Question
Find the exact value of sin2(theta) if cos(theta) = -sqrt 5/3 and 180 < theta < 270.
A)-1/9
B)-4 sqrt 5/9
C)1/9
D)4 sqrt 5/9
B?
sin^2(theta) + cos^2(theta) = 1
sin^2(theta) = 1 - cos^2(theta)
sin^2(theta) = 1 - (sqrt 5/3)^2
sin^2(theta) = 1 - (sqrt 25/9)
A)-1/9
B)-4 sqrt 5/9
C)1/9
D)4 sqrt 5/9
B?
sin^2(theta) + cos^2(theta) = 1
sin^2(theta) = 1 - cos^2(theta)
sin^2(theta) = 1 - (sqrt 5/3)^2
sin^2(theta) = 1 - (sqrt 25/9)
Answers
Reiny
You are getting a lot of these wrong lately.
(I am going to use sinx for your sin(theta)
since cosx = √5 /3
using Pythagoras I found the other side ot the triangle to be 2.
But we are in the third quadrant so sinx =-2/3
then sin 2x = 2sinxcosx
= 2(-2/3)(-√5/3)
= 4√5/3 which is D
with a calculator, you could have easily checked that your choice and the others beside D would not work.
(I am going to use sinx for your sin(theta)
since cosx = √5 /3
using Pythagoras I found the other side ot the triangle to be 2.
But we are in the third quadrant so sinx =-2/3
then sin 2x = 2sinxcosx
= 2(-2/3)(-√5/3)
= 4√5/3 which is D
with a calculator, you could have easily checked that your choice and the others beside D would not work.
Reiny
oops, sorry type
make 4√5/3 which is D
read : 4√5/9 which is D
make 4√5/3 which is D
read : 4√5/9 which is D
Jon
I really thought I had that one.
My 1st thought was D but I figured since sqrt 5/3 was negative then so would my final answer.
My 1st thought was D but I figured since sqrt 5/3 was negative then so would my final answer.
Reiny
Are you familiar with the CAST rule, that is, do you know which ratios are negative in which quadrants??
Jon
No