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Trig proving (1 Viewer)

Thread starter crex
Start date Aug 17, 2013

C

crex

New Member

#1

Hey guys,
how do you prove this?
$sin(a-b) = 2 sin\frac{(a-b)} {2} cos \frac {(a-b)}2$
Thanks!

Last edited: Aug 17, 2013

Solution

BoS Revolutionary

#2

A little rusty eheh but...

Sin2x = 2SinxCosx

So if you consider:

Sin2[(a-b)/2] = 2Sin[(a-b)/2]Cos[(a-b)/2]

Therfore:

Sin(a-b) = (as above)

C

crex

New Member

#3

Solution said:
A little rusty eheh but...

Sin2x = 2SinxCosx

So if you consider:

Sin2[(a-b)/2] = 2Sin[(a-b)/2]Cos[(a-b)/2]

Therfore:

Sin(a-b) = (as above)

oh tyty

B

Bobbo1

Member

#4

yep simple double angle

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