Double-Angle Formulas

• Nov 29th 2009, 09:21 AM
iluvmathbutitshard
Double-Angle Formulas
Find the exact values of sin 2u, cos 2u, and tan 2u using the double angle formulas.:
cos u = -2/3, pi/2 < u < pi

So I know that the formulas are:
sin 2u = 2 sin u cos u
cos 2u = cos^2 u - sin^2 u
tan 2u = (2 tan u) / (1 - tan^2 u)

Now I am having trouble figuring out what sin u is. I believe that it is
sqrt(3), but I am not sure.
If I am right, the first formula would be:
sin 2u = (2) (sqrt3) (-2/3)
but I'm not sure how to multiply a square root with a fraction.

Any help is appreciated.
• Nov 29th 2009, 09:27 AM
Jhevon
Quote:

Originally Posted by iluvmathbutitshard
Find the exact values of sin 2u, cos 2u, and tan 2u using the double angle formulas.:
cos u = -2/3, pi/2 < u < pi

So I know that the formulas are:
sin 2u = 2 sin u cos u
cos 2u = cos^2 u - sin^2 u
tan 2u = (2 tan u) / (1 - tan^2 u)

Now I am having trouble figuring out what sin u is. I believe that it is
sqrt(3), but I am not sure.
If I am right, the first formula would be:
sin 2u = (2) (sqrt3) (-2/3)
but I'm not sure how to multiply a square root with a fraction.

Any help is appreciated.

nope, that's not the right value for sin(u). there are several ways you can get to sin(u). Two conventional ways are:

Use the fact that $\sin^2 u + \cos^2 u = 1$

Solve for $\sin u$. Note that we are in the second quadrant so you will need to take the positive square root. As sine is positive in that quadrant.

Another way: Draw a right-triangle with an acute angle $u$. The adjacent side to $u$ should be labeled $2$ while the hypotenuse should be labeled $3$. Use Pythagoras' Theorem to find the opposite side and use sine = opposite/hypotenuse
• Nov 29th 2009, 09:35 AM
iluvmathbutitshard
So sin u would be sqrt5/3?

If so, how would I multiply this by -2/3?

• Nov 29th 2009, 09:39 AM
Jhevon
Quote:

Originally Posted by iluvmathbutitshard
So sin u would be sqrt5/3?

yes, $\sin u = \frac {\sqrt 5}3$

Quote:

If so, how would I multiply this by -2/3?

the same way you multiply any two fractions... multiply the numerators together to get the new numerator and the denominators together to get the new denominator. $\frac {\sqrt 5}3 \cdot \frac {-2}3 = \frac {-2 \sqrt 5}9$ :p
• Nov 29th 2009, 09:43 AM
iluvmathbutitshard
Ok, great :)

SO then for cos 2u:
cos^2 u - sin^2 u
(4/9) - (5/9) = (-1/9)

Is this correct?
• Nov 29th 2009, 09:47 AM
Jhevon
Quote:

Originally Posted by iluvmathbutitshard
Ok, great :)

Quote:

SO then for cos 2u:
cos^2 u - sin^2 u
(4/9) - (5/9) = (-1/9)

Is this correct?
yup
• Nov 29th 2009, 09:58 AM
iluvmathbutitshard
My answer for sin u was
(-4sqrt5) / (9)

And one last thing, is my tan correct:
(2 tan u) / (1 - tan^2 u)
(2 (-sqrt5/2)) / (1 - (5/4))

So this would simplify, I think, to
[(-2 sqrt5) / (2)] / [-1/4]
Is this correct?
• Nov 29th 2009, 10:03 AM
Jhevon
Quote:

Originally Posted by iluvmathbutitshard
My answer for sin u was
(-4sqrt5) / (9)

ok, good.

Quote:

And one last thing, is my tan correct:
(2 tan u) / (1 - tan^2 u)
(2 (-sqrt5/2)) / (1 - (5/4))

So this would simplify, I think, to
[(-2 sqrt5) / (2)] / [-1/4]
Is this correct?
that's not simplified :p

for this last question, I would use the fact that $\tan 2 u = \frac {\sin 2u}{\cos 2u}$

just divide the first two answers you got...
• Nov 29th 2009, 10:07 AM
iluvmathbutitshard
So the answer would be 4 sqrt5?
• Nov 29th 2009, 10:11 AM
Jhevon
Quote:

Originally Posted by iluvmathbutitshard
So the answer would be 4 sqrt5?

yup (Sun)
• Nov 29th 2009, 10:14 AM
iluvmathbutitshard
Thank you so much for your help.
(Rofl)