Trigonometry, Prove the Identity and more

by rocomath
Tags: identity, prove, trigonometry
 P: 1,754 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 $$\theta = 45^\circ$$?
 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.
 P: 154 11) You should know the identity $$1-cos(2 \phi) = 2 sin^2 (\phi)$$, that should help.
P: 1,754
Trigonometry, Prove the Identity and more

 Quote by bel 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
 Sci Advisor HW Helper P: 11,915 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.
P: 1,754
 Quote by 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.
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.
 Sci Advisor HW Helper P: 11,915 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.
P: 1,754
 Quote by 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.
how did you get the sin/cos to the 4th power?

idk if my handwriting is clear, but $$\sin\theta = \frac{I_1\cos\phi}{\sqrt{(I_1\cos\phi)^2 + (I_2\sin\phi)^2}}$$
 Sci Advisor HW Helper P: 11,915 Okay. Raise to the second power and do all multiplications. What do you get ?
 P: 1,294 from where you got these questions ;P? thanks.
P: 1,754
 Quote by rootX 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
P: 1,754
 Quote by dextercioby 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

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

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?
 Sci Advisor HW Helper P: 11,915 It's okay so far. Now cross multiply.
P: 1,294
 Quote by rocophysics 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.
 P: 1,294 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
P: 1,754
 Quote by dextercioby 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

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

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

right so far?
HW Helper
P: 11,915
 Quote by 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
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.
P: 1,754
 Quote by 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
WTF!!! two steps? lol ...

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