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

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.

 

[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).