# TrigWatch

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#1
Question 15. I attempted to use simultaneous equations to do this as I couldn't see another way at the time. However, my answer is wrong. I am trying to figure out what went wrong here.

Is it impossible to use simultaneous equations here for some reason unbeknownst to me?
Did I make a mistake with the algebra?

I'm just trying to figure out what went wrong by doing this.

Photos of the textbook, and my attempt, are included below.
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1 year ago
#2
(Original post by Illidan2)
Question 15. I attempted to use simultaneous equations to do this as I couldn't see another way at the time. However, my answer is wrong. I am trying to figure out what went wrong here.

Is it impossible to use simultaneous equations here for some reason unbeknownst to me?
Did I make a mistake with the algebra?

I'm just trying to figure out what went wrong by doing this.

Photos of the textbook, and my attempt, are included below.
Can't see the photos.
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#3
My apologies. Here you go:

https://imgur.com/a/Qy1HH7A
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#4
I am fairly confident that my method was probably an extremely unnecessarily arduous one, but it was the only one I could think of at the time and i'd hoped it would work nonetheless. Can simultaneous equations not be used in this way? My intention was to find a value for sin x and cos x that is consistent with both original equations, then plug this into the double-angle formulae cos^2(x)-sin^2(x). Unless my algebra manipulation is the problem, my method does not appear to be valid for a problem such as this.
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1 year ago
#5
cos(x) should be (m+n)/2.
sign wrong

You could just have done a difference of two squares
mn = cos^2 - sin^2
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#6
(Original post by mqb2766)
cos(x) should be (m+n)/2.
sign wrong
Ahh. I notice this now. Where I have it as cos x= (3m+n/2), I should have cos x=(m+n)/2, yeah. However, the solution to this problem is apparently "cos 2x=mn", meaning i'm still not anywhere close, especially as I have m^2 as part of my answer.
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#7
(Original post by mqb2766)
cos(x) should be (m+n)/2.
sign wrong

You could just have done a difference of two squares
mn = cos^2 - sin^2
Yes, I could have done a difference of two squares, but I did not notice at the time that I could. I am more concerned whether or not the method I thought of using is even valid in the first place, though.
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1 year ago
#8
(Original post by Illidan2)
Ahh. I notice this now. Where I have it as cos x= (3m+n/2), I should have cos x=(m+n)/2, yeah. However, the solution to this problem is apparently "cos 2x=mn", meaning i'm still not anywhere close, especially as I have m^2 as part of my answer.
The DOTS is the way to go as you can just write the answer down. However, your approach should work as well. You'd have
((m+n)/2)^2 - ((m-n)/2)^2
I think that gives mn?
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#9
(Original post by mqb2766)
The DOTS is the way to go as you can just write the answer down. However, your approach should work as well. You'd have
((m+n)/2)^2 - ((m-n)/2)^2
I think that gives mn?
I may be mistaken but after expanding those brackets I was left with 3mn/2. It may be a mistake of my own doing , but in squaring the second pair of brackets I obtained a spare mn/2 which was not cancelled out.
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1 year ago
#10
(Original post by Illidan2)
I may be mistaken but after expanding those brackets I was left with 3mn/2. It may be a mistake of my own doing , but in squaring the second pair of brackets I obtained a spare mn/2 which was not cancelled out.
Either I could type out the expression or you could show what you've done. Can you do the latter pls?
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#11
I can, and I would have already if my phone wasn't causing so many problems right now, sending a simple image takes some time. I'll let you know as soon as the imgur link has been updated
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1 year ago
#12
(Original post by Illidan2)
I can, and I would have already if my phone wasn't causing so many problems right now, sending a simple image takes some time. I'll let you know as soon as the imgur link has been updated
Ok, I'll wait :-). Make sure you get your negative signs right, that's what caused the problem in the first place.
Also, you could just type it in, the expression is relatively simple.
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#13
The imgur link is updated with a new image containing the expression. I've probably make an incredibly obvious mistake, but I would certainly benefit from having it pointed out. Thanks
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1 year ago
#14
(Original post by Illidan2)
The imgur link is updated with a new image containing the expression. I've probably make an incredibly obvious mistake, but I would certainly benefit from having it pointed out. Thanks
The first mn term should be mn/2 (when you expand ((m+n)/2)^2
1
#15
You're right. I have no idea why I wrote it that way. I see it now, though. Thank you!
0
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