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Using half angle identities?

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Using half-angle identites, find the exact value for sin(195degrees).

I think I got an answer, so just an answer would be okay, but shown work would be great too!
thanks in advance
asked Jan 11, 2013 in TRIGONOMETRY by angel12 Scholar

1 Answer

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Using half-angle identites, find the exact value for sin(195degrees).

=sin (195)

{Note : sin^2(A) = (1-cos2A) / 2  ,  sinA = + √[ (1-cos2A) / 2 ]}

sin( 195 ) = + √[ (1-cos2(195)) / 2 ]

Simplify

sin( 195 ) = + √[ (1-cos390) / 2 ]

{Note : cos 390 = cos (360+30 )     ( in quadrant is I )

                       = cos 30 = (√3)/2 }

sin( 195 ) = + √[ (1-(√3)/2) / 2]

[ Note : sin (195) = sin(180+15) = - sin (15)

Select negative because of sin195 is in quardrant III where sin is negative ]

sin( 195 ) = -  √[ (1-(√3)/2) / 2]

sin( 195 ) = -  √[ (2-√3)/2) / 2]    here 1 , 2 LCM is 2

Simplify

sin( 195 ) = -  √[ (2-√3)/4)]

Simplify

sin( 195 ) = -  √[ (2-√3)] / 2                          here  √4 = 2

There fore

sin( 195 ) = -  √[ (2-√3)] / 2 

answered Jan 11, 2013 by richardson Scholar

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