How to find sin x and tan x?

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1 Answer
Jim G.
Feb 27, 2018

#sinx=-4/5" and "tanx=-4/3#

Explanation:

#"using the "color(blue)"trigonometric identities"#

#•color(white)(x)tanx=sinx/cosx#

#•color(white)(x)sin^2x+cos^2x=1#

#rArrsinx=+-sqrt(1-cos^2x)#

#"since "(3pi)/2 < x<2pi#

#"this places x in the fourth quadrant where both"#

#sinx<0" and "tanx<0#

#rArrsinx=-sqrt(1-(3/5)^2)#

#color(white)(rArrsinx)=-sqrt(16/25)=-4/5#

#rArrtanx=(-4/5)/(3/5)=-4/5xx5/3=-4/3#

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