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)

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- Nov 7th 2009, 09:26 PMJohnny Walker BlackNeed 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) - Nov 7th 2009, 09:45 PMDebsta
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. - Nov 8th 2009, 06:50 AMGrandad
Hello Johnny Walker BlackHere's the method for each one:

1) Use the identities below to expand the LHS:$\displaystyle \sin(x+y) = \sin x \cos y +\cos x \sin y$Simplify the result. Then take out a common factor of $\displaystyle \sin x$. Then use:

$\displaystyle \cos(x+y) = \cos x \cos y - \sin x \sin y$

$\displaystyle \cos^2y + \sin^2y = 1$and you're done.

2) Use the identities:$\displaystyle \cos 3x = 4\cos^3x - 3\cos x$Simplify the result, and factorise. Then use:

$\displaystyle \sin 3x = 3\sin x - 4\sin^3 x$

$\displaystyle \cos^2x = 1 - \sin^2 x$in the numerator. Simplify; cancel, and you're there.

3) Get rid of $\displaystyle \sec B$ and $\displaystyle \csc B$ (horrible things!) by multiplying top-and-bottom of the fraction by $\displaystyle \sin B\cos B$, using the fact that:$\displaystyle \csc B\sin B = 1$ and $\displaystyle \sec B \cos B = 1$Then use:$\displaystyle \sin B = 2 \sin\tfrac12B\cos\tfrac12B$Simplify. Then use $\displaystyle \tan\tfrac12B = \frac{\sin\tfrac12B}{\cos\tfrac12B}$, and you're there.

$\displaystyle \cos B = 2\cos^2\tfrac12B - 1$

Grandad