Trig Volume Pi - 10.3 Exercises

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Trig Volume Pi - 10.3 Exercises

djoconno43
Dear Dr. Stitz and Dr. Zeager

Thanks for a great way to re-learn trig!

Question for you. I am working through the identities for the circular functions (Chapter 10.3), and I am encountering roadblocks starting on Question 117.

Here's Question 117 as it appears in the book, which seems unsolvable:

[cos(x) / 1 - tan(x) ]+ [sin(x) / 1 - cot(x) ] = sin(x) + cos(x)

Is this solvable?

I was so stuck that I had to "google" Question 117 - and Google gave a similar but different (and solvable) identity for "Cos x + Sin X"

[sin x / 1-cot x] - [cos x / tan x - 1] = sin x + cos x

Is the Google question the intended question? Is Question 117 as written solvable?

Thanks again for providing such an accessible resource.

Dan
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Re: Trig Volume Pi - 10.3 Exercises

djoconno43
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Re: Trig Volume Pi - 10.3 Exercises

Carl Stitz
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In reply to this post by djoconno43
Hello!

Sorry I'm just seeing this now...

The second identity is the same as the first!

if you factor a (-1) out of the denominator of the second term in the second identity, you both identities are identical:
[sin x / (1-cot x)] - [cos x / (tan x - 1)] =[sin x / (1-cot x)] - [cos x / (-1) (-tan x + 1)]
                                                          = =[sin x / (1-cot x)] + [cos x /  (1-tan x)]

Hope that helps!
CS
Dan
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Re: Trig Volume Pi - 10.3 Exercises

Dan
Thanks for replying and thanks for opening up this great resource
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Re: Trig Volume Pi - 10.3 Exercises

djoconno43
In reply to this post by Carl Stitz
Do you mind if I ask another Question?

Example 11.6.1, Question 2. page 296 of the Trig Text

21x(sq) +10xy(rt 3) + 31(y)(sq)=144

I can reduce, by sweeping through a rt 3, to:

(36)(x prime)(sq) + (16)(y prime)(sq) + 5(x prime)(y prime)(rt 3) - 5(x prime)(y prime) =144

Does the "troublesome" (5(x prime)(y prime) go away if I multiply that by (rt 3) and then cancel it with the + 5(x prime)(y prime)(rt 3)?

Thanks,

Dan
[hidden email]

On Sun, Sep 13, 2015 at 8:59 PM, Carl Stitz [via Stitz Zeager Open Source Mathematics] <[hidden email]> wrote:
Hello!

Sorry I'm just seeing this now...

The second identity is the same as the first!

if you factor a (-1) out of the denominator of the second term in the second identity, you both identities are identical:
[sin x / (1-cot x)] - [cos x / (tan x - 1)] =[sin x / (1-cot x)] - [cos x / (-1) (-tan x + 1)]
                                                          = =[sin x / (1-cot x)] + [cos x /  (1-tan x)]

Hope that helps!
CS


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Re: Trig Volume Pi - 10.3 Exercises

Carl Stitz
Administrator
Hello again!

Be careful - in order to obtain an equivalent equation, you'd need to multiply both sides of the equation (and hence, each term) by sqrt(3)... so based on what I see there, I don't think your plan will work.

I'd double check the substitution - since in the "prime" axes, the cross terms should cancel out.

Good luck!
Carl
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