Trigonometry, Prove the Identity and more

[rocomath]

Homework Statement

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:
\sin\theta = \frac{I_1\cos\phi}{\sqrt{(I_1\cos\phi)^2 + (I_2\sin\phi)^2}}. It can happen that I_1 = I_2. Assuming that this happens, simplify the equation.

Homework Equations

The Attempt at a Solution

11. Idk what I'm doing wrong

50. GRRR!

53. Can I make this assumption that \phi = 45^\circ and that \theta = 45^\circ?

 

[nicktacik]

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]

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

 

[rocomath]

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

lol omg ... i see it now, I'm so exhausted! thanks

 

[dextercioby]

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]

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]

You wrote 52 but i think you meant 53. You have

I_{1}^{2}\cos^{4}\theta=I_{2}^{2}\sin^{4}\theta

You can't take sqrt like that.

 

[rocomath]

You wrote 52 but i think you meant 53. You have

I_{1}^{2}\cos^{4}\theta=I_{2}^{2}\sin^{4}\theta

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 \sin\theta = \frac{I_1\cos\phi}{\sqrt{(I_1\cos\phi)^2 + (I_2\sin\phi)^2}}

 

[dextercioby]

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

 

[rootX]

from where you got these questions ;P?
thanks.

 

[rocomath]

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]

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 please ... i can't think straight anymore

\sin\theta = \frac{I_1\cos\phi}{\sqrt{(I_1\cos\phi)^2 + (I_2\sin\phi)^2}}

squaring both sides ...

\sin^2\theta = \frac{I_1^2\cos^2\phi}{I_1^2\cos^2\phi + I_2^2\sin^2\phi}

is my denominator right, or did i screw it up?

 

[dextercioby]

It's okay so far. Now cross multiply.

 

[rootX]

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]

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]

It's okay so far. Now cross multiply.

\sin^2\theta I_1^2\cos^2\phi + \sin^2\theta I_2^2\sin^2\phi = I_1^2\cos^2\phi

factoring

I_1^2\cos^2\phi(\sin^2\theta-1) = -I_2^2 \sin^2\theta\sin^2\phi<br /> <br /> (\sin^2\theta-1) = -\cos^2\theta

I_1^2\cos^2\phi = \frac{-I_2^2\sin^2\theta\sin^2\phi}{-\cos^2\theta}

right so far?

 

[dextercioby]

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 f_{1}&#039;(x)=f_{2}&#039;(x) \nRightarrow f_{1}(x)=f_{2}(x).

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]

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 ...

 

[rocomath]

Thanks you.
Yes I do. Just finished the first one.

wow, I almost missed these ones!

i love their synthesis problems, they're pretty good

 

[rocomath]

EDIT at EDIT: OOOps, i didn't see there were 2 angles, theta and phi. Then disregard my comments about the 4-th power.

lol! i told you that on post #8, is all good! this is good latex practice

EDIT ... oops i didn't say it, i just made it neater to show there were 2 angles, argh! my fault.

 

[rootX]

how about this one:
\frac{1}{{sin\left( t\right) }^{2}}=\frac{{I}_{2}^{2}\,{tan\left( p\right) }^{2}}{{I}_{1}^{2}}+1

 

[rootX]

take 1 to the left side, and this further simplifies the relationship :D

so cot(t) = [I2/I1]tan(p)

 

[nicktacik]

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

Nope, otherwise you could prove that 1=2.

 

[rocomath]

how about this one:
\frac{1}{{sin\left( t\right) }^{2}}=\frac{{I}_{2}^{2}\,{tan\left( p\right) }^{2}}{{I}_{1}^{2}}+1

is it okay to take the square root of both sides ...

I_1^2\cos^2\phi = \frac{I_2^2\sin^2\phi\sin^2\theta}{\cos^2\theta}

then it would become

\frac{I_1\cos\phi}{\sin\phi} = \frac{I_2\sin\theta}{\cos\theta}

i'm looking over yours ...

 

[rootX]

it gives the same answer:
cot(t) = [I2/I1]tan(p)

 

[learningphysics]

For 53, just substitute I_2 = I_1, then simplify. Factor the I_1^2 in the denominator... then take the square root of I_1^2, you have sin^2 + cos^2 in the denominator... you can simplify...

 

[rootX]

For 53, just substitute I_2 = I_1, then simplify. Factor the I_1^2 in the denominator... then take the square root of I_1^2, you have sin^2 + cos^2 in the denominator... you can simplify...

it says this only true sometimes...not always

 

[learningphysics]

it says this only true sometimes...not always

The question says, "assuming that this happens, simplify the equation"

 

[rootX]

The question says, "assuming that this happens, simplify the equation"

oh yes, it sounds correct

 

[learningphysics]

oh wow, very nice. thank you! i didn't even think about taking the e of both sides.

Be sure to begin with the identity and work towards the result you're trying to prove... ie: begin with sec^2x - tan^2x = 1 (don't know if you need to prove this also... it's a simple proof).

