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).
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.
$\displaystyle 1 + \tan^2{\beta} = \sec^2{\beta}$
$\displaystyle 1 + 6^2 = \sec^2{\beta}$
$\displaystyle 37 = \sec^2{\beta}$
$\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