Find sin2x, cos2x, and tan2x if sinx = -3/sqrt10 and x terminates in quadrant 3.

Question

Find sin2x, cos2x, and tan2x if sinx = -3/sqrt10 and x terminates in quadrant 3. I've been stuck on this and don't know how to go about it.

Answer 1

First note that: sin(2X) = 2*sin(x)*cos(x) and cos(x) = sqrt(1-sin²(x)) and also that cos(x) is a negative value in quadrant 3.

Therefore, if sin(x) = -3/√10 , we may obtain: cos(x) = sqrt(1 - 9/10) = -1/10.

This means sin(2X) = 2*sin(x)*cos(x) = 2 (√3)/10 = (√3)/5.

Now, note that cos(2x) = sin²(x) - cos²(x) = (-3/√10)² - (-1/10)² = 8/100.

Finally, tan(2x) = sin(2x)/cos(2x) = 5 (√3)/2.

Answer 2

Because sinx = -3/sqrt10, then we can think of a right triangle with opposite side 3 to the angle x and hypotenuse sqrt(10). Then, by the Pythagorean theorem, the adjacent side = sqrt(10-9)=1

Then, cosx=-1/sqrt10 (since x is in quadrant 3).

sin(2x)=2sinxcosx=2(-3/sqrt10)(-1/sqrt10)= 3/5

cos(2x)=1-2sin²(x)=1-9/5=-4/5

tan(2x)=-¾

I hope this helps! Please let me know if you have any questions. Good luck!

Answer 3

isn't the angle described by point (-1,-3)?

√((√10)²-3²)=1 and horizontal coordinate needs to be negative

for angle to be in quadrant 3

cos(x)=-1/√10

sin(2x)=2sin(x)cos(x)=2(3/10)

cos(2x)=cos²(x)-sin²(x)=(1/10)-(9/10)

tan(2x)=sin(2x)/cos(2x)