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]?
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.
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.
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.
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]
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
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?
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.
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
[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?
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.
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.