Trigonometry, Prove the Identity and more


by rocomath
Tags: identity, prove, trigonometry
rocomath
rocomath is offline
#1
Aug24-07, 03:06 PM
rocomath's Avatar
P: 1,757
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]?
Phys.Org News Partner Science news on Phys.org
SensaBubble: It's a bubble, but not as we know it (w/ video)
The hemihelix: Scientists discover a new shape using rubber bands (w/ video)
Microbes provide insights into evolution of human language
nicktacik
nicktacik is offline
#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
bel is offline
#3
Aug24-07, 03:20 PM
P: 155
11) You should know the identity [tex] 1-cos(2 \phi) = 2 sin^2 (\phi) [/tex], that should help.

rocomath
rocomath is offline
#4
Aug24-07, 03:23 PM
rocomath's Avatar
P: 1,757

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
dextercioby is offline
#5
Aug24-07, 03:25 PM
Sci Advisor
HW Helper
P: 11,866
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
rocomath is offline
#6
Aug24-07, 03:37 PM
rocomath's Avatar
P: 1,757
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
dextercioby is offline
#7
Aug24-07, 03:44 PM
Sci Advisor
HW Helper
P: 11,866
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
rocomath is offline
#8
Aug24-07, 04:01 PM
rocomath's Avatar
P: 1,757
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
dextercioby is offline
#9
Aug24-07, 04:05 PM
Sci Advisor
HW Helper
P: 11,866
Okay. Raise to the second power and do all multiplications. What do you get ?
rootX
rootX is offline
#10
Aug24-07, 04:08 PM
rootX's Avatar
P: 1,295
from where you got these questions ;P?
thanks.
rocomath
rocomath is offline
#11
Aug24-07, 04:19 PM
rocomath's Avatar
P: 1,757
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
rocomath is offline
#12
Aug24-07, 04:24 PM
rocomath's Avatar
P: 1,757
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
dextercioby is offline
#13
Aug24-07, 04:26 PM
Sci Advisor
HW Helper
P: 11,866
It's okay so far. Now cross multiply.
rootX
rootX is offline
#14
Aug24-07, 04:32 PM
rootX's Avatar
P: 1,295
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
rootX is offline
#15
Aug24-07, 04:35 PM
rootX's Avatar
P: 1,295
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
rocomath is offline
#16
Aug24-07, 04:37 PM
rocomath's Avatar
P: 1,757
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
dextercioby is offline
#17
Aug24-07, 04:38 PM
Sci Advisor
HW Helper
P: 11,866
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
rocomath is offline
#18
Aug24-07, 04:38 PM
rocomath's Avatar
P: 1,757
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 ...


Register to reply

Related Discussions
Please help me prove this identity Advanced Physics Homework 0
Prove Heron's Formula (Trigonometry) Precalculus Mathematics Homework 4
could someone please prove this identity Precalculus Mathematics Homework 2
How to prove this identity? Introductory Physics Homework 1
Trigonometry identity proof trouble. (Help much appreciated) Introductory Physics Homework 15