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### Trigonometric Identity

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Date: 02/12/98 at 16:02:07
From: Toby Smith
Subject: Trigonometric Identity

I have to prove that

sin 7x = 7(sin x) - 56(sin x)^3 + 112(sin x)^5 - (sin x)^7

But I have reached the point where I have:

sin 7x = 7(sin x) - 56(sin x)^3 + 112(sin x)^5 - 64(sin x)^7

I have graphed both of these on a Sharp EL-9300 Graphics Calculator,
and I think they are equal. How can this be?

But, more importantly, how can mine be changed to make it identical to
the one I am after?

Regards,
Toby Smith
```

```
Date: 02/13/98 at 07:47:26
From: Doctor Anthony
Subject: Re: Trigonometric Identity

You don't mention which method you used to find sin(7x).

One method is to evaluate  sin(6x + x) = sin(6x)cos(x) + cos(6x)sin(x)
and work on down with similar expressions for sin(6x), sin(5x) and so
on. This method is laborious and likely to end up with mistakes in the
heavy algebra.

A better method is using DeMoivre's theorem.

cos(7x) + i.sin(7x) = (cos(x) + i.sin(x))^7

Now expand the right hand side using the binomial theorem and take the
imaginary part to get sin(7x)

Using c to represent cos(x) and s to represent sin(x)

(c + i.s)^7 = c^7 + 7.c^6.(i.s) + 21.c^5.(i.s)^2 + 35.c^4.(i.s)^3
+ 35.c^3.(i.s)^4 + 21.c^2.(i.s)^5 + 7.c.(i.s)^6 + (i.s)^7

Take out the odd powers on the right as these will be the imaginary
part. Remember that i^3 = -i and i^4 = +1 so i^5 = i, i^7 = -i

We then get

sin(7x) = 7cos^6(x).sin(x) - 35cos^4(x).sin^3(x) + 21cos^2(x).sin^5(x)
- sin^7(x)

The last stage is to convert the various cos terms to sin terms using
the identity  cos^2(x) =  1 - sin^2(x)

cos^4(x) = 1 - 2sin^2(x) + sin^4(x)

cos^6(x) = 1 - 3sin^2(x) + 3sin^4(x) - sin^6(x)

So putting in these expressions, we get

sin(7x)=

7sin(x)[1-3sin^2(x)+3sin^4(x)-sin^6(x)] - 35sin^3(x)[1-2sin^2(x) +
sin^4(x)] + 21sin^5(x)[1-sin^2(x)] - sin^7(x)

= 7sin(x) - 56sin^3(x) + 112sin^5(x) - 64sin^7(x)

The second version that you give is the correct one. The two versions
will not be equal except for small values of sin(x), where for example
.01^7 = 10^(-14)

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Trigonometry

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