Small question on Pythagoras theorem












3












$begingroup$


The projections-of-the-legs over the hypotenuse should add up to the hypotenuse $c$.

Is there any alternative way to prove below?
$$acos alpha + bsin alpha = sqrt{a^2+b^2}$$
enter image description here






algebra-precalculus







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    The projections-of-the-legs over the hypotenuse should add up to the hypotenuse $c$.

    Is there any alternative way to prove below?
    $$acos alpha + bsin alpha = sqrt{a^2+b^2}$$
    enter image description here






    algebra-precalculus







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      The projections-of-the-legs over the hypotenuse should add up to the hypotenuse $c$.

      Is there any alternative way to prove below?
      $$acos alpha + bsin alpha = sqrt{a^2+b^2}$$
      enter image description here






      algebra-precalculus







      share|cite|improve this question









      $endgroup$




      The projections-of-the-legs over the hypotenuse should add up to the hypotenuse $c$.

      Is there any alternative way to prove below?
      $$acos alpha + bsin alpha = sqrt{a^2+b^2}$$
      enter image description here






      algebra-precalculus




      algebra-precalculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked 2 hours ago









      rsadhvikarsadhvika

      1,6951228




      1,6951228






















          1 Answer
          1






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          5












          $begingroup$

          Well, $$cos alpha = frac acquad &quad sin alpha = frac bc$$



          so $$acos alpha +bsin alpha = frac 1c times (a^2+b^2)=frac {c^2}c=c$$



          Of course, that last step requires the Pythagorean Theorem ("PT"). It is worth remarking that, without using PT the argument shows $$acos alpha +bsin alpha = frac {a^2+b^2}c$$ and since the OP has shown that $$acos alpha +bsin alpha = c$$ 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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ahh how did I miss using the big outer triangle! Looks silly now haha thank you so much :)
            $endgroup$
            – rsadhvika
            2 hours ago










          • $begingroup$
            @palaeomathematician that's right, but I can rearrange that step as below (next comment)
            $endgroup$
            – rsadhvika
            2 hours ago










          • $begingroup$
            $$a*cos alpha +b*sin alpha = a*frac{a}{c} + b*frac{b}{c} = frac 1c times (a^2+b^2)$$
            $endgroup$
            – rsadhvika
            2 hours ago












          • $begingroup$
            Then set that equal to $c$ : $$ frac 1c times (a^2+b^2) = c implies a^2+b^2=c^2$$
            $endgroup$
            – rsadhvika
            2 hours ago










          • $begingroup$
            Does that still feel like circular proof @palaeomathematician
            $endgroup$
            – rsadhvika
            2 hours ago











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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          Well, $$cos alpha = frac acquad &quad sin alpha = frac bc$$



          so $$acos alpha +bsin alpha = frac 1c times (a^2+b^2)=frac {c^2}c=c$$



          Of course, that last step requires the Pythagorean Theorem ("PT"). It is worth remarking that, without using PT the argument shows $$acos alpha +bsin alpha = frac {a^2+b^2}c$$ and since the OP has shown that $$acos alpha +bsin alpha = c$$ 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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ahh how did I miss using the big outer triangle! Looks silly now haha thank you so much :)
            $endgroup$
            – rsadhvika
            2 hours ago










          • $begingroup$
            @palaeomathematician that's right, but I can rearrange that step as below (next comment)
            $endgroup$
            – rsadhvika
            2 hours ago










          • $begingroup$
            $$a*cos alpha +b*sin alpha = a*frac{a}{c} + b*frac{b}{c} = frac 1c times (a^2+b^2)$$
            $endgroup$
            – rsadhvika
            2 hours ago












          • $begingroup$
            Then set that equal to $c$ : $$ frac 1c times (a^2+b^2) = c implies a^2+b^2=c^2$$
            $endgroup$
            – rsadhvika
            2 hours ago










          • $begingroup$
            Does that still feel like circular proof @palaeomathematician
            $endgroup$
            – rsadhvika
            2 hours ago
















          5












          $begingroup$

          Well, $$cos alpha = frac acquad &quad sin alpha = frac bc$$



          so $$acos alpha +bsin alpha = frac 1c times (a^2+b^2)=frac {c^2}c=c$$



          Of course, that last step requires the Pythagorean Theorem ("PT"). It is worth remarking that, without using PT the argument shows $$acos alpha +bsin alpha = frac {a^2+b^2}c$$ and since the OP has shown that $$acos alpha +bsin alpha = c$$ 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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Ahh how did I miss using the big outer triangle! Looks silly now haha thank you so much :)
            $endgroup$
            – rsadhvika
            2 hours ago










          • $begingroup$
            @palaeomathematician that's right, but I can rearrange that step as below (next comment)
            $endgroup$
            – rsadhvika
            2 hours ago










          • $begingroup$
            $$a*cos alpha +b*sin alpha = a*frac{a}{c} + b*frac{b}{c} = frac 1c times (a^2+b^2)$$
            $endgroup$
            – rsadhvika
            2 hours ago












