Help me with this proof if ABC is a right triangle

Andrax

1. Homework Statement
prove that sin^2(a)=sin^2(b)+sin^2(c) if and only if ABC is a right triangle in A

i worked really hard on this one i'm really confused why i didn' get the answer
2. Homework Equations

3. The Attempt at a Solution
a+b+c=pi
tried turning everythng to cos 2x didn't helpi really couldn't do this one, can't use complex numbers by the way...please help

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Snark1994

Well, $a+b+c = \pi$ is going to be true of all triangles, so that's not necessarily going to be much help.

What does ABC being a right-angled triangle in A imply for the value of a? What does that imply for the value of b+c? What does that imply for the value of $\sin^2 a$?

Andrax

Well, $a+b+c = \pi$ is going to be true of all triangles, so that's not necessarily going to be much help.

What does ABC being a right-angled triangle in A imply for the value of a? What does that imply for the value of b+c? What does that imply for the value of $\sin^2 a$?
i tried everything you said also transforming $\sin^2 a$ won't help me since i want to keep sin a

Snark1994

transforming $\sin^2 a$ won't help me since i want to keep sin a
Transforming $\sin^2 a$ will help you. In fact, the easiest way I can see of proving the result from right-to-left (remember for an "if and only if" you need to prove it both ways) is to show that $\sin^2 a = k = \sin^2 b + \sin^2 c$ for some specific number k whose value you'll have to work out.

Like I asked, if the triangle is right-angled in A, what can you say about the value of 'a'? You can be very specific!

SammyS

Staff Emeritus
Homework Helper
Gold Member
$\sin((\pi/2) - \theta)=\cos(\theta)$

Also, for this to be true, a must be the right angle of the right triangle.

Andrax

$\sin((\pi/2) - \theta)=\cos(\theta)$

Also, for this to be true, a must be the right angle of the right triangle.
solved using Pythagoras's theorem --> let abc be a triangle wehave sin a^2 = AB^2 etc..
thank everyone..

"Help me with this proof if ABC is a right triangle"

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