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The projections-of-the-legs over the hypotenuse should add up to the hypotenuse $c$.
Is there any alternative way to prove below? $$a\cos \alpha + b\sin \alpha = \sqrt{a^2+b^2}$$ enter image description here

AgentS
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1 Answers1

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Well, $$\cos \alpha = \frac ac\quad \&\quad \sin \alpha = \frac bc$$

so $$a\cos \alpha +b\sin \alpha = \frac 1c \times (a^2+b^2)=\frac {c^2}c=c$$

Of course, that last step requires the Pythagorean Theorem ("PT"). It is worth remarking that, without using PT the argument shows $$a\cos \alpha +b\sin \alpha = \frac {a^2+b^2}c$$ and since the OP has shown that $$a\cos \alpha +b\sin \alpha = c$$ we can combine the two arguments to get $$\frac {a^2+b^2}c=c \implies a^2+b^2=c^2$$ so the two arguments together yield a proof of PT.

lulu
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  • Ahh how did I miss using the big outer triangle! Looks silly now haha thank you so much :) – AgentS Jan 26 '19 at 19:16
  • @palaeomathematician that's right, but I can rearrange that step as below (next comment) – AgentS Jan 26 '19 at 19:24
  • $$a\cos \alpha +b\sin \alpha = a\frac{a}{c} + b\frac{b}{c} = \frac 1c \times (a^2+b^2)$$ – AgentS Jan 26 '19 at 19:24
  • Then set that equal to $c$ :

    $$ \frac 1c \times (a^2+b^2) = c \implies a^2+b^2=c^2$$

     – AgentS Jan 26 '19 at 19:26
  • Does that still feel like circular proof @palaeomathematician – AgentS Jan 26 '19 at 19:26
  • @palaeomathematician I'm not sure I understand where the controversy is, or if there is controversy.
    What I wrote is not a proof of the Pythagorean Theorem,, nor was it intended to be; indeed it uses that Theorem. Of course, the argument provided by the OP is also not a proof of the Pythagorean Theorem. That argument does show that $a\cos \alpha +b\sin \alpha =c$ but the OP then uses the PT to get to $\sqrt {a^2+b^2}$.     – lulu Jan 26 '19 at 19:30
  • @lulu I think the OP wants to prove the Pythagorean theorem using that $a\cos \alpha + b\sin \alpha = c$ –  Jan 26 '19 at 19:33
  • @palaeomathematician Why do you think that? The way I read the question, the OP was asking for a proof of the given formula that doesn't rely on the addition of the new perpendicular. In any event, I wasn't trying to prove the PT. – lulu Jan 26 '19 at 19:35
  • @palaeomathematician Worth remarking, that without using PT my argument shows $a\cos \alpha +b\sin \alpha = \frac {a^2+b^2}c$ and since the OP's argument shows that $a\cos \alpha +b\sin \alpha = c$ we can combine the two arguments to get a proof of PT. – lulu Jan 26 '19 at 19:38
  • @lulu By using elementary trigonometry the OP concluded that $a\cos \alpha + b\sin \alpha = c$. But he also knows that according to the PT we should have $c = \sqrt{a^2+b^2}$. So he wants to prove that $c = a\cos \alpha + b\sin \alpha = \sqrt{a^2+b^2} $, that is that the PT holds. But maybe we should wait to see what the OP says, since only he knows what he wanted. –  Jan 26 '19 at 19:40
  • You both are right. In the post I didn't mention it as a proof for PT because I was just looking for another connection between $a\cos \alpha + b\sin\alpha$ and $c$. Thanks again :) – AgentS Jan 26 '19 at 19:49