Higher maths problem

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Rox17
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#1
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#1
You have to solve for cosTPQ. Idk what to do, I did cosTPQ=cos(2x + 90) and did the expansion but I got a negative number? Please helppp
https://ibb.co/s58fHt4
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mqb2766
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#2
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#2
(Original post by Rox17)
You have to solve for cosTPQ. Idk what to do, I did cosTPQ=cos(2x + 90) and did the expansion but I got a negative number? Please helppp
https://ibb.co/s58fHt4
Can you uplood the original question and your attempt.
Assuming the diagram is correct and Im understanding the original problem, why not just calculate x, then use it to get the actual angle you want.
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Rox17
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#3
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#3
I can't get you the original question bc it's in a textbook and I don't have it with me. I also should've mentioned that this is non calculator
https://ibb.co/XsDqcrV
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mqb2766
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#4
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#4
(Original post by Rox17)
I can't get you the original question bc it's in a textbook and I don't have it with me. I also should've mentioned that this is non calculator
https://ibb.co/XsDqcrV
Probably best to post the original question when you can get to it. Are you expected to calculate the trig value or the actual angle?
Note that as its > 90, youd expect cos() to be negative. However, you seem to double the trig value rather than doubling the angle, then taking the trig of that.
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Rox17
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#5
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#5
How do I double just the angle? I think I'm supposed to calculate the exact value, not the actual angle. I did cos90=0 since I took the 90 from the right angle at the top
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mqb2766
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#6
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#6
(Original post by Rox17)
How do I double just the angle? I think I'm supposed to calculate the exact value, not the actual angle. I did cos90=0 since I took the 90 from the right angle at the top
The question (whereever it comes from) is about trig identites. So maybe have a read about those and write out your identity expansions carefully/clearly.
Last edited by mqb2766; 1 month ago
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Rox17
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#7
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#7
Okk thanks
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Naan102
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#8
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#8
(Original post by Rox17)
Okk thanks
Did u manage to work it out btw
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Rox17
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#9
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#9
No I didnt😭😭it's something to do w that cos90 i think bc it makes the answer negative, ill ask my teacher od Wednesday if I still can't figure it out
Last edited by Rox17; 1 month ago
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mqb2766
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#10
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#10
(Original post by Rox17)
No I didnt😭😭it's something to do w that cos90 i think bc it makes the answer negative, ill ask my teacher od Wednesday if I still can't figure it out
As above, why do you think that cos(2x+90) being negative is wrong?
Have you written the identity out clearly for cos(2x + 90)? I.e. try getting the sin(2x) part correct.
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Rox17
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#11
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#11
(Original post by mqb2766)
As above, why do you think that cos(2x+90) being negative is wrong?
Have you written the identity out clearly for cos(2x + 90)? I.e. try getting the sin(2x) part correct.
Bc it's a triangle and I'm pretty sure it can't be a negative angle right? I've rewritten it with the correct 2x part but the cos90 makes the first part 0 so the second part becomes negative? Or am I being stupid😭
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mqb2766
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#12
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#12
(Original post by Rox17)
Bc it's a triangle and I'm pretty sure it can't be a negative angle right? I've rewritten it with the correct 2x part but the cos90 makes the first part 0 so the second part becomes negative? Or am I being stupid😭
sin(x) and/or cos(x) is certainly not negative. But the angle is 2x + 90, so what must the sign of cos(2x + 90) be?
In your identity, the cos(90) is indeed 0, so the value of cos(2x+90) is determined by the second part.
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Rox17
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#13
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#13
(Original post by mqb2766)
sin(x) and/or cos(x) is certainly not negative. But the angle is 2x + 90, so what must the sign of cos(2x + 90) be?
In your identity, the cos(90) is indeed 0, so the value of cos(2x+90) is determined by the second part.
But the second part is taken away from the first part and there's no negatives in it so it's always gonna be negative?
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mqb2766
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#14
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#14
(Original post by Rox17)
But the second part is taken away from the first part and there's no negatives in it so it's always gonna be negative?
2x+90 > 90
so cos(2x+90) is indeed negative.
https://sites.google.com/view/tlmath...e3-trig-graphs
If you don't have your textbook handy where the question comes from, its really worth waiting until you do have it to go over the bits properly rather than being led too much to the answer.
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Naan102
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#15
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#15
(Original post by Rox17)
No I didnt😭😭it's something to do w that cos90 i think bc it makes the answer negative, ill ask my teacher od Wednesday if I still can't figure it out
Can u post the question when u have it. Thanks
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Rox17
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#16
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#16
(Original post by Naan102)
Can u post the question when u have it. Thanks
Yeah it's in the leckie and leckie higher maths textbook if u have it, around pg41 exercise 2e I think q9
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Rox17
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#17
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#17
(Original post by mqb2766)
2x+90 > 90
so cos(2x+90) is indeed negative.
https://sites.google.com/view/tlmath...e3-trig-graphs
If you don't have your textbook handy where the question comes from, its really worth waiting until you do have it to go over the bits properly rather than being led too much to the answer.
OHH OK I THINK I GOT IT NOW I thought I found the angle but I realised I was finding what cos = 🤦*♀️🤦*♀️🤦*♀️🤦*♀️
I put the inverse cos of the negative answer I got through a calculator just to check and it turned out positive
Oops don't mind the surded 13 at the end I forgot to scribble it out
https://ibb.co/ThbPNby
Last edited by Rox17; 1 month ago
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mqb2766
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#18
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#18
(Original post by Rox17)
OHH OK I THINK I GOT IT NOW I thought I found the angle but I realised I was finding what cos = 🤦*♀️🤦*♀️🤦*♀️🤦*♀️
I put the inverse cos of the negative answer I got through a calculator just to check and it turned out positive
https://ibb.co/ThbPNby
Looks about right. Note for the cos() being 12/13, it imples sin() is 5/13 as 5: 12:13 is a pythagorean triple
Last edited by mqb2766; 1 month ago
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Rox17
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#19
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#19
(Original post by mqb2766)
Looks about right. Note for the cos() being 12/13, it imples sin() is 5/13 as 5: 12:13 is a pythagorean triple
Okk thank you!
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