# Homework Help: Trigonometry, Prove the Identity and more

1. Aug 24, 2007

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

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 $$\phi = 45^\circ$$ and that $$\theta = 45^\circ$$?

Last edited: Aug 24, 2007
2. Aug 24, 2007

### 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.

3. Aug 24, 2007

### bel

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

Last edited: Aug 24, 2007
4. Aug 24, 2007

### rocomath

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

5. Aug 24, 2007

### 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.

6. Aug 24, 2007

### 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.

Last edited: Aug 24, 2007
7. Aug 24, 2007

### 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.

8. Aug 24, 2007

### rocomath

how did you get the sin/cos to the 4th power?

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

9. Aug 24, 2007

### dextercioby

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

10. Aug 24, 2007

### rootX

from where you got these questions ;P?
thanks.

11. Aug 24, 2007

### 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

12. Aug 24, 2007

### 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

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

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?

Last edited: Aug 24, 2007
13. Aug 24, 2007

### dextercioby

It's okay so far. Now cross multiply.

14. Aug 24, 2007

### 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.

15. Aug 24, 2007

### 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

16. Aug 24, 2007

### rocomath

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

factoring

$$I_1^2\cos^2\phi(\sin^2\theta-1) = -I_2^2 \sin^2\theta\sin^2\phi (\sin^2\theta-1) = -\cos^2\theta$$

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

right so far?

17. Aug 24, 2007

### dextercioby

Unfortunately $$f_{1}'(x)=f_{2}'(x) \nRightarrow f_{1}(x)=f_{2}(x)$$.

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.

Last edited: Aug 24, 2007
18. Aug 24, 2007

### rocomath

WTF!!! two steps? lol ...

19. Aug 24, 2007

### rocomath

i love their synthesis problems, they're pretty good

20. Aug 24, 2007

### rocomath

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.

21. Aug 24, 2007

### rootX

$$\frac{1}{{sin\left( t\right) }^{2}}=\frac{{I}_{2}^{2}\,{tan\left( p\right) }^{2}}{{I}_{1}^{2}}+1$$

22. Aug 24, 2007

### rootX

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

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

23. Aug 24, 2007

### nicktacik

Nope, otherwise you could prove that 1=2.

24. Aug 24, 2007

### rocomath

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 ...

25. Aug 24, 2007