How do I solve for tan (theta) and theta?

if #sin(theta)=0.5# and #cos(theta)=-sqrt (3/2)#, evaluate:
a) #tan(theta)#
b) #theta#

2 Answers
landon3
Feb 5, 2018

#tan(theta)=-sqrt(6)#

#theta=ilog((sqrt(2))*(1+sqrt(3))/2)#

Explanation:

First,Solve for #tan(theta)# By using the fact That #tan(theta)=sin(theta)/cos(theta)#

and Since You said that #sin(theta)=0.5=1/2#

and #cos(theta)=-sqrt(3)/2#

That means #tan(theta)=sin(theta)/cos(theta)=(1/2)/(-sqrt(3)/2)=##-1/sqrt(3)=-sqrt(3)/3#

Lets Use the #cos(theta)# equation to find #theta#

#theta=cos^-1(-sqrt(6)/2)=ilog(u)#

when #u=(sqrt(6)+sqrt(2))/2=(sqrt(2))(1+sqrt(3))/2#

Aviv S.
Feb 5, 2018

#tan(theta)=1/-sqrt(3)~~-0.5773...#

#theta=150º#

Explanation:

I think you meant to write:

#sin(theta)=0.5#
#cos(theta)=color(red)(-sqrt(3)/2)#

For part (a), we can use the definition that #tan# is just #sin/cos#:

#tan(theta)=sin(theta)/cos(theta)#

#=>0.5/(-sqrt(3)/2)=1/-sqrt(3)#

For part (b), think about where on the unit circle the #y#-coordinate of a point is #0.5#; it's at #30º# and #150º#. Here are those points:

desmos.com/calculator

Similarly, also think about where on the unit circle the #x#-coordinate of a point is #-sqrt(3)/2#; it's at #150º# and #210º#

desmos.com/calculator

The point that both of these criteria share is #(-sqrt(3)/2,0.5)#, or the #150º# rotation.

Therefore, #theta# is be #150º#.

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