Trigonometric identity

  • Thread starter Ry122
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  • #1
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In the following question I figure that i need to prove that h holds true with the trigonometric identity subbed into the denominator.
I'm not sure how to simplify the equation any further after that though.
Can someone provide any insight?

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Answers and Replies

  • #2
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Forget about the given formula for a moment. Try to find a good expression for the height h.

You could start by drawing the perpendicular on the baseline through the other point of the triangle. This will divide the baseline in section c' and c''.

Now, can you express h in function of c' and alpha?
And can you express h in function of c'' and beta?
 
  • #3
565
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tanb=h/c'
tana=h/c

Substituting I end up with tana/tanb = c'/c

But what does this show me?
 
  • #4
22,089
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tanb=h/c'
tana=h/c

Substituting I end up with tana/tanb = c'/c

But what does this show me?

Indeed, so

[tex]c^\prime=\frac{h}{\tan{\beta}}~\text{and}~c=\frac{h}{\tan{\alpha}}[/tex]

Now we also know that [itex]c=c^\prime+c^{\prime \prime}[/itex]

So what do you get if you plug in these values of c' and c''??
 
  • #5
565
2


Using the equations for c, c' and c'', this is as close as I got to the original formula:

h = c*sin(b)/cos(b)
+ c*sin(a)/cos(a)

What further steps do I need to take?
 
  • #6
dynamicsolo
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Please check your algebra: you should have

[itex]c = h (\frac{cos\alpha}{sin\alpha} + \frac{cos\beta}{sin\beta} ) [/itex] ,

which you would then solve for h . (It will help to add the ratios first.)
 
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  • #7
565
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that's exactly what I have, but with c as the subject.
Just wondering how to get it into a similar form to the original formula.
 
  • #8
dynamicsolo
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What you have is not equivalent to c = c' + c'' . (It is not correct to add those terms.)

Using the terms micromass has in post #4 , you have h = c' tan(beta) and, separately,
h = c'' tan(alpha) [he has a typo] . (And I had to edit my typo in post #6.)

The equation in post #6 is what the suggestions in the thread bring you to.

You will also need the "angle-addition" formula for sine at some point.
 
  • #9
565
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Can you please re-arrange your equation in post 6 so that h is the subject and tell me how it's any different from mine?
 
  • #10
dynamicsolo
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[itex]h = \frac{c}{ (\frac{cos\alpha}{sin\alpha} + \frac{cos\beta}{sin\beta} )} [/itex]

c has to be divided by the sum of the two ratios.
 
  • #11
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Thanks. Is the angle-addition formula you're referring to the trigonometric identity shown in my first post? I'm not sure how I would apply that here.
 
  • #12
dynamicsolo
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Add the two ratios together and you'll see where it needs to be applied.
 

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