simple trig question.. (1 Viewer)

[skillz]

11) If Tan x= 1/5 and pi<x<3pi/2,
find the exact value of
a) sin x, b)cos x

13)
solve for the following for x [0,2pi]

a)tan(3x - pi/6) = sqrt3/3

c)2tan(x/2)+2=0

for c. i get up to here,

tan(x/2)=-1 = -pi/4
pi- pi/4, 2pi-pi/4, 3pi-pi/4, 4pi-pi/4

3pi/4, 7pi/4, 11pi/4, 15pi/4

3pi/8, 7pi/8, 11pi/8, 15pi/8

but that's not the correct answer..

thanks for any help.

 

[Raginsheep]

For 11, just use the tanx=1/5 to draw a right angled triangle and pythagoras to find the missing side. Then read sinx and cosx off the triangle but I don't understand u'r domain restriction. Is it suppose to be pi< x <3pi/2?

For 13c, if 0< x <2pi, then for tan(x/2)=-1, the domain is 0<x/2<pi.
.:.x/2=pi/4, NOT -pi/4 remeber the ASTC circle thingy.
.:.x=pi/2


Hmm......just realised that typing:
pi< x <3pi/2
withot the spaces makes the forums display pi<x<3pi/2....

 

[Mountain.Dew]

for 13(a), realise that sqrt3 / 3 = 1/sqrt 3

so 3x-pi/6 = pi/6, etc etc etc...so go from there.