If sinx= -1/4 and x terminates in the third quadrant, find the exact value of sin2x
My answer is sqrt-15/8
I get
cos(x)
=sqrt(1-sin^2(x))
=sqrt(1-(1/4)^2)
=sqrt(15/16)
so
sin(2x)
=+2sin(x)cos(x)
=+2*(1/4)sqrt(15/16)
=sqrt(15) / 8
wait i don't understand how you did your math?
dont you use the formula sin2 theta =2sintheta cos theta?
I used that formula sin^2x+cos2x=1
-1/4 squared minus that by one and squareroot it square root of 15 over 4
you can do these a number of ways...
Awesome thanks mathmate!
You're welcome!
Hello from the future
To find the value of sin(2x), we can use the double-angle formula for sine:
sin(2x) = 2sin(x)cos(x)
Given that sin(x) = -1/4 in the third quadrant, we can determine the value of cos(x) using the Pythagorean identity:
sin^2(x) + cos^2(x) = 1
(-1/4)^2 + cos^2(x) = 1
1/16 + cos^2(x) = 1
cos^2(x) = 1 - 1/16
cos^2(x) = 15/16
Since x is in the third quadrant, cos(x) is negative:
cos(x) = -√(15/16) = -√15/4 = -√15/4
Now, substituting the values of sin(x) and cos(x) into the double-angle formula:
sin(2x) = 2(-1/4)(-√15/4)
= 2(1/4)(√15/4)
= (√15)/8
The exact value of sin(2x) is (√15)/8, which matches your answer of sqrt-15/8.
sin2x=2sinx cos x
= 2*-.25*cos x but
sin^2x+cos^2x=1 so
cos^2x=.75
cosx=sqrt 3/4
so now you have it.
check sign: if x is in the third quadrant, so 2x will be in the second quadrant, where sine is +