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Fatts13
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#21
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#21
(Original post by 13 1 20 8 42)
The first solution set is redundant because all such solutions are contained in the second. The last is incorrect. pi/6 does not work, for instance. For the other solutions you should get cos(x/2) + cos(5x/2) = 0 so 2 * cosx * cos(3x/2) = 0.
Ok so 2npi/5 is the valid one for the solution only?
And for the second solution 2cos(x)cos(3x/2) = 0
=> cos(x)cos(3x/2) = 0 so cosx = 0 and cos3x/2 = 0?
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math42
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#22
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#22
(Original post by Fatts13)
Ok so 2npi/5 is the valid one for the solution only?
And for the second solution 2cos(x)cos(3x/2) = 0
=> cos(x)cos(3x/2) = 0 so cosx = 0 and cos3x/2 = 0?
Well 4npi/5 is valid but unnecessary
Indeed. Well, either can be 0 (just clarifying that it's or, not and)
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Fatts13
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#23
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#23
(Original post by 13 1 20 8 42)
Well 4npi/5 is valid but unnecessary
Indeed. Well, either can be 0 (just clarifying that it's or, not and)
OMD thank you so very very much, you've been a massive help Sorry for being a hassle.

I'll rep you once TSR let me XD
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math42
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#24
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#24
(Original post by Fatts13)
OMD thank you so very very much, you've been a massive help Sorry for being a hassle.

I'll rep you once TSR let me XD
No worries. Not really a hassle as I am just sitting around today anyway (uni starts tomorrow).
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Fatts13
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#25
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#25
(Original post by 13 1 20 8 42)
No worries. Not really a hassle as I am just sitting around today anyway (uni starts tomorrow).
If thats the case, here's another question Name:  Screen Shot 2016-10-02 at 20.04.30.png
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Size:  20.7 KB
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math42
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#26
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#26
(Original post by Fatts13)
If thats the case, here's another question Name:  Screen Shot 2016-10-02 at 20.04.30.png
Views: 38
Size:  20.7 KB
first one I got a weird general solution but threw in n = 17 and it worked so I'll assume I did it right. Express everything in terms of sin and cos, remove the denominators, and it should all become clear..

Second one I have an expression in tan(theta), it's only slightly simpler. Play around with it, express everything in terms of sintheta and costheta again.
edit: lol it's way simpler just needed to cancel stuff
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Fatts13
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#27
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#27
(Original post by 13 1 20 8 42)
first one I got a weird general solution but threw in n = 17 and it worked so I'll assume I did it right. Express everything in terms of sin and cos, remove the denominators, and it should all become clear..

Second one I have an expression in tan(theta), it's only slightly simpler. Play around with it, express everything in terms of sintheta and costheta again.
edit: lol it's way simpler just needed to cancel stuff
So tan3theta would become sin3theta/cos3theta and cot(7theta +pi/7) would become cos(7theta+pi/7)/sin(7theta + pi/7)?
And then from there I did:
cos(7theta +pi/7) = cos(7theta)cos(pi/7) - sin(7theta)sin(pi/7) and then opened up sin as well and I'm at loss XD
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math42
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#28
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#28
(Original post by Fatts13)
So tan3theta would become sin3theta/cos3theta and cot(7theta +pi/7) would become cos(7theta+pi/7)/sin(7theta + pi/7)?
And then from there I did:
cos(7theta +pi/7) = cos(7theta)cos(pi/7) - sin(7theta)sin(pi/7) and then opened up sin as well and I'm at loss XD
No need to expand like that. Just cross-multiply, get everything on one side and you should recognise the resulting structure.
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Fatts13
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#29
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#29
(Original post by 13 1 20 8 42)
No need to expand like that. Just cross-multiply, get everything on one side and you should recognise the resulting structure.
so sin(3theta)sin(7theta+pi/7)/cos(3theta)cos(7theta+pi/7)?
then does it become tan(3theta)tan(7theta +pi/7)?
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math42
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#30
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#30
(Original post by Fatts13)
so sin(3theta)sin(7theta+pi/7)/cos(3theta)cos(7theta+pi/7)?
then does it become tan(3theta)tan(7theta +pi/7)?
I dunno what is going on there..

