1. Need 3 Homework problems

Verify these: [i need help getting started and other help you can throw at me]

1) sin(x + y)cos y - cos (x + y)sin y = sin x

2) cot x = (cos3x + cos x) / (sin 3x - sin x)

3) tan(B/2) = sec B / (sec B csc B + csc B)

2. Look up (and learn) the expansion rules for sin(a+b), cos(a+b), tan(a+b), sin (a-b),
sin2a, etc. And apply them here. Have a go or you'll never never know.

3. Hello Johnny Walker Black
Originally Posted by Johnny Walker Black
Verify these: [i need help getting started and other help you can throw at me]

1) sin(x + y)cos y - cos (x + y)sin y = sin x

2) cot x = (cos3x + cos x) / (sin 3x - sin x)

3) tan(B/2) = sec B / (sec B csc B + csc B)
Here's the method for each one:

1) Use the identities below to expand the LHS:
$\sin(x+y) = \sin x \cos y +\cos x \sin y$

$\cos(x+y) = \cos x \cos y - \sin x \sin y$
Simplify the result. Then take out a common factor of $\sin x$. Then use:
$\cos^2y + \sin^2y = 1$
and you're done.

2) Use the identities:
$\cos 3x = 4\cos^3x - 3\cos x$

$\sin 3x = 3\sin x - 4\sin^3 x$
Simplify the result, and factorise. Then use:
$\cos^2x = 1 - \sin^2 x$
in the numerator. Simplify; cancel, and you're there.

3) Get rid of $\sec B$ and $\csc B$ (horrible things!) by multiplying top-and-bottom of the fraction by $\sin B\cos B$, using the fact that:
$\csc B\sin B = 1$ and $\sec B \cos B = 1$
Then use:
$\sin B = 2 \sin\tfrac12B\cos\tfrac12B$

$\cos B = 2\cos^2\tfrac12B - 1$
Simplify. Then use $\tan\tfrac12B = \frac{\sin\tfrac12B}{\cos\tfrac12B}$, and you're there.