I am pretty sure this can be done completely through identities. My LaTeX isn't working on TSR so:(Original post by 13 1 20 8 42)
I haven't made a proper go at it (should be revising not going back to try tricky a level stuff lol) but on playing around with stuff I find it depends on the values of theta which seems to not be what was intended..Spoiler:Showsquare both sides to get 2 + 2root(1  sin^2(2theta)) = lambda^2sin^2(theta) + mu^2cos^2(theta) + 2lambdamucos(theta)sin(theta)
but then the root on the lhs is either cos(2theta) or cos(2theta) depending on theta, and then the whole LHS expression is either 2(1 + cos(2theta)) or 2(1  cos(2theta)), giving us either 4cos^2(theta) or 4sin^2(theta), then we can compare coefficients and get different results in each case..
or maybe you want to check back into the original and see if they work, didn't bother with that lol
Spoiler:Show
I'm inclined to say, Lambda = 2 and Mu = 0 or Lambda = 0 and Mu = 2
But surely that leads to sin(x)=cos(x) => tan(x)=1 => x = pi/4 only...
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 11062016 14:56
Last edited by Euclidean; 11062016 at 14:58. 
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 11062016 15:04
(Original post by Euclidean)
Find real values of λ and μ such that:
If anyone has a go, please spoiler solutions 
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 11062016 15:09
(Original post by Craig1998)
I know this isn't that good a solution, but could you substitute some values ( = 0, ) and get some simulatenous equations which you can use to work out and . Another solution I tried didn't seem to work when I substituted values for theta back in.
Here's what I thought initially:
(Original post by Euclidean)
I am pretty sure this can be done completely through identities. My LaTeX isn't working on TSR so:Spoiler:Show
I'm inclined to say, Lambda = 2 and Mu = 0 or Lambda = 0 and Mu = 2
But surely that leads to sin(x)=cos(x) => tan(x)=1 => x = pi/4 only... 
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 11062016 15:15
(Original post by Euclidean)
I am pretty sure this can be done completely through identities. My LaTeX isn't working on TSR so:Spoiler:Show
I'm inclined to say, Lambda = 2 and Mu = 0 or Lambda = 0 and Mu = 2
But surely that leads to sin(x)=cos(x) => tan(x)=1 => x = pi/4 only...
Spoiler:Showlambda = +/2, mu = 0 or mu = +/2, lambda = 0, so yours with a bit extra. Didn't really check what worked but for the positive 2 in each case I checked out random ones in the appropriate range I thought it could work in and they worked so I dunno. When taking those square roots should be modulus right since a range of theta is not specified and you thus have loads of critical points when the signs of stuff change. So their probably needs to be some fairly extensive case analysis.

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 11062016 15:25
(Original post by Euclidean)
I am not sure, my proposed solution seems incredibly flawed at the moment so I'm not sure whether my question actually has any real solutions...
Here's what I thought initially:
Note yet another editSpoiler:ShowOn [0,pi/4], I am quite sure that lambda = 0, mu = 2 works.
Here we have the LHS equal to costheta + sintheta + costheta  sintheta (since costheta > sintheta on this interval) which is of course equal to the RHS.
Then on [pi/4, pi/2], lambda = 2, mu = 0 works, I think, because now sintheta > costheta
And we can probably categorise all the intervals if we care enough..
Actually I think it is intervals of size pi/2 which are key, seems [pi/4, pi/4] has lambda = 0, mu = 2, then [pi/4,3pi/4] has lambda = 2, mu = 0, then by educated pattern guessing though it'll be easy to check [3pi/4,5pi/4] probably has lambda = 0, mu = 2 and [5pi/4,7pi/4] mu = 0, lambda = 2, and then round and round we go again..Last edited by math42; 11062016 at 15:34. 
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 11062016 15:45
Guys anyone know if this question is even in our syllabus lol C3 Solomon paper K, question 4b.