You have  \displaystyle \frac{ \sin 3 \theta}{ \cos 3 \theta} = \frac{ \cos (7 \theta + \frac{\pi}{7})}{ \sin (7 \theta + \frac{\pi}{7})} just cross multiply to get everything off the denominator
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Fatts13
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#31
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#31
(Original post by 13 1 20 8 42)
I dunno what is going on there..

You have  \displaystyle \frac{ \sin 3 \theta}{ \cos 3 \theta} = \frac{ \cos (7 \theta + \frac{\pi}{7})}{ \sin (7 \theta + \frac{\pi}{7})} just cross multiply to get everything off the denominator
OMD sorry I meant equals not divide.
So sin(3theta)sin(7theta+pi/7) = cos(3theta)cos(7theta+pi/7)
I have no clue what to do next
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math42
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#32
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#32
(Original post by Fatts13)
OMD sorry I meant equals not divide.
So sin(3theta)sin(7theta+pi/7) = cos(3theta)cos(7theta+pi/7)
I have no clue what to do next
Cos(A + B) = ?
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Fatts13
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#33
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#33
(Original post by 13 1 20 8 42)
Cos(A + B) = ?
Oh so now we expand? cos(7theta +pi/7) = cos(7theta)cos(pi/7) - sin(7theta)sin(pi/7)
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math42
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#34
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#34
(Original post by Fatts13)
Oh so now we expand? cos(7theta +pi/7) = cos(7theta)cos(pi/7) - sin(7theta)sin(pi/7)
Noo; don't complicate, simplify. Try getting everything on one side of the equation (i.e. something = 0) and it might click..
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Fatts13
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#35
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#35
(Original post by 13 1 20 8 42)
Noo; don't complicate, simplify. Try getting everything on one side of the equation (i.e. something = 0) and it might click..
My head is gone, it's not clicking anything, been deprived of maths for too long.
I got sin(3theta)sin(7theta+pi/7)/cos(3theta)cos(7theta+pi/7) =1
I honestly don't know how to use cos(A+B) without expanding.
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math42
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#36
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#36
(Original post by Fatts13)
My head is gone, it's not clicking anything, been deprived of maths for too long.
I got sin(3theta)sin(7theta+pi/7)/cos(3theta)cos(7theta+pi/7) =1
I honestly don't know how to use cos(A+B) without expanding.
You had sin(3theta)sin(7theta+pi/7) = cos(3theta)cos(7theta+pi/7)
so cos(3theta)cos(7theta+pi/7) - sin(3theta)sin(7theta+pi/7) = 0
Isn't this structure familiar?
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Fatts13
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#37
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#37
(Original post by 13 1 20 8 42)
You had sin(3theta)sin(7theta+pi/7) = cos(3theta)cos(7theta+pi/7)
so cos(3theta)cos(7theta+pi/7) - sin(3theta)sin(7theta+pi/7) = 0
Isn't this structure familiar?
Nope I did not know we could do that. I presumed because everything was multiplication so you could only divide and multiply.
So from there we then have:
cos(3theta + 7theta + pi/7) = 0
cos(10theta + pi/7) = 0
cos^-1(0) = pi/2
10theta + pi/7 = pi/2 + 2npi
so then the general solutions are: +/- pi/28 + pin/5? Correct?
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math42
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#38
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#38
(Original post by Fatts13)
Nope I did not know we could do that. I presumed because everything was multiplication so you could only divide and multiply.
So from there we then have:
cos(3theta + 7theta + pi/7) = 0
cos(10theta + pi/7) = 0
cos^-1(0) = pi/2
10theta + pi/7 = pi/2 + 2npi
so then the general solutions are: +/- pi/28 + pin/5? Correct?
Correct up to cos(10theta + pi/7) = 0 but your solution is incomplete. Note that cosx = 0 whenever x = n*pi + pi/2 (look at the graph).
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