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Trigonometry, Prove the Identity and more

by rocomath
Tags: identity, prove, trigonometry
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rocomath
#1
Aug24-07, 03:06 PM
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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]?
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nicktacik
#2
Aug24-07, 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.
bel
#3
Aug24-07, 03:20 PM
P: 154
11) You should know the identity [tex] 1-cos(2 \phi) = 2 sin^2 (\phi) [/tex], that should help.

rocomath
#4
Aug24-07, 03:23 PM
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Trigonometry, Prove the Identity and more

Quote Quote by bel View Post
11) You should know the identity [tex] 1-cos(2 \phi) = 2 sin^2 (\phi) [/tex], that should help.
lol omg ... i see it now, i'm so exhausted! thanks
dextercioby
#5
Aug24-07, 03:25 PM
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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.
rocomath
#6
Aug24-07, 03:37 PM
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Quote Quote by nicktacik View Post
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.
oh wow, very nice. thank you! i didn't even think about taking the e of both sides.

anyone for 53? can i make that assumption or did i screw up? i did it about 3x.
dextercioby
#7
Aug24-07, 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.
rocomath
#8
Aug24-07, 04:01 PM
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Quote Quote by dextercioby View Post
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.
how did you get the sin/cos to the 4th power?

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]
dextercioby
#9
Aug24-07, 04:05 PM
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Okay. Raise to the second power and do all multiplications. What do you get ?
rootX
#10
Aug24-07, 04:08 PM
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from where you got these questions ;P?
thanks.
rocomath
#11
Aug24-07, 04:19 PM
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Quote Quote by rootX View Post
from where you got these questions ;P?
thanks.
do you like them?

it's from Algebra and Trigonometry, Beecher (Calc-prep book) ... i love it! i didn't learn Trigonometry properly b/c i took it in a 3-week mini-mester ... i love Math now so i'm studying harddd
rocomath
#12
Aug24-07, 04:24 PM
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Quote Quote by dextercioby View Post
Okay. Raise to the second power and do all multiplications. What do you get ?
ok i'm getting the same answer ... i even substituted b/c the damn I's were making me dizzy

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?
dextercioby
#13
Aug24-07, 04:26 PM
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It's okay so far. Now cross multiply.
rootX
#14
Aug24-07, 04:32 PM
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Quote Quote by rocophysics View Post
do you like them?

it's from Algebra and Trigonometry, Beecher (Calc-prep book) ... i love it! i didn't learn Trigonometry properly b/c i took it in a 3-week mini-mester ... i love Math now so i'm studying harddd
Thanks you.
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.
rootX
#15
Aug24-07, 04:35 PM
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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
rocomath
#16
Aug24-07, 04:37 PM
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Quote Quote by dextercioby View Post
It's okay so far. Now cross multiply.
[tex]\sin^2\theta I_1^2\cos^2\phi + \sin^2\theta I_2^2\sin^2\phi = I_1^2\cos^2\phi[/tex]

factoring

[tex]I_1^2\cos^2\phi(\sin^2\theta-1) = -I_2^2 \sin^2\theta\sin^2\phi

(\sin^2\theta-1) = -\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?
dextercioby
#17
Aug24-07, 04:38 PM
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Quote Quote by rootX View Post
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
Unfortunately [tex]f_{1}'(x)=f_{2}'(x) \nRightarrow f_{1}(x)=f_{2}(x) [/tex].

EDIT: At Rocophysics. Yes. Now do you see the 4-th 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 4-th power.
rocomath
#18
Aug24-07, 04:38 PM
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Quote Quote by rootX View Post
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
WTF!!! two steps? lol ...


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