Question about identities

**superduper1** · July 22nd 2010, 10:22 PM

Please explain how:
$-2sinx - \sqrt{3}sin^2 x - \sqrt{3}cos^2 x$

='s

$-2sinx - \sqrt{3}$

**pencil09** · July 22nd 2010, 10:29 PM

Originally Posted by superduper1

Please explain how:
$-2sinx - \sqrt{3}sin^2 x - \sqrt{3}cos^2 x$

='s

$-2sinx - \sqrt{3}$

$-2sinx - \sqrt{3}sin^2 x - \sqrt{3}cos^2 x$

since, $sin^2 x + cos^2 x = 1$

$-2sinx - \sqrt{3}sin^2 x - \sqrt{3}(1 - sin^2 x)$

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-- hope it'll help --
...pertamax gan....

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