FP2 De Moivre's theorem / trig

(a) Show that sin5@ = sin@(16sin^4@-20sin^2@+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?

Could you let @=pi/5 ?

Reply 2

PapercutMagnet

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 sin@ for which sin5@ = 0

And then I guess you do 5@ = 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.

(edited 11 years ago)

Reply 3

thers

OP

5

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?

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) ?

Reply 5

thers

OP

5

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

Reply 6

PapercutMagnet

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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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. (: