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Trigonometry, Prove the Identity and more 
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#1
Aug2407, 03:06 PM

P: 1,755

1. The problem statement, all variables and given/known data
Prove the identity 11. 1  co5xcos3x  sin5xsin3x = 2sin^2x 50. ln secx + tanx = ln secx  tanx 52. The following equation occurs in the study of mechanics: [tex]\sin\theta = \frac{I_1\cos\phi}{\sqrt{(I_1\cos\phi)^2 + (I_2\sin\phi)^2}}[/tex]. It can happen that [tex]I_1 = I_2[/tex]. Assuming that this happens, simplify the equation. 2. Relevant equations 3. The attempt at a solution 11. Idk what I'm doing wrong 50. GRRR!!! 53. Can I make this assumption that [tex]\phi = 45^\circ[/tex] and that [tex]\theta = 45^\circ[/tex]? 


#2
Aug2407, 03:18 PM

P: 100

For 50
ln(secx + tanx) = ln(secx  tanx) ln(secx + tanx) = ln(1 / (secx  tanx)) secx + tanx = 1 / (secx  tanx) Cross multiply, and you're good. 


#3
Aug2407, 03:20 PM

P: 154

11) You should know the identity [tex] 1cos(2 \phi) = 2 sin^2 (\phi) [/tex], that should help.



#4
Aug2407, 03:23 PM

P: 1,755

Trigonometry, Prove the Identity and more



#5
Aug2407, 03:25 PM

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P: 11,896

So you'd know, 11 is almost done, in case you didn't bother to write the whole computation, if you have already made it. Just cut 1 from both sides of the equation, multiply the result by 1 and then use the fact that sin^2 +cos^2 =1.



#6
Aug2407, 03:37 PM

P: 1,755

anyone for 53? can i make that assumption or did i screw up? i did it about 3x. 


#7
Aug2407, 03:44 PM

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You wrote 52 but i think you meant 53. You have
[tex]I_{1}^{2}\cos^{4}\theta=I_{2}^{2}\sin^{4}\theta [/tex] You can't take sqrt like that. 


#8
Aug2407, 04:01 PM

P: 1,755

idk if my handwriting is clear, but [tex]\sin\theta = \frac{I_1\cos\phi}{\sqrt{(I_1\cos\phi)^2 + (I_2\sin\phi)^2}}[/tex] 


#9
Aug2407, 04:05 PM

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Okay. Raise to the second power and do all multiplications. What do you get ?



#11
Aug2407, 04:19 PM

P: 1,755

it's from Algebra and Trigonometry, Beecher (Calcprep book) ... i love it! i didn't learn Trigonometry properly b/c i took it in a 3week minimester ... i love Math now so i'm studying harddd 


#12
Aug2407, 04:24 PM

P: 1,755

check my work up to this point plz ... i can't think straight anymore [tex]\sin\theta = \frac{I_1\cos\phi}{\sqrt{(I_1\cos\phi)^2 + (I_2\sin\phi)^2}}[/tex] squaring both sides ... [tex]\sin^2\theta = \frac{I_1^2\cos^2\phi}{I_1^2\cos^2\phi + I_2^2\sin^2\phi}[/tex] is my denominator right, or did i screw it up? 


#13
Aug2407, 04:26 PM

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It's okay so far. Now cross multiply.



#14
Aug2407, 04:32 PM

P: 1,294

Yes I do. Just finished the first one. wow, I almost missed these ones! I only did examples from 6.3, and nothing exciting was there, so I moved to the next exercise also because I had less time that day, and was tired from the work. I am on the last chapter of that book. I am doing this because I thought I need to brush up my basic skills before I start my first year of university(in EE). In calculus, I got to integration by parts, and then got reluctant. 


#15
Aug2407, 04:35 PM

P: 1,294

for #50:
can I just take derivatives of both sides, and then prove it? that way it takes only two steps to solve all the problem 


#16
Aug2407, 04:37 PM

P: 1,755

factoring [tex]I_1^2\cos^2\phi(\sin^2\theta1) = I_2^2 \sin^2\theta\sin^2\phi (\sin^2\theta1) = \cos^2\theta[/tex] [tex]I_1^2\cos^2\phi = \frac{I_2^2\sin^2\theta\sin^2\phi}{\cos^2\theta}[/tex] right so far? 


#17
Aug2407, 04:38 PM

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EDIT: At Rocophysics. Yes. Now do you see the 4th powers coming up ? You have obtained what i had already written b4. EDIT at EDIT: OOOps, i didn't see there were 2 angles, theta and phi. Then disregard my comments about the 4th power. 


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