# cosine of tangent

• Feb 28th 2009, 10:30 AM
estex198
cosine of tangent
The problem is to find Tan(B / 2)

What is the cosine of Tan(B) when (pi/2 < B < pi) which would be quad 2.

tanB = (sqrt(11) / 5)

Seems to me that since tan = sine / cosine, the cosine would be 5, but Coursecompass.com (the site where I do all my homework) insists this is incorrect.

TIA

- Rusty
• Feb 28th 2009, 11:18 AM
skeeter
Quote:

Originally Posted by estex198
The problem is to find Tan(B / 2)

What is the cosine of Tan(B) when (pi/2 < B < pi) which would be quad 2.

tanB = (sqrt(11) / 5)

Seems to me that since tan = sine / cosine, the cosine would be 5, but Coursecompass.com (the site where I do all my homework) insists this is incorrect.

TIA

- Rusty

cosine of any angle must be between -1 and 1 ... it cannot equal 5.

$B$ is in quadrant II

$\tan{B} = \frac{-\sqrt{11}}{-5} = \frac{opposite \, side}{adjacent \, side}$

$(-\sqrt{11})^2 + (-5)^2 = (hypotenuse)^2$

$11 + 25 = 36$

$hypotenuse = 6$

$\cos{B} = \frac{adjacent \, side}{hypotenuse} = -\frac{5}{6}$

or, using an identity ...

$\tan{B} = \frac{\sqrt{11}}{5}
$

$\sec^2{B} = 1 + \tan^2{B}$

$\sec^2{B} = 1 + \frac{11}{25} = \frac{36}{25}$

$\sec{B} = -\frac{6}{5}$

$\cos{B} = -\frac{5}{6}$
• Feb 28th 2009, 11:39 AM
skeeter
Quote:

Originally Posted by skeeter
cosine of any angle must be between -1 and 1 ... it cannot equal 5.

$B$ is in quadrant II

correction ... messed up my signs

$\tan{B} = -\frac{\sqrt{11}}{5}$

$\tan{B} = \frac{\sqrt{11}}{-5} = \frac{opposite \, side}{adjacent \, side}$

$(\sqrt{11})^2 + (-5)^2 = (hypotenuse)^2$

$11 + 25 = 36$

$hypotenuse = 6$

$\cos{B} = \frac{adjacent \, side}{hypotenuse} = -\frac{5}{6}$

or, using an identity ...

$\tan{B} = -\frac{\sqrt{11}}{5}
$

$\sec^2{B} = 1 + \tan^2{B}$

$\sec^2{B} = 1 + \frac{11}{25} = \frac{36}{25}$

$\sec{B} = -\frac{6}{5}$

$\cos{B} = -\frac{5}{6}$

.
• Feb 28th 2009, 05:20 PM
estex198
thanks
Skeeter, you rock man! Thanks for the help and for such a quick response. Peace.

- Rusty