 

[Gib Z]

Unfortunately f_{1}&#039;(x)=f_{2}&#039;(x) \nRightarrow f_{1}(x)=f_{2}(x).

True, but it does mean that the function may only differ by a constant = ]

So you in fact can use the method, but at the end you must try a test value of x, say 0, and find the constant.

 

[rootX]

but that would require more number of steps lol.

 

[rootX]

50. ln |secx + tanx| = -ln |secx - tanx|
dy/dx of both sides = sec x

and ln |sec0 + tan0| = -ln |sec0 - tan0|
0=0

hmm.. only three steps!

 

[rocomath]

new problem, Prove the Identity

\tan^{-1}x+cot^{-1}x=\frac{\pi}{2}

i used the sum and difference identity of tan to prove that both sides were undefined ...

\tan^{-1}x+cot^{-1}x=\frac{\pi}{2}

\tan(\tan^{-1}x+cot^{-1}x)=\tan(\frac{\pi}{2})

\frac{\tan(tan^{-1}x)+\tan(\cot^{-1}x)}{1-\tan(\tan^{-1}x)\tan(cot^{-1}x)}=\tan\frac{\pi}{2}

latex isn't working ...

\frac{x+1/x}\{1-1} = tan(UND)

UND = UND

help with my logic please!

 

[rootX]

you know tan(pi/2) = UD
but cot(pi/2) = 0

so, take cot of both sides
and you can write it as
1/tan(all side)

 

[rocomath]

you know tan(pi/2) = UD
but cot(pi/2) = 0

so, take cot of both sides
and you can write it as
1/tan(all side)

well wouldn't i need the sum and difference identity for cot?

i proofed it the way i got the sum/diff for tan and got 0 = 0

EDIT: crap, i messed up, let me re-work this ... my handwriting is very bad, lol

by luck i arrived at the same answer, 0 in my numerator.

 

[rootX]

well wouldn't i need the sum and difference identity for cot?

no, you don't
cot\left( atan\left( x\right) +acot\left( x\right) \right) =0
\frac{1}{tan\left( atan\left( x\right) +acot\left( x\right) \right) }=0

 

[Gib Z]

\tan (\pi/2 -x) = \cot x.

 

[rocomath]

my teacher just showed me how to work 52, he did it in less than 5 steps ... amazing

 

[rootX]

he squared them, or used that assumption?

 

[cristo]

my teacher just showed me how to work 52, he did it in less than 5 steps ... amazing

I know I'm jumping into a thread I've probably not read properly here, but in question 52 it says that "I_1 can equal I_2 in some situations in mechanics. Assuming this is the case, simplify." Therefore, one simply sets I_1=I_2, which happens to simplify the equation a great deal.

 

[P.O.L.A.R]

here is a good one prove tan(1/n) = tan(1/[n+1])+tan(1/[n^2+1])

note: I don't have the problem with me I have make sure it is addition which I am ppositive it is.

 

[matadorqk]

1=2

Nope, otherwise you could prove that 1=2.

Which is clearly true.

 

[rootX]

here is a good one prove tan(1/n) = tan(1/[n+1])+tan(1/[n^2+1])

note: I don't have the problem with me I have make sure it is addition which I am ppositive it is.

after, spending an hour, I plugged in some random number:
f(n) = -tan\left( [\frac{1}{{n}^{2}+1}]\right) -tan\left( [\frac{1}{n+1}]\right) +tan\left( \frac{1}{n}\right)

f(6)
-tan\left( [\frac{1}{7}]\right) -tan\left( [\frac{1}{37}]\right) +tan\left( \frac{1}{6}\right)

I guess that really doesn't give a 0!

 

[P.O.L.A.R]

after, spending an hour, I plugged in some random number:
f(n) = -tan\left( [\frac{1}{{n}^{2}+1}]\right) -tan\left( [\frac{1}{n+1}]\right) +tan\left( \frac{1}{n}\right)

f(6)
-tan\left( [\frac{1}{7}]\right) -tan\left( [\frac{1}{37}]\right) +tan\left( \frac{1}{6}\right)

I guess that really doesn't give a 0!

You have to remember that when working with proofs you can only work one side of the proof not both. Usually it is best to start with the side that has the most variables.

 

[Gib Z]

If rootX is correct is that his expression doesn't not give zero, his *proof* is always a disproof by form of the counterexample n=6.

 

[P.O.L.A.R]

here is a good one prove tan(1/n) = tan(1/[n+1])+tan(1/[n^2+1])

note: I don't have the problem with me I have make sure it is addition which I am ppositive it is.

arctan(1/n) = arctan(1/[n+1])+arctan(1/[n^2+n+1])

forgot the n sorry. also I thought I put arctan but I guess not wow I really messed that one up I aologize. A little tired I guess.

 

[dextercioby]

Apply tangent on both sides of the equation. What do you get ?

 

[P.O.L.A.R]

Apply tangent on both sides of the equation. What do you get ?

I already proved it myself people were asking proof questions and I found this one to be a good one.

Starting that way didnt work for me. I couldn't figure out where to go after taking the tan.