Question #139918
If sinx = 4/5 and siny= -12/13 , 0<x<Pi over 2, 3pi over 2 < y < 2pi, evaluate

a) cos (x+y)
b) tan (x-y)
1
Expert's answer
2020-11-01T17:15:03-0500


cosa=±1sin2a,tana=sina/cosa\cos a = \pm \sqrt{1 - sin^2 a}, \tan a = \sin a / \cos a


Precalculate:

cosx=1(4/5)2=3/5,tanx=4/3\cos x = \sqrt{1 - (4/5)^2} = 3/5, \tan x = 4/3

cosy=1(12/13)2=5/13,tany=12/5\cos y = \sqrt{1 - (-12/13)^2} = 5/13, \tan y = -12/5


Answer:

a)cos(x+y)=cosxcosysinxsiny=3/55/134/5(12/13)=63/65\begin{aligned}a) \cos(x+y) &= \cos x \cdot \cos y - \sin x \cdot \sin y \\ &= 3/5 \cdot 5/13 - 4/5 \cdot (-12/13) \\ &= 63/65\end{aligned}


b)tan(xy)=(tanxtany)/(1+tanxtany)=(4/3(12/5))/(1+4/3(12/5))=56/(1548)=56/33\begin{aligned} b) \tan(x-y) &= (\tan x - \tan y) / (1 + \tan x \cdot \tan y) \\ &= (4/3 - (-12/5)) / (1 + 4/3 \cdot (-12/5)) \\ &= 56 / (15 - 48 ) = -56 / 33\end{aligned}





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