Trigonometry identity

Johann
Visitor
2016-04-30 07:11:45
#1
Hi,
I need some help with this exercicse:
sin2 (a + b) = cos2a + cos2b - 2cos(a)cos(b)cos(a + b)
I have to show that this is true.

RE: Trigonometry identity

SteveC
Member
2016-04-30 07:16:13
#2
So, you have this
sin2 (a + b) = cos2a + cos2b - 2cos(a)cos(b)cos(a + b)

We have to use this formula

$\dpi{100} cos(a+b) = cos\, a \cdot cos \, b - sin\, a\cdot sin\, b$

I'm gonna start with the member from the right. So though, we have:

$\dpi{100} cos^{2}a + cos^{2}b - 2cos\, a \cdot cos\, b \cdot cos(a+b) = cos^{2}a + cos^{2}b - 2cos\, a \cdot cos\, b \cdot(cos\, a \cdot cos\, b - sin\, a \cdot sin\, b ) = cos^{2}a + cos^{2}b - 2\cdot cos^{2}\, a \cdot cos^{2}\, b +2\cdot cos\, a \cdot cos\, b \cdot sin\, a \cdot sin\, b$

$\dpi{100} = cos^{2}a(1-cos^{2}b) + cos^{2}b(1-cos^{2}a) +2\cdot cos\, a \cdot cos\, b \cdot sin\, a \cdot sin\, b$
And then we know the formula: sin2 a + cos2a = 1 => cos2 a = 1 - sin2a

and we have
$\dpi{100} = cos^{2}a\cdot sin^{2}b + cos^{2}b\cdot sin^{2}a +2\cdot cos\, a \cdot cos\, b \cdot sin\, a \cdot sin\, b =$

$\dpi{100} = (cos \, a\cdot sin \, b + cos \, b\cdot sin \, a)^{2} = sin^{2}(a+b)$