If tan x = -3/4 and pi/2<x<pi find values of cos x and sin x?

Question

#tan x = -3/4 #and#pi/2 <x<pi # find the value of #cos x# and #sin x#

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Answer 1 · 2018-06-22T16:28:37

#cosx=-4/5 and sinx=3/5#

Here,

#tanx=-3/4 < 0 and pi/2 < x < pi=>II^(nd)Quadrant#

So,

#color(red)(sinx > 0 and cosx < 0#

Now,

#sec^2x=1+tan^2x=1+9/16=25/16#

#=>cos^2x=1/sec^2x=16/25#

#=>color(blue)(cosx=-4/5to[becausecosx < 0]#

We know that,

#sin^2x=1-cos^2x=1-16/25=9/25#

#=>color(blue)(sinx=+3/5to[becausesinx > 0]#

Hence,

#cosx=-4/5 and sinx=3/5#

Answer 2 · 2018-06-22T16:31:59

Second quadrant, positive sine and negative cosine,

#cos x= - 4/5 #

#sin x= 3/5 #

We're looking for #x# in the second quadrant, so we want a positive sine and negative cosine.

In general, the multivalued

#arctan (a/b)#

refers to a right triangle, opposite #a#, adjacent #b# so hypotenuse #sqrt{a^2+b^2}.# The sign on the square root is always ambiguous in these, so

#cos arctan(a/b) = pm b/sqrt{a^2+b^2}#

#sin arctan(a/b) = pm a/sqrt{a^2+b^2}#

Of course #3^2+4^2=5^2# so the hypotenuse is #5#.

#cos arctan({-3}/4) = pm 4/5 #

#sin arctan({-3}/4) = pm 3/5 #

We want a positive sine and negative cosine.

#cos x= - 4/5 #

#sin x= 3/5 #