Trigonometry, Prove the Identity and more

rocomath

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]?

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 [tex] 1-cos(2 \phi) = 2 sin^2 (\phi) [/tex], that should help.

rocomath

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

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

[tex]I_{1}^{2}\cos^{4}\theta=I_{2}^{2}\sin^{4}\theta [/tex]

You can't take sqrt like that.

rocomath

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

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

rootX

from where you got these questions ;P?
thanks.

rocomath

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

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

It's okay so far. Now cross multiply.

rootX

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

[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

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.