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 11062016 15:52
(Original post by 13 1 20 8 42)
What I did led to I think sole potential solutionsSpoiler:Showlambda = +/2, mu = 0 or mu = +/2, lambda = 0, so yours with a bit extra. Didn't really check what worked but for the positive 2 in each case I checked out random ones in the appropriate range I thought it could work in and they worked so I dunno. When taking those square roots should be modulus right since a range of theta is not specified and you thus have loads of critical points when the signs of stuff change. So their probably needs to be some fairly extensive case analysis.
(Original post by 13 1 20 8 42)
Incidentally this is very reminiscent of a problem I had in my analysis paper yesterday which involved as a subpart showing that there is only one root of cosx = sinx in [0,pi/2]
Note yet another editSpoiler:ShowOn [0,pi/4], I am quite sure that lambda = 0, mu = 2 works.
Here we have the LHS equal to costheta + sintheta + costheta  sintheta (since costheta > sintheta on this interval) which is of course equal to the RHS.
Then on [pi/4, pi/2], lambda = 2, mu = 0 works, I think, because now sintheta > costheta
And we can probably categorise all the intervals if we care enough..
Actually I think it is intervals of size pi/2 which are key, seems [pi/4, pi/4] has lambda = 0, mu = 2, then [pi/4,3pi/4] has lambda = 2, mu = 0, then by educated pattern guessing though it'll be easy to check [3pi/4,5pi/4] probably has lambda = 0, mu = 2 and [5pi/4,7pi/4] mu = 0, lambda = 2, and then round and round we go again.. 
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 11062016 15:58
(Original post by Euclidean)
This is true, I hadn't considered that sin(2x)=/=1 etc. Although, where I initially modified the problem from (M4 paper) the angle theta is acute.
It's quite interesting that mu and lambda are now functions of theta rather than constants, that also completely skipped my mind. 
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 11062016 16:10
(Original post by mathsmann)
Guys anyone know if this question is even in our syllabus lol C3 Solomon paper K, question 4b.
Spoiler:Show
(sorry latex isn't working properly)
Assume true, therefore log base 2 of 3 = p/q, where p and q are integers.
2^p/q = 3 (note that q cannot equal 0)
(2^p)^1/q = 3
2^p = 3^q
2^n is always even when n>1 (1, 2, 4, 8..), 3^n is always odd (1, 3, 9, 27...)
So that means the only solution is p = 0 and q = 0, but since q does not equal 0, we have a contradiction. This leads to log base 2 of 3 being irrational. 
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 11062016 17:06
(Original post by Craig1998)
Not sure, but I think the solution for this is quite easy anyway (plus I love proof).
Spoiler:Show
(sorry latex isn't working properly)
Assume true, therefore log base 2 of 3 = p/q, where p and q are integers.
2^p/q = 3 (note that q cannot equal 0)
(2^p)^1/q = 3
2^p = 3^q
2^n is always even when n>1 (1, 2, 4, 8..), 3^n is always odd (1, 3, 9, 27...)
So that means the only solution is p = 0 and q = 0, but since q does not equal 0, we have a contradiction. This leads to log base 2 of 3 being irrational. 
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 11062016 17:08
when sin(theta) =1/2 , can anyone tell me what the theta values will be because the theta values im getting are wrong apparently

particlestudent
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 11062016 17:24
(Original post by mathsmann)
when sin(theta) =1/2 , can anyone tell me what the theta values will be because the theta values im getting are wrong apparently 
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 11062016 17:31
(Original post by particlestudent)
Should get 210 and 330 as the first 2 values on the positive x side. 
pineneedles
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 11062016 18:03
(Original post by Lilly1234567890)
Can someone explain to me what this question means..
Given that the equation f(x) = k where k is a constant, has exactly two roots.
state the range of possible values of k.
What does that mean?
f(x) = k
So f(x)  k = 0
K transforms f(x) by shifting it down by k units. We need to find the value of k which will mean f(x) will intersect with the x axis in two places. In the graph I've attached, if k is 1, the graph will have one root, meaning that as long as k > 1, the function will have two roots.
Another poster said you could do it using the discriminant, and you could, but this is how you would do it if you don't actually have f(x), and you just have a graph as some questions I've seen. I'll try and find an example and edit this post.
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sabahshahed294
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 11062016 18:10
Hello, would someone help me out in Q9?
http://www.madasmaths.com/archive/iy...apers/c3_h.pdf 
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 11062016 18:15
(Original post by sabahshahed294)
Hello, would someone help me out in Q9?
http://www.madasmaths.com/archive/iy...apers/c3_h.pdf 
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 11062016 19:43
(Original post by 13 1 20 8 42)
I presume you're fine on the relationship between cot and tan. Use the formula for tan(2theta) to derive a formula for tan(4theta). Then you plug the specific values into everything and the result falls out 
pineneedles
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 11062016 21:28
I'm having trouble with question 7:
http://www.madasmaths.com/archive/iy...apers/c3_u.pdf
I understand that sin1x = arcsinx but I'm not sure how we can manipulate the equation given. I thought maybe we can take the sin and cosine of both sides simultaneously but I'm pretty sure you can't do that.
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Kevin De Bruyne
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 11062016 21:35
(Original post by pineneedles)
I'm having trouble with question 7:
http://www.madasmaths.com/archive/iy...apers/c3_u.pdf
I understand that sin1x = arcsinx but I'm not sure how we can manipulate the equation given. I thought maybe we can take the sin and cosine of both sides simultaneously but I'm pretty sure you can't do that.
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pineneedles
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 11062016 21:55
(Original post by SeanFM)
Not simultaneously  try applying them both to the original to the equation, one at a time.
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