- #1
Maria
How do I solve these two:
cos 2v = cos v
sin 2v = sin v
cos 2v = cos v
sin 2v = sin v
Solving for cos 2v and sin 2v: Quick Guide
- Thread starter Maria
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AI Thread Summary
To solve the equations cos 2v = cos v and sin 2v = sin v, double angle formulas are applied. For cos 2v, using the identity cos 2v = cos²v - sin²v leads to the conclusion that v must equal 0, as this satisfies the equation. The second equation, sin 2v = sin v, can also be approached similarly, but it is noted to be more complex. The discussion emphasizes the importance of selecting values for t within the range of [-1, 1]. Overall, both equations can be solved through trigonometric identities and algebraic manipulation.
- #2
plover
Homework Helper
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Double angle formulas, e.g. sin 2x = 2(sin x)(cos x)
So your second equation becomes:
2(sin x)(cos x) = sin x
So your second equation becomes:
2(sin x)(cos x) = sin x
- #3
Motifs
- 43
- 0
cos2v=2(cosv)^2-1
t=cosv and solve equation.
The second I remmember you already asked and were properly answered by matt and mods, so why "repeat" it ?
[dont forget to choose only t's in the range [-1,1] only]
t=cosv and solve equation.
The second I remmember you already asked and were properly answered by matt and mods, so why "repeat" it ?
[dont forget to choose only t's in the range [-1,1] only]
- #4
wm
- 164
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Maria said:
Does it go like this?
(1) cosv = cos2v = cos( v + v) = cosv cos v - sinv sinv
= cosv [cos v] - [sinv sinv] = cosv[ 1] - [0] if it is to equal cosv, as given.
Therefore need: [cosv] = 1 and [sinv] = 0; so v = 0.
[PS: Your teacher may prefer Motifs' suggestion where you solve a standard quadratic equation.]
(2) is similar, but more interesting. Good luck.
Last edited:
Kindly see the attached pdf. My attempt to solve it, is in it.
I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction.
I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance)
My thoughts for the first case , one of them must be 12 volt while the other is 0
The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Here is what I tried
This question was actually asked in one of our engineering entrances.
The answer was 1D.
My teachers say that we have to use μₘ/fₘ to get to this answer. I cannot understand why. I'll be really glad if you could tell me the exact definition of power (numerically) that works in all scenarios.
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