# Thread: Proof of an equation

1. ## Proof of an equation

Hello everyone!

I'm new to this forum, I stumbled upon this domain because I really needed some help proving this equation. I'm generally horrible at this, even after practising for a bit and none of my friends are any better off than I am, so I couldn't get help anywhere.
Anyway, I need step by step proof that sin²B + sin²C -sin²A = 2cosA * sinB * sinC
If any of you could figure this out, that would be amazing.

SSG

2. ## Re: Proof of an equation

Originally Posted by SillyStudentGuy
I'm new to this forum, I stumbled upon this domain because I really needed some help proving this equation. I'm generally horrible at this, even after practising for a bit and none of my friends are any better off than I am, so I couldn't get help anywhere. Anyway, I need step by step proof that sin²B + sin²C -sin²A = 2cosA * sinB * sinC
Have you even tested these?
Look at this sum. Then look Here.
Are they equal?

Have you stated the question corrected?

3. ## Re: Proof of an equation

Thank you very much for replying.

I forgot to mention that A, B and C represent corners of a triangle. Upon testing, using 60° for A, 80° for B and 40°, I found that both sides were equal (0.63302222)

SSG

4. ## Re: Proof of an equation

Originally Posted by SillyStudentGuy;933537[COLOR="#FF0000"
I forgot to mention that A, B and C represent corners of a triangle[/COLOR]. Upon testing, using 60° for A, 80° for B and 40°, I found that both sides were equal (0.63302222)
Don't you think that is important? Now that is an identity. In fact at some point I worked out something similar. I am not spending the time again.

Here are hint aids. $(A+B+C)=\pi$, $\sin(A+B+C)=\cos(B) \cos(C) \sin(A) + \cos(A) \cos(C) \sin(B) + \cos(A) \cos(B) \sin(C) - \sin(A) \sin(B) \sin(C)$ See Here.

Because $\sin(X+Y)=\sin(X)\cos(Y)+\sin(Y)\cos(X)$ what does $\sin(\pi-A)=~?$

Good luck!

5. ## Re: Proof of an equation

Originally Posted by SillyStudentGuy
Hello everyone!

I'm new to this forum, I stumbled upon this domain because I really needed some help proving this equation. I'm generally horrible at this, even after practising for a bit and none of my friends are any better off than I am, so I couldn't get help anywhere.
Anyway, I need step by step proof that sin²B + sin²C -sin²A = 2cosA * sinB * sinC
If any of you could figure this out, that would be amazing.

SSG
The first thing I would do is notice that is $A= B= C= \frac{\pi}{2}$, the left hand side is 1+ 1- 1= 1 while the right hand side is 2(0)(1)(1)= 0. The two sides are not equal- what you are trying to prove is not true!

6. ## Re: Proof of an equation

Originally Posted by HallsofIvy
The first thing I would do is notice that is $A= B= C= \frac{\pi}{2}$, the left hand side is 1+ 1- 1= 1 while the right hand side is 2(0)(1)(1)= 0. The two sides are not equal- what you are trying to prove is not true!
Hi, Prof G Ivy did you read this whole thread before posting this? Here is the salient addition:
Originally Posted by SillyStudentGuy
I forgot to mention that A, B and C represent corners of a triangle.
SillyStudentGuy clearly means by "corners of a triangle" the angles of a triangles. In which case your example does not apply. With that addendum the equation is an identity.

7. ## Re: Proof of an equation

searching the web I found 1/2*twelve solutions.
Here is one that I liked

$a^2=b^2+c^2-2 b c \cos A$

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

replace $a,b,c$ from the second equation into the first and simplify