Given that tan¤=3/4 and ¤ is in the second quadrant find sin2¤ (¤=theta)
2 answers
see other post.
cos theta = + OR - 1 / sqrt ( 1 + tan ^ 2 theta )
In Quadrant II, cosine are negative so :
cos theta = - 1 / sqrt ( 1 + tan ^ 2 theta )
cos theta = - 1 / sqrt [ 1 + ( 3 / 4 ) ^ 2 theta ]
cos theta = - 1 / sqrt ( 1 + 9 / 16 )
cos theta = - 1 / sqrt ( 16 / 16 + 9 / 16 )
cos theta = - 1 / sqrt ( 25 / 16 )
cos theta = - 1 / ( 5 / 4 )
cos theta = - 4 / 5
sin ^ 2 theta + cos ^ 2 theta = 1
sin ^ 2 theta = 1 - cos ^ 2 theta
sin theta = + OR - sqrt ( 1 - cos ^ 2 theta )
In Quadrant II, sine are positive so :
sin theta = sqrt ( 1 - cos ^ 2 theta )
sin theta = sqrt [ 1 - ( - 4 / 5 ) ^ 2 ]
sin theta = sqrt ( 1 - 16 / 25 )
sin theta = sqrt ( 25 / 25 - 16 / 25 )
sin theta = sqrt ( 9 / 25 )
sin theta = 3 / 5
sin 2 theta = 2 sin theta cos theta
sin 2 theta = 2 * 3 / 5 * - 4 / 5
sin 2 theta = - 24 / 25
In Quadrant II, cosine are negative so :
cos theta = - 1 / sqrt ( 1 + tan ^ 2 theta )
cos theta = - 1 / sqrt [ 1 + ( 3 / 4 ) ^ 2 theta ]
cos theta = - 1 / sqrt ( 1 + 9 / 16 )
cos theta = - 1 / sqrt ( 16 / 16 + 9 / 16 )
cos theta = - 1 / sqrt ( 25 / 16 )
cos theta = - 1 / ( 5 / 4 )
cos theta = - 4 / 5
sin ^ 2 theta + cos ^ 2 theta = 1
sin ^ 2 theta = 1 - cos ^ 2 theta
sin theta = + OR - sqrt ( 1 - cos ^ 2 theta )
In Quadrant II, sine are positive so :
sin theta = sqrt ( 1 - cos ^ 2 theta )
sin theta = sqrt [ 1 - ( - 4 / 5 ) ^ 2 ]
sin theta = sqrt ( 1 - 16 / 25 )
sin theta = sqrt ( 25 / 25 - 16 / 25 )
sin theta = sqrt ( 9 / 25 )
sin theta = 3 / 5
sin 2 theta = 2 sin theta cos theta
sin 2 theta = 2 * 3 / 5 * - 4 / 5
sin 2 theta = - 24 / 25
2 answers