          • $begingroup$
            Then set that equal to $c$ : $$ frac 1c times (a^2+b^2) = c implies a^2+b^2=c^2$$
            $endgroup$
            – rsadhvika
            2 hours ago










          • $begingroup$
            Does that still feel like circular proof @palaeomathematician
            $endgroup$
            – rsadhvika
            2 hours ago














          5












          5








          5





          $begingroup$

          Well, $$cos alpha = frac acquad &quad sin alpha = frac bc$$



          so $$acos alpha +bsin alpha = frac 1c times (a^2+b^2)=frac {c^2}c=c$$



          Of course, that last step requires the Pythagorean Theorem ("PT"). It is worth remarking that, without using PT the argument shows $$acos alpha +bsin alpha = frac {a^2+b^2}c$$ and since the OP has shown that $$acos alpha +bsin alpha = c$$ 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.






          share|cite|improve this answer











          $endgroup$



          Well, $$cos alpha = frac acquad &quad sin alpha = frac bc$$



          so $$acos alpha +bsin alpha = frac 1c times (a^2+b^2)=frac {c^2}c=c$$



          Of course, that last step requires the Pythagorean Theorem ("PT"). It is worth remarking that, without using PT the argument shows $$acos alpha +bsin alpha = frac {a^2+b^2}c$$ and since the OP has shown that $$acos alpha +bsin alpha = c$$ 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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 2 hours ago









          lulululu

          40.2k24778




          40.2k24778












          • $begingroup$
            Ahh how did I miss using the big outer triangle! Looks silly now haha thank you so much :)
            $endgroup$
            – rsadhvika
            2 hours ago










          • $begingroup$
            @palaeomathematician that's right, but I can rearrange that step as below (next comment)
            $endgroup$
            – rsadhvika
            2 hours ago










          • $begingroup$
            $$a*cos alpha +b*sin alpha = a*frac{a}{c} + b*frac{b}{c} = frac 1c times (a^2+b^2)$$
            $endgroup$
            – rsadhvika
            2 hours ago












          • $begingroup$
            Then set that equal to $c$ : $$ frac 1c times (a^2+b^2) = c implies a^2+b^2=c^2$$
            $endgroup$
            – rsadhvika
            2 hours ago










          • $begingroup$
            Does that still feel like circular proof @palaeomathematician
            $endgroup$
            – rsadhvika
            2 hours ago


















          • $begingroup$
            Ahh how did I miss using the big outer triangle! Looks silly now haha thank you so much :)
            $endgroup$
            – rsadhvika
            2 hours ago










          • $begingroup$
            @palaeomathematician that's right, but I can rearrange that step as below (next comment)
            $endgroup$
            – rsadhvika
            2 hours ago










          • $begingroup$
            $$a*cos alpha +b*sin alpha = a*frac{a}{c} + b*frac{b}{c} = frac 1c times (a^2+b^2)$$
            $endgroup$
            – rsadhvika
            2 hours ago












          • $begingroup$
            Then set that equal to $c$ : $$ frac 1c times (a^2+b^2) = c implies a^2+b^2=c^2$$
            $endgroup$
            – rsadhvika
            2 hours ago










          • $begingroup$
            Does that still feel like circular proof @palaeomathematician
            $endgroup$
            – rsadhvika
            2 hours ago
















          $begingroup$
          Ahh how did I miss using the big outer triangle! Looks silly now haha thank you so much :)
          $endgroup$
          – rsadhvika
          2 hours ago




          $begingroup$
          Ahh how did I miss using the big outer triangle! Looks silly now haha thank you so much :)
          $endgroup$
          – rsadhvika
          2 hours ago












          $begingroup$
          @palaeomathematician that's right, but I can rearrange that step as below (next comment)
          $endgroup$
          – rsadhvika
          2 hours ago




          $begingroup$
          @palaeomathematician that's right, but I can rearrange that step as below (next comment)
          $endgroup$
          – rsadhvika
          2 hours ago












          $begingroup$
          $$a*cos alpha +b*sin alpha = a*frac{a}{c} + b*frac{b}{c} = frac 1c times (a^2+b^2)$$
          $endgroup$
          – rsadhvika
          2 hours ago






          $begingroup$
          $$a*cos alpha +b*sin alpha = a*frac{a}{c} + b*frac{b}{c} = frac 1c times (a^2+b^2)$$
          $endgroup$
          – rsadhvika
          2 hours ago














          $begingroup$
          Then set that equal to $c$ : $$ frac 1c times (a^2+b^2) = c implies a^2+b^2=c^2$$
          $endgroup$
          – rsadhvika
          2 hours ago




          $begingroup$
          Then set that equal to $c$ : $$ frac 1c times (a^2+b^2) = c implies a^2+b^2=c^2$$
          $endgroup$
          – rsadhvika
          2 hours ago












          $begingroup$
          Does that still feel like circular proof @palaeomathematician
          $endgroup$
          – rsadhvika
          2 hours ago




          $begingroup$
          Does that still feel like circular proof @palaeomathematician
          $endgroup$
          – rsadhvika
          2 hours ago


















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