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# FP2 De Moivre's theorem / trig Watch

1. (a) Show that [email protected] = [email protected](16sin^[email protected]^[email protected]+5) done this

(b) Find exact values of x for 16x^4-20x^2+5=0 done this and got plus or minus sqrt(5/8 + or - sqrt(5)/8) which is correct

(c) Deduce exact values of sin(pi/5) and sin(2pi/5), explaining clearly the reasons for your answers.

I'm struggling to make a connection between the earlier parts of the question to do (c). Any hints?
2. Could you let @=pi/5 ?
3. I remember doing this but I can't remember why...

Right, judging by what's in the textbook and in my notes, the roots in b are the values for [email protected] for which [email protected] = 0

And then I guess you do [email protected] = sin^-1(0)
Divide those by five to get @. That's where you get 2pi/five.
And then you put that angle in
Sin2pi/5= the roots to b

I'm not sure that's right but that's how I would tackle it.
4. Ok I set @ = pi/5 and got 0 = 16sin^4(pi/5) -20sin^2(pi/5) + 5 used my answer to (b) to find sin(pi/5). I typed into my calc sin(pi/5) to find the approx value and then determined which root from (b) corresponded to the calc value for sin(pi/5). so got sqrt(5/8 - sqrt(5)/8). Did the same for sin(2pi/5). Now I don't know whether my solution was justified enough as it says "explain clearly" so is there a proper method for determining which out of the 4 roots was correct, rather than just using the calc?
5. (Original post by thers)
Ok I set @ = pi/5 and got 0 = 16sin^4(pi/5) -20sin^2(pi/5) + 5 used my answer to (b) to find sin(pi/5). I typed into my calc sin(pi/5) to find the approx value and then determined which root from (b) corresponded to the calc value for sin(pi/5). so got sqrt(5/8 - sqrt(5)/8). Did the same for sin(2pi/5). Now I don't know whether my solution was justified enough as it says "explain clearly" so is there a proper method for determining which out of the 4 roots was correct, rather than just using the calc?
Sketch y = sin x and indicate the approximate value of sin (pi/5) and sin (2pi/5) ?
6. (Original post by Mr M)
Sketch y = sin x and indicate the approximate value of sin (pi/5) and sin (2pi/5) ?
Ok that should suffice. Thanks for the help
7. (Original post by thers)
Ok I set @ = pi/5 and got 0 = 16sin^4(pi/5) -20sin^2(pi/5) + 5 used my answer to (b) to find sin(pi/5). I typed into my calc sin(pi/5) to find the approx value and then determined which root from (b) corresponded to the calc value for sin(pi/5). so got sqrt(5/8 - sqrt(5)/8). Did the same for sin(2pi/5). Now I don't know whether my solution was justified enough as it says "explain clearly" so is there a proper method for determining which out of the 4 roots was correct, rather than just using the calc?
I'm not sure if a calc answer would suffice. I'm doing that chapter soon so if I figure out another way asides from what I've said before I'll let you know. (:

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8. (Original post by PapercutMagnet)
I'm not sure if a calc answer would suffice. I'm doing that chapter soon so if I figure out another way asides from what I've said before I'll let you know. (:
It's almost like I'm invisible.
9. (Original post by Mr M)
It's almost like I'm invisible.
I'm sorry.
I'm not entirely sure what you meant by your first post.
Diagrams are always a good option. Whether it'd be valid enough just based on an approximation I'm not sure though. :/

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10. (Original post by PapercutMagnet)
Whether it'd be valid enough just based on an approximation I'm not sure though. :/
I am and it is.

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