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Help me with this proof if ABC is a right triangle

  • Thread starter Andrax
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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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Well, [itex]a+b+c = \pi[/itex] 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 [itex]\sin^2 a[/itex]?
 
117
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Well, [itex]a+b+c = \pi[/itex] 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 [itex]\sin^2 a[/itex]?
i tried everything you said also transforming [itex]\sin^2 a[/itex] won't help me since i want to keep sin a
 
18
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transforming [itex]\sin^2 a[/itex] won't help me since i want to keep sin a
Transforming [itex]\sin^2 a[/itex] 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 [itex]\sin^2 a = k = \sin^2 b + \sin^2 c[/itex] 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

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[itex]\sin((\pi/2) - \theta)=\cos(\theta)[/itex]

Also, for this to be true, a must be the right angle of the right triangle.
 
117
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[itex]\sin((\pi/2) - \theta)=\cos(\theta)[/itex]

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..
 

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