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C2 - Logarithms

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Original post by Chittesh14
Nvm, sorry I already done it again in the time - it was 1 mistake lol but I done it all over again haha!


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Is that mechanics?
Original post by RDKGames
Is that mechanics?


Nah C2 lol.
Chapter 2 sine and cosine rule

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Original post by RDKGames
I noticed they were symmetric hence why I chose my method. Anything else and I would've struggled as I've never solved these high degree polynomials before. If there a faster way of solving them/different methods? Like using matrices or something?

Also I just noticed my other solution is not correct, so I think I'll just leave it at that.


Well polynomials with degree 5 or above are not necessarily solvable, at least not in a closed form using only radicals. So you have to pick them carefully.
Original post by RDKGames
Is that mechanics?


Help please. I got the answer, but in a different way to the actual answer. I still got it correct though I guess....

The answer had the denominator as 2+2 2 + \sqrt 2, whereas I had it as sin y = 12\frac {1}{\sqrt 2}
(edited 7 years ago)
Q5 Sine and Cosine.png6.0KB
Original post by Chittesh14
Help please. I got the answer, but in a different way to the actual answer. I still got it correct though I guess....

The answer had the denominator as 2+2 2 + \sqrt 2, whereas I had it as sin y = 12\frac {1}{\sqrt 2}


Rationalise the denominator.
Original post by RDKGames
Rationalise the denominator.


I rationalised it by multiplying it by root 2 and then it became the numerator / 1?
How do you get to 2 + root 2?
Original post by Chittesh14
I rationalised it by multiplying it by root 2 and then it became the numerator / 1?
How do you get to 2 + root 2?


You don't have to rationalise the denominator of sin(y). Use the sine rule and then mutiply top and bottom by root 2 on one of the fractions which you divide by 1/root2
(edited 7 years ago)
Original post by RDKGames
You don't have to rationalise the denominator of sin(y). Use the sine rule and then mutiply top and bottom by root 2 on one of the fractions which you divide by 1/root2


Dw, I'm too confused. That's what I did anyway ^ I just multiplied everything by root 2 and eventually got to the answer.
Original post by RDKGames
I noticed they were symmetric hence why I chose my method. Anything else and I would've struggled as I've never solved these high degree polynomials before. If there a faster way of solving them/different methods? Like using matrices or something?

Also I just noticed my other solution is not correct, so I think I'll just leave it at that.


Try this question if you get a chance.

It is given that tan⁡kπ20,k=1,5,9,13,17 \tan {\frac{k\pi}{20}}, k=1,5, 9, 13, 17 are all solutions to the equation
x5−5x4−10x3+10x2+5x−1=0 \displaystyle x^5-5x^4-10x^3+10x^2+5x-1=0 .

Find the exact values of tan⁡kπ20 for k=1,5,9,13,17 \tan \frac{k\pi}{20} \text{ for } k=1, 5, 9, 13, 17 .
[You may not use a calculator.]
Original post by RDKGames
Rationalise the denominator.


Help please.
Q5
image.jpeg62.8KB
Original post by RDKGames
Rationalise the denominator.


That's what I did.
image.jpeg274.5KB
Original post by Chittesh14
That's what I did.


Well firstly you might want to use 30 degrees as stated in the question, not 36.
Original post by RDKGames
Well firstly you might want to use 30 degrees as stated in the question, not 36.


Oops lol. Fk me I was playing games, will start in a bit again -_- :angry:!!!! I hate when this happens.
Original post by Ano123
Try this question if you get a chance.

It is given that tan⁡kπ20,k=1,5,9,13,17 \tan {\frac{k\pi}{20}}, k=1,5, 9, 13, 17 are all solutions to the equation
x5−5x4−10x3+10x2+5x−1=0 \displaystyle x^5-5x^4-10x^3+10x^2+5x-1=0 .

Find the exact values of tan⁡kπ20 for k=1,5,9,13,17 \tan \frac{k\pi}{20} \text{ for } k=1, 5, 9, 13, 17 .
[You may not use a calculator.]


Been wondering for quite some time but I just can't get anywhere :frown:

I can get as far as showing that these are the roots with De Moivre's Theorem but anything after that and I'm clueless. Would be nicer is the roots followed
tan⁡θ=−tan⁡(π−θ) \tan \theta = -\tan(\pi-\theta) but at this stage I can't see any methods apart from perhaps some long winded double/half angle formulae to seperate the angles.

But hey, I can tell you that for k=5, tan⁡kπ20=1 \tan \frac{k\pi}{20} = 1 :biggrin:
Original post by RDKGames
Been wondering for quite some time but I just can't get anywhere :frown:

I can get as far as showing that these are the roots with De Moivre's Theorem but anything after that and I'm clueless. Would be nicer is the roots followed
tan⁡θ=−tan⁡(π−θ) \tan \theta = -\tan(\pi-\theta) but at this stage I can't see any methods apart from perhaps some long winded double/half angle formulae to seperate the angles.

But hey, I can tell you that for k=5, tan⁡kπ20=1 \tan \frac{k\pi}{20} = 1 :biggrin:


Factor theorem, factorise leaving a quartic, then solve quartic similarly to question I posted the other day.
Original post by RDKGames
Well firstly you might want to use 30 degrees as stated in the question, not 36.


Can you find the answer to the question I posted and post them plz. I am confused on this stupid question lol.


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Original post by Chittesh14
Can you find the answer to the question I posted and post them plz. I am confused on this stupid question lol.


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I would if I understood what the question is asking. "Two triangles ABC"... ?
Original post by Ano123
Factor theorem, factorise leaving a quartic, then solve quartic similarly to question I posted the other day.


Of course it had to be something as simple as that :/

tan(kπ20)=1+5−5+25[br]=1[br]=1+5+5+25[br]=1−5−5−25[br]=1−5+5−25 tan(\frac{k\pi}{20}) = 1+\sqrt5-\sqrt{5+2\sqrt5}[br]=1[br]=1+\sqrt5+\sqrt{5+2\sqrt5}[br]=1-\sqrt5-\sqrt{5-2\sqrt5}[br]=1-\sqrt5+\sqrt{5-2\sqrt5}

for k=1,5,9,13,17k=1, 5, 9, 13, 17 respectively.
Original post by RDKGames
I would if I understood what the question is asking. "Two triangles ABC"... ?


Idek, split the two triangles into two separate triangles like I did in my working out lol.
I think it refers to this example...


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Original post by RDKGames
I would if I understood what the question is asking. "Two triangles ABC"... ?


This example...
image.jpeg375.2KB

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