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# trigonometric identity

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Use the formula

sinθ=1/2i(e^(iθ)-e^(-iθ))

To obtain the identity

sin^5θ=1/16(sin5θ−5sin3θ +10sinθ)

Apr 14, 2015

#3
+890
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To save on typing, let e^(i.theta) = z.

Then, using the binomial expansion,

$$\sin^{5}\theta=\\ \left[(z-1/z)/2\imath\right]^{5}=(z^{5}-5z^{3}+10z-10/z+5/z^{3}-1/z^{5})/32\imath$$

so,

$$16\sin^{5}\theta=\left[(z^{5}-1/z^{5})-5(z^{3}-1/z^{3})+10(z-1/z)\right]/2\imath \\ =\sin5\theta-5sin3\theta + 10sin\theta$$

Apr 17, 2015

#1
+112008
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I have not obtained the identity yet .

I did check your formula.  It seems to me that the 2i needs to be in brackets.

Apr 15, 2015
#2
+5

i have started it that way:

sin^5θ=1/(2i)^5(e^(iθ)-e^(-iθ))^5

Apr 15, 2015
#3
+890
+5

To save on typing, let e^(i.theta) = z.

Then, using the binomial expansion,

$$\sin^{5}\theta=\\ \left[(z-1/z)/2\imath\right]^{5}=(z^{5}-5z^{3}+10z-10/z+5/z^{3}-1/z^{5})/32\imath$$

so,

$$16\sin^{5}\theta=\left[(z^{5}-1/z^{5})-5(z^{3}-1/z^{3})+10(z-1/z)\right]/2\imath \\ =\sin5\theta-5sin3\theta + 10sin\theta$$

Bertie Apr 17, 2015
#4
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Thanks Bertie,

Thanks anon for asking this great question :))

I took about another 10 steps to get to the end but that is quite neat.

(I used yours as a guide.)

Thank you

Oh Bertie,

I saved some of my birthday cake for you.  I shall get it from the fridge

Chris and the regular foum users cooked it for me and it is delicious.

Also it is not as fattening as it looks and I can have an infinite number of big slices from the one cake.

It defies the laws of physics, matter is not conserved, it is endlessly reproduced.

Enjoy :)

Apr 17, 2015
#5
+890
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Thanks for the cake Melody, it looks delicious.

I'm not going to ask you how many candles were on it.

Apr 17, 2015