or height.
They're not found in my textbook except the you have to use either the Law of Sine or Cosine.
 
I have three altitudes and all three sides and I'm able to find all three angles with the help of SohCahToa, but I unable to find a general equation with the help of pythagorean identities for side of \(\displaystyle c^2\) without using one of the altitudes or height.

You seem to be trying to solve some problem, but you haven't stated it. What are you given, and what do you want to find?

As for the equations you wrote, I think they are all reasonably easy to derive, so they can't be "new" in the sense that no one has ever stated them.
 
Here is the Law of Cosine, and with Sin you don't have to use the law either to find the lenght of triangles given three angles.

You still haven't stated clearly what you are trying to do. Please state what you are given (e.g. the side lengths), what you want to find (e.g. the altitudes), and any constraints (e.g. you aren't allowed to use the law of cosines -- and why). The more you tell us about your needs, the quicker we can help.
 
You still haven't stated clearly what you are trying to do. Please state what you are given (e.g. the side lengths), what you want to find (e.g. the altitudes), and any constraints (e.g. you aren't allowed to use the law of cosines -- and why). The more you tell us about your needs, the quicker we can help.

I wanted to know wether It was new, although it may seem very easy for mathematicians as what you've sayed in your comment "was too easy". The equations are corrected and make sense now, and I have looked Pythagoreans identities even though it's hard to explain.

\(\displaystyle (c\times\sin B)^2+(b\times\cos C)^2=b^2\)

\(\displaystyle (c\times\sin A)^2+( a\times\cos C)^2=a^2\)

\(\displaystyle c\times\sin B)^2+(c\times\cos B)^2=c^2\)

\(\displaystyle (c\times\sin A)^2+(c\times\cos A)^2=c^2\)

\(\displaystyle (b\times\cos C)+(c\times\cos B)=a\)

\(\displaystyle (a\times\cos C)+(c\times\cos A)=b\)

\(\displaystyle (a\times\cos B)+(b \times\cos A)=c\)

\(\displaystyle (c\times\cos B)+(b \times\cos C)=a\)

\(\displaystyle (c\times\cos A)+(a\times\cos C)=b\)

and to find one of the altitude:

\(\displaystyle h_a=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}\)
\(\displaystyle h_b=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2b}\)
\(\displaystyle h_c=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2c}\)

Thanks for your help.
 
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I wanted to know wether It was new, although it may seem very easy for mathematicians as what you've sayed in your comment "was too easy". The equations are corrected and make sense now, and I have looked Pythagoreans identities even though it's hard to explain.

$(c\times\sin B)^2+(c\times b\times\cos C)^2=b^2$

$(c\times\sin A)^2+(c\times a\times\cos C)^2=a^2$

$(c\times\sin B)^2+(c\times\cos B)^2=c^2$

$(c\times\sin A)^2+(c\times\cos A)^2=c^2$

$(b\times\cos C)+(c\times\cos B)=a$

$(a\times\cos C)+(c\times\cos A)=b$

$(a\times\cos B)+(b \times\cos A)=c$

$(c\times\cos B)+(b \times\cos C)=a$

$(c\times\cos A)+(a\times\cos C)=b$

and to find one of the altitude:

$h_a=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}$
$h_b=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2b}$
$h_c=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2c}$
 
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I wanted to know wether It was new, although it may seem very easy for mathematicians as what you've sayed in your comment "was too easy". The equations are corrected and make sense now, and I have looked Pythagoreans identities even though it's hard to explain.

$(c\times\sin B)^2+(b\times\cos C)^2=b^2$

$(c\times\sin A)^2+(a\times\cos C)^2=a^2$

$(c\times\sin B)^2+(c\times\cos B)^2=c^2$

$(c\times\sin A)^2+(c\times\cos A)^2=c^2$

$(b\times\cos C)+(c\times\cos B)=a$

$(a\times\cos C)+(c\times\cos A)=b$

$(a\times\cos B)+(b \times\cos A)=c$

$(c\times\cos B)+(b \times\cos C)=a$

$(c\times\cos A)+(a\times\cos C)=b$

and to find one of the altitude:

$h_a=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}$
$h_b=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2b}$
$h_c=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2c}$
All equations of this form, if valid, have been known for a long time. Also many of the equations are duplicated. For example, if you have an equation and you replace A with B, B with C and C with A then you really have the same equation!! Do you see that???
 
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To make it easier(do you see the black cat)

All equations of this form, if valid, have been known for a long time. Also many of the equations are duplicated. For example, if you have an equation and you replace A with B, B with C and C with A then you really have the same equation!! Do you see that???

larrousse.jpg
 
No, i do not see the black cat. Actually I am blind. You should think before you type. In any case, your picture is extremely unclear since it is unclear exactly which length is sin A, Sin B, Cos A and CosB. It also looks to be that you are labeling the angles as Cos A, Cos B and Cos C. You do know that when you compute the Cosine of an angle you (usually) get a number, NOT an angle. What is that all about?
 
According to Pythagoreans theorem these are an unexplained properties of all triangles without using the Law of Sine and Cosine.


\(\displaystyle (c\times\sin A)^2+(a\times\cos C)^2=a^2\)

\(\displaystyle (c\times\sin B)^2+(b\times\cos C)^2=b^2\)



For side c, the base is not only 1 the equations are:

\(\displaystyle (c\times\sin B)^2+(c\times\cos B)^2=c^2\)

\(\displaystyle (c\times\sin A)^2+(c\times\cos A)^2=c^2\)

\(\displaystyle (a\times\cos B)+(b\times\cos A)=c\)

\(\displaystyle (b\times\cos C)+(c\times\cos B)=a\)

\(\displaystyle (a\times\cos C)+(c\times\cos A)=b\)

\(\displaystyle (a\times\cos B)+(b \times\cos A)=c\)

\(\displaystyle (c\times\cos B)+(c\times b \times\cos C)=a\)

\(\displaystyle (c\times\cos A)+(c \times a\times\cos C)=b\)

and to find three of the altitudes given three sides:

\(\displaystyle h_a\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}\)

\(\displaystyle h_b\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2b}\)

\(\displaystyle h_c\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2c}\)

I want to know if it is new ? and are there more explanation for the 3th line?[/QUOTE]
 

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According to Pythagoreans theorem these are an unexplained properties of all triangles without using the Law of Sine and Cosine.

I want to know if it is new ? and are there more explanation for the 3th line?

I don't think "unexplained" means what you think it means. You've never told us what you mean by it.

Also, it's irrelevant whether you (directly) use the laws of sines and cosines to prove something, since they can themselves be proved, and you can just use the same ideas as in those proofs in order to prove facts that would otherwise be proved using them.

None of this is new in any significant sense.

What is the "third line" (we never write "3th") that you are asking about, and what do you want to know about it?
 
I don't think "unexplained" means what you think it means. You've never told us what you mean by it.

Also, it's irrelevant whether you (directly) use the laws of sines and cosines to prove something, since they can themselves be proved, and you can just use the same ideas as in those proofs in order to prove facts that would otherwise be proved using them.

None of this is new in any significant sense.

What is the "third line" (we never write "3th") that you are asking about, and what do you want to know about it?

\(\displaystyle (\sin A)^2+(\sin B)^2+(\sin C)^2+(\cos A)^2+(\cos B)^2+(\cos C)^2=3\) apply for all triangles.

\(\displaystyle c^2=a^2+b^2-2ab\cos \theta\).
Since we know, sides of the triangle, a,b,c, we can use the formula for all three altitudes or above formula to calculate \(\displaystyle \cos\theta\).

We can use the Law of Cosine or three altitudes of. \(\displaystyle \triangle ABC \)to get the three angles,not in degrees, with precision.
 
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\(\displaystyle (\sin A)^2+(\sin B)^2+(\sin C)^2+(\cos A)^2+(\cos B)^2+(\cos C)^2=3\) apply for all triangles.

\(\displaystyle c^2=a^2+b^2-2ab\cos \theta\).
Since we know a,b,c, we can use the formula for all three altitudes or above formula to calculate \(\displaystyle \cos\theta\).

We can use the Law of Cosine or three altitudes of. \(\displaystyle \triangle ABC \)to get the three angles,not in degrees, with precision.
For a \(\displaystyle \triangle ABC\) there exist three sides, a, b, c and by using the formulas of the altitudes,I am able to find all 3 angles with the help of trigonometric identities SohCahToa.
At the same time, by using three angles I'm able to find the lenghts of all triangle without the help of the Law of sine.

The following equations represent the schematic of all triangles to obtain the result needed to find 3 sides and three angles of triangles.

\(\displaystyle (c\times\sin B)^2+(b\times\cos C)^2=b^2\)

\(\displaystyle (c\times\sin A)^2+(a\times\cos C)^2=a^2\)

\(\displaystyle (c\times\sin B)^2+(c\times\cos B)^2=c^2\)

\(\displaystyle (c\times\sin A)^2+(c\times\cos A)^2=c^2\)

\(\displaystyle (b\times\cos C)+(c\times\cos B)=a\)

\(\displaystyle (a\times\cos C)+(c\times\cos A)=b\)


\(\displaystyle (a\times\cos B)+(b \times\cos A)=c\)


\(\displaystyle (c\times\cos B)+(b \times\cos C)=a\)

\(\displaystyle (c\times\cos A)+(a\times\cos C)=b\)

and to find the altitudes for all triangles:

\(\displaystyle h_a=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}\)

\(\displaystyle h_b=\frac{\sqrt{(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2b}\)

\(\displaystyle h_c=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2c}\)

I have three altitudes and all three sides and I'm able to find all three angles with the help of SohCahToa.

And the following equation apply for all triangles

\(\displaystyle (\sin A)^2+(\sin B)^2+(\sin C)^2+(\cos A)^2+(\cos B)^2+(\cos C)^2=3\)



And the law of sine states when one of the base of the triangle is one then:


\(\displaystyle \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}=\) where the reciprocal of the diameter is \(\displaystyle \Angle C\)

\(\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=d \) where d is the diameter


Let \(\displaystyle \triangle ABC\) be sides a, b, and c. I am using that the length of the side opposite vertex A is called a, B for b and C for c.

If \(\displaystyle \theta=\angle C\). Then the Cosine says that

\(\displaystyle c^2=a^2+b^2-2ab\cos \theta.\)
Since we know a, b, and c, we can use the formula for all three altitudes or above formula to calculate \(\displaystyle \cos\theta.\)

We can use the Law of Cosine for all three angles.


Another way to obtain \(\displaystyle \cos C \):

\(\displaystyle \cos C=\frac{ b-c\times\cos A}{ a}\)

You can refer to the picture attached here.
 
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