sinA

=

cosA-coA.cos2A-(1/2)sin2A

sin2A-cosA

ALSO

8sinA.cosA.cos2A.cos4A = sin8A

how do you do these? and can u explain the 2nd one as well ... thanks

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- April 21st 2008, 12:09 PMJono Mathematicianmath is a PROBLEM!
sinA

=

__cosA-coA.cos2A-(1/2)sin2A__

sin2A-cosA

ALSO

8sinA.cosA.cos2A.cos4A = sin8A

how do you do these? and can u explain the 2nd one as well ... thanks - April 21st 2008, 01:27 PMwingless
Use and .

- April 21st 2008, 02:19 PMJono Mathematician
yeah i know that silly, but how i do it. haha

- April 21st 2008, 07:20 PMSoroban
Hello, Jono Mathematician!

Wingless suggested a few identities:

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We must find ways to apply them . . .

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