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### Trigonometry Identities

```
Date: 01/07/97 at 20:03:43
From: Stephen A Barder
Subject: Trig Identities

I'm a junior in high school and got this problem for homework today:

Verify the following identity:

sin(x+y+z) = sinxcosycosz + cosxsinycosz + cosxcosysinz - sinxsinysinz

I know how to evaluate stuff like sin(x+y), but I don't even know
where to start with three variables.  Your help would be really
appreciated.

Tim Barder
```

```
Date: 01/08/97 at 00:54:54
From: Doctor Pete
Subject: Re: Trig Identities

Hi,

It's really not as difficult as you may think:  If you know that

Sin[w+x] = Sin[w]Cos[x] + Cos[w]Sin[x]

then it seems perfectly natural to ask what this would be if w = y+z.
This gives:

Sin[(y+z)+x] = Sin[y+z]Cos[x] + Cos[y+z]Sin[x]

Now, I leave it to you to further expand this expression, which is
fairly straightfoward.

In a similar vein, note that:

Cos[x]  = Cos[x]

Cos[2x] = Cos[x]^2 - Sin[x]^2
= Cos[x]^2 - (1-Cos[x]^2)
= 2 Cos[x]^2 - 1

Cos[3x] = Cos[x + 2x]
= Cos[x]Cos[2x] - Sin[x]Sin[2x]
= Cos[x](2 Cos[x]^2 - 1) - Sin[x](2 Sin[x]Cos[x])
= 2 Cos[x]^3 - Cos[x] - 2 Sin[x]^2 Cos[x]
= 2 Cos[x]^3 - Cos[x] - 2 (1-Cos[x]^2)Cos[x]
= 4 Cos[x]^3 - 3 Cos[x]

Cos[4x] = Cos[2x + 2x]
= 2 Cos[2x]^2 - 1
= 2 (2 Cos[x]^2 - 1)^2 - 1
= 2 (4 Cos[x]^4 - 4 Cos[x]^2 + 1) - 1
= 8 Cos[x]^4 - 8 Cos[x]^2 + 1

Apparently, we can always express Cos[kx] for any positive integer k
as some combination of powers of Cos[x].

Challenge 1 (easy):  Find Cos[5x], Cos[6x], Cos[7x] in terms of powers
of Cos[x].

Challenge 2 (medium):  Find Sin[kx] in terms of powers of Sin[x] for
integer values of k from 1 to 7.

Challenge 3 (hard):  Note that we can view Cos[kx] and Sin[kx] as
polynomials in terms of Cos[x] and Sin[x],
respectively.  From what you found in parts 1, 2,
is there a pattern in the coefficients of these
polynomials?  (Don't forget terms with
coefficient 0.)  Can you guess the pattern?

Challenge 4 (hard):  If you guessed a pattern in part 3, prove that
this pattern works; that is, find the polynomial
C[k,x] in Cos[x] such that C[k,x] = Cos[kx], and
the polynomial S[k,x] in Sin[x] such that
S[k,x] = Sin[kx].

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

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