Another: theta is a second quadrant angle in standard position and tan theta = - (sqrt5)/(2) .
Find the exact value of cos theta.
I assume I use the equation 1 / cos(theta)^2 = 1 + tan (theta)^2
However how does this work out algebraically?
tanØ= -√5/2 , so Ø must be in II
make a sketch of a right-angled triangle with sides √5 and 2, the hypotenuse is √9 or 3
since tanØ = -√5/2 = y/x
then
in quadrant II, x = -2, y = √5 and r = 3
cosØ = -2/3
thanks!
To find the exact value of cos theta, we can use the equation:
1 / cos(theta)^2 = 1 + tan(theta)^2
First, let's substitute the given value of tan theta into the equation:
1 / cos(theta)^2 = 1 + [-(sqrt5)/(2)]^2
Squaring the value of tan theta:
1 / cos(theta)^2 = 1 + 25/4
Now, let's simplify the right side of the equation:
1 / cos(theta)^2 = 4/4 + 25/4 = 29/4
To eliminate the denominator, we can multiply both sides by cos(theta)^2:
cos(theta)^2 * (1 / cos(theta)^2) = cos(theta)^2 * (29/4)
Simplifying this:
1 = 29 * cos(theta)^2 / 4
Now, multiply both sides by the reciprocal of 29/4, which is 4/29:
1 * (4/29) = (29 * cos(theta)^2 / 4) * (4/29)
Simplifying gives us:
4/29 = cos(theta)^2
Finally, to find the exact value of cos theta, we can take the square root of both sides:
√(4/29) = √[cos(theta)^2]
Thus, the exact value of cos theta is:
cos(theta) = ± √(4/29)
Note: Since theta is in the second quadrant, where cosine is negative, we take the negative square root. So:
cos(theta) = -√(4/29)