# Thread: tanget converting to cosine

1. ## tanget converting to cosine

if tan β = 6 then what is cos β equal to?

i really don't know to approach this problem. i don't even know where to begin

please and thankyou to anyone who can help

2. Originally Posted by luckystar001
if tan β = 6 then what is cos β equal to?

i really don't know to approach this problem. i don't even know where to begin

please and thankyou to anyone who can help
$\displaystyle 0^o<\beta<\frac{{\pi}}{2}$

$\displaystyle tan\beta=\frac{6x}{x}$

The hypotenuse of the right-angled triangle is $\displaystyle \sqrt{x^2+36x^2}=x\sqrt{37}$

$\displaystyle cos\beta=\frac{x}{x\sqrt{37}}=\frac{1}{\sqrt{37}}$

$\displaystyle {\pi}<\beta<\frac{3{\pi}}{2}$

$\displaystyle cos\beta=-\frac{1}{\sqrt{37}}$

You can also work with the unit circle and let the radius be 1 (hypotenuse=1).

3. wow you are amazing! i would have never thought of using a right triangle to solve this problem; i was trying to find an answer using algebraic means and i only got as far as tan β = sin β / sin β

but why is β limited to (pi, 2pi) ?

4. Originally Posted by luckystar001
wow you are amazing! i would have never thought of using a right triangle to solve this problem; i was trying to find an answer using algebraic means and i only got as far as tan β = sin β / sin β

but why is β limited to (pi, 2pi) ?
It isn't, but $\displaystyle tan\beta$ gives the slope of a line through the origin

and if that's positive, then the line is increasing.
If you draw a unit-circle (radius=1, centred at the origin)
and draw a line through the origin with slope=6,
you will see this corresponds to two angles,
as there are two points on the circumference of the circle.

If we go beyond 360 degrees, we only wind up where we started!

When you have a geometric or trigonometric representation of the situation,
it can be very clear.

5. Originally Posted by luckystar001

but why is β limited to (pi, 2pi) ?
Sorry, I didn't answer that correctly!

If the angle was in either of the other two quadrants, tan(angle) would be negative.

I should have written

$\displaystyle 0<\beta<\frac{{\pi}}{2}$

and

$\displaystyle {\pi}<\beta<\frac{3{\pi}}{2}$

6. Originally Posted by luckystar001
if tan β = 6 then what is cos β equal to?
$\displaystyle 1 + \tan^2{\beta} = \sec^2{\beta}$

$\displaystyle 1 + 6^2 = \sec^2{\beta}$

$\displaystyle 37 = \sec^2{\beta}$

$\displaystyle \displaystyle \frac{1}{37} = \cos^2{\beta}$

$\displaystyle \displaystyle \pm \frac{1}{\sqrt{37}} = \cos{\beta}$

(+) if $\displaystyle \beta$ is in quad I , (-) if $\displaystyle \beta$ is in quad III

7. i think i finally understand this problem!

A big Thank you to all who helped! you've shown me so many ways to approach this problem!