[SOLVED] double argument properties

Sevens · December 20th 2006, 04:36 PM

Need some urgent help solving double argument properties for Calc/Trig course. Anybody out there that can help?

If cos A=3/5 and 0<A<90, find sin 2A and cos 1/2 A

Any help on this?

**Soroban** · December 20th 2006, 08:32 PM

Hello, Sevens!

You're expected to know the necessary identities:

. . $\sin2A\:=\:2\sin A\cos A$

. . $\cos\frac{A}{2}\:=\:\pm\sqrt{\frac{1 + \cos A}{2}}$

If $\cos A = \frac{3}{5}$ and $0^o < A < 90^o$ , find $[1]\;\sin2A$ and $[2]\;\cos\frac{A}{2}$

[1] Since $\cos A = \frac{3}{5}$ , we have a 3-4-5 right triangle. . Hence: . $\sin A = \frac{4}{5}$

Therefore: . $\sin2A \;=\;2\sin A\cos A \;=\;2\left(\frac{4}{5}\right)\left(\frac{3}{5}\ri ght)\;=\;\boxed{\frac{24}{25}}$

[2] $\cos\frac{A}{2} \;=\;\sqrt{\frac{1 + \cos A}{2}} \;=\;\sqrt{\frac{1 + \frac{3}{5}}{2}} \;=\;\sqrt{\frac{8}{10}}\;=\;\sqrt{\frac{4}{5}} \;=\;\frac{2}{\sqrt{5}}\;=\;\boxed{\frac{2\sqrt{5} }{5}}$

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