# Thread: Trig Help

1. ## Trig