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September 1, 2014

September 1, 2014

Posted by **James** on Tuesday, September 17, 2013 at 10:25am.

(sinA cosA)

show that

A^2=(cos2A sin2A)

(sin2A cos2A)

- further Mathematics -
**Steve**, Tuesday, September 17, 2013 at 2:01pmA^2 = (cosA-sinA)^2 (sinAcosA)^2

= (cos^2A-2sinAcosA+sin^2A)(sin^2Acos^2A)

= (1-2sinAcosA)(1/4)(2sinAcosA)^2

= (1/4)(1-sin2A)(sin2A)(sin2A)

Hmmm. What if A = -pi/4?

(1/√2 + 1/√2)(-1/√2 * 1/√2) =? (0)(0)

-1/√2 =? 0

Nope. I don't think it's true. Typo?

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