Expert: Steve Holleran - 10/16/2007

Question
I am having some trouble with a couple of identity problems. I appreciate any help I can get from you.
Here they are:

1. sin(a+b+c)+sin(a-b-c)/cos(a+b+c)-cos(a-b-c)=tan(b)tan(c)-1/tan(b)+tanc


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

3. 1+sin(2x)/cos(x)+sin(x)=cos(2x)/cos(x)-sin(x)

Thank you


Answer
Hi Cherry,

Boy, #1 and #3 are pretty tough!

1.  On this one, group things this way:

sin(a+b+c) = sin(a + (b+c))
sin(a-b-c) = sin(a - (b+c))
cos(a+b+c) = cos(a + (b+c))
cos(a-b-c) = cos(a - (b+c))

Then
sin(a + (b+c)) = sin a cos(b+c) + cos a sin(b+c)

sin(a - (b+c)) = sin a cos(b+c) - cos a sin(b+c)

Adding, you get 2 sin a cos(b+c)

For the bottom,

cos(a + (b+c)) = cos a cos(b+c) - sin a sin(b+c)

cos(a - (b+c)) = cos a cos(b+c) + sin a sin(b+c)

Subtracting these, you get

    -2 sin a sin(b+c)

So you have

   [2 sin a cos(b+c)] / [-2 sin a sin(b+c)]

Cancel the 2's and the sin a 's and you have:

     cos(b+c) / -sin(b+c)

=[cos b cos c - sin b sin c] / [-(sin b cos c + cos b sin c]

or

[cos b cos c - sin b sin c]/[-sin b cos c - cos b sin c]

Now divide each term, top and bottom, by cos b cos c.

You will get :

     [ 1 - tan b tan c] / [-tan b - tan c]

Factor -1 out of the top and bottom and cancel them, and you have

   tan b tan c - 1 / tan b + tan c


2.  2 sin^2 (2x) + cos (4x)

=   2 sin^2 (2x) + cos 2(2x)

=   2 sin^2 (2x) + (1 - 2 sin^2 (2x))  Double-angle identity
                                      for cosine

= 1, so this is a tautology--true for all values of x.


3.  Top:

  1 + sin 2x = 1 + 2 sin x cos x
So

[1 + 2 sin x cos x] / [cos x + sin x}

Multiply top and bottom by (cos x - sin x) and get

[cos x - sin x + 2 sin x cos^2 x - 2 sin^2 x cos x]

over [(cos x + sin x)(cos x - sin x)]

Now, on top, group the cos x with the -2sin^2 x cos x and factor out cos x:

 cos x (  - 2 sin^2 x)

Group the other two terms and factor out sin x:

 sin x (2 cos^2 x - 1)

Each parenthesis is a double angle formula for cosine, so

[cos x(cos 2x) + sin x (cos 2x)] is the numerator.

Now factor out cos 2x and the top is:

 cos 2x (cos x + sin x)

Cancel the cos x + sin x with the one in the bottom and you have

  cos 2x / (cos x - sin x)

Hope you could follow this okay.

It took awhile to figure these out, and even longer to try to type them in!  I hope this helps

Steve