Asked by Dana
-Find sin(x+y), cos(x-y), tan(x+y) and the quadrant of x+y
sinx=-3/5, cosy=-7/25, x and y in Q3
sinx=-3/5, cosy=-7/25, x and y in Q3
Answers
Answered by
Reiny
In III , if sinx = -3/5, cosx = -4/5 , tanx = 3/4
in III, if cosy = -7/25, siny = -24/25 , tany = 24/7
( You should recognize the right-angled triangles with sides
3-4-5 and 7-24-25 )
sin(x+y)
= sinxcosy + cosxsiny
= (-3/5)(-7/25) + (-4/5)(-24/25)
= (21 + 96)/125
= 117/125
cos(x-y) = cosxcosy + sinxsiny
= (-4/5)(-7/25) + (-3/5)(-24/25)
= 28/125 + 72/125
= 100/125
= 4/5
tan(x+y) = (tanx + tany)/(1 - tanxtany)
= (3/4 + 24/7)/(1 - (3/4)(24/7))
= (117/28) / (1 - 18/7)
= (117/28) / (-11/7)
= (117/28)(-7/11)
= -117/44
sin(x+y) is positive and tan(x+y) is negative
the only quadrant where the sine is (+) and the tangent is (-) is quadrant II
x+y is in the second quadrant.
in III, if cosy = -7/25, siny = -24/25 , tany = 24/7
( You should recognize the right-angled triangles with sides
3-4-5 and 7-24-25 )
sin(x+y)
= sinxcosy + cosxsiny
= (-3/5)(-7/25) + (-4/5)(-24/25)
= (21 + 96)/125
= 117/125
cos(x-y) = cosxcosy + sinxsiny
= (-4/5)(-7/25) + (-3/5)(-24/25)
= 28/125 + 72/125
= 100/125
= 4/5
tan(x+y) = (tanx + tany)/(1 - tanxtany)
= (3/4 + 24/7)/(1 - (3/4)(24/7))
= (117/28) / (1 - 18/7)
= (117/28) / (-11/7)
= (117/28)(-7/11)
= -117/44
sin(x+y) is positive and tan(x+y) is negative
the only quadrant where the sine is (+) and the tangent is (-) is quadrant II
x+y is in the second quadrant.
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