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### Proving Trig Identities

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Date: 11/15/98 at 20:53:06
From: Margi Moscoe
Subject: Trig Identities

Dear Dr. Math,

I would really appreciate it if you would show me how to prove these
identities:

1. sin^6x + cos^6x = 1 - 3sin^2xcos^2x (where x represents the angle)

2. cos3x = 4cos^3x - 3cosx (where x represents the angle)

3. 1 + cosx + sinx / 1 - cosx + sinx = 1 + cosx/sinx

Thank you very much for taking the time to help me with these questions
- your assistence is very much appreciated.

Long live Dr. Math!
```

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Date: 12/02/98 at 15:08:54
From: Doctor Dianna
Subject: Re: Trig Identities

Dear Margi,

I hope this helps:

1st Question: sin^6x + cos^6x = 1 - 3sin^2xcos^2x

We will manipulate the left side to look like the right side. Begin by
converting sin^6x to (sin^2x)^3 and likewise for cos^6x. It is easier
to work with sines and cosines when they are squared because there are
more identities to choose from. Your problem will then look like this:

sin^6x + cos^6x = (sin^2x)^3 + (cos^2x)^3

Now you will want to factor the right side as the sum of two cubes. You
may remember the formula for that is:

(a+b)^3 = (a+b)(a^2-ab+b^2)

To finish up this proof you will need to use the identity

sin^2x+cos^2x = 1.

You may need to be creative when using the identity, for example,

1-cos^2x = sin^2x.

Also, you can use the fact that sin^4x = (sin^2x)^2. As a final hint,
you will need to do a lot of foiling, factoring, and combining of like
terms. Good luck.

Next question: cos3x = 4cos^3x - 3cosx

The best way to begin is by manipulating the left side.  Note that

cos3x = cos(2x+x).

Now use a sum/difference formula, namely

cos(a+b) = cosacosb - sinasinb.

This will provide you with an excellent start.

As you finish this proof you may want to use the identities for sin2x
and cos2x. Also note that you are trying to make cos3x equivalent to
4cos^3x-3cosx. Notice there is no use of the sine function. Therefore,
as you work through the proof, look for ways to change sines into
cosines. For example the identity sin^2x = 1-cos^2x is one such way.

A final thought for this problem is that you will need to do a lot of
distribution. Be careful, it sometimes gets tricky with all those trig
functions and squares. Good luck.

Final question:  (1+sinx+cosx)/(1-cosx+sinx) = (1+cosx)/sinx

The best way to begin is with the left side, and rearrange the terms:

[(1+sinx)+cosx]/[(1+sinx)-cosx].

The idea here is to multiply this fraction by the conjugate. In other
words, multiply top and bottom by:

[(1+sinx)+cosx]

The result should be:

[(1+sinx)+cosx]^2 / [(1+sinx)^2-cos^2x]

It is best now to expand this expression by performing a lot of foil.
Once you carry out all that multiplication, you should see several
opportunities to use the identity sin^2x+cos^2x = 1.

This problem will require you to combine like terms and factor a lot.
Some things you will need to factor include:

1. Factor out a 2 in both numerator and denominator.
2. Factor out a sinx in the denominator.
3. Factor out a cosx in the last 2 terms of the numerator.
4. Factor out a (1+sinx) in the numerator.

After you factor and combine like terms you should be very close to the
final result. Thanks for your question and good luck with all your
other trig proofs.

- Doctor Dianna, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Trigonometry

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