Which trigonometric identity do I use?

**doodofnood** · November 20th 2010, 09:54 AM

cosx - cosy = -2sin(x+y / 2)sin(x-y / 2)

I can't seem to find any that would work for it.
Thanks in advance!

**TheCoffeeMachine** · November 20th 2010, 09:59 AM

I realised that what you wanted was to prove it.

$\displaystyle -2\sin\left(\frac{x+y}{ 2}\right)\sin\left(\frac{x-y}{ 2}\right) = -2\sin\left(\frac{x}{2}+\frac{y}{2}\right)\sin\left (\frac{x}{2}-\frac{y}{2}\right)$

Using the identity $\sin\left(\alpha \pm \beta\right) = \sin{\alpha}\cos{\beta}\pm \cos{\alpha}\sin{\beta}$ , we have:

$\displaystyle -2\sin\left(\frac{x}{2}+\frac{y}{2}\right)\sin\left (\frac{x}{2}-\frac{y}{2}\right)= -2\left(\sin{\frac{x}{2}}\cos{\frac{y}{2}}+\cos{\fr ac{x}{2}}\sin{\frac{y}{2}}\right)\left(\sin{\frac{ x}{2}}\cos{\frac{y}{2}}-\cos{\frac{x}{2}}\sin{\frac{y}{2}}\right)$

$\displaystyle = -2\sin^2{\frac{x}{2}}\cos^2{\frac{y}{2}}\right)+2\c os^2{\frac{x}{2}}\sin^2{\frac{y}{2}} = -2\left(1-\cos^2{\frac{x}{2}}\right)\cos^2{\frac{y}{2}}\righ t)+2\cos^2{\frac{x}{2}}\left(1-\cos^2{\frac{y}{2}}\right)$

$\displaystyle = -2\cos^2{\frac{y}{2}}+2\cos^2{\frac{x}{2}}\cos^2{\f rac{y}{2}}+2\cos^2{\frac{x}{2}}-2\cos^2{\frac{x}{2}}\cos^2{\frac{y}{2}} = -2\cos^2{\frac{y}{2}}+2\cos^2{\frac{x}{2}}$

$\displaystyle = -2\left(\frac{\cos{y}+1}{2}\right)+2\left(\frac{\co s{x}+1}{2}\right) = -\left(\cos{y}+1}\right)+\left(\cos{x}+1\right) = \cos{x}-\cos{y}. \ \ \ \square$

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