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How do you find the exact trig value for this and how do you know whether its negative or positive given -1290 degrees?

1 Answer

Aug 18, 2015

$sin(-1290^o)=1/2$ $color(white)("XXXX")$ $cos(-1290^o)= -sqrt(3)/2$ $color(white)("XXXX")$ $tan(-1290^o)= -1/sqrt(3)$

Explanation:

$-1290^o$ represents $3$ complete cycles
$color(white)("XXXX")$ plus half a cycle
$color(white)("XXXX")$ plus $30^o$
$color(white)("XXXXXXXX")$ (all in clockwise direction)

So $-1290^o$
$color(white)("XXXX")$ has a reference angle of $30^o$
$color(white)("XXXX")$ and is in Quadrant II ( $rArr$ only $sin$ is positive based on CAST system)

$30^o$ is one of the standard angles but see image (below) for clarification if necessary.
enter image source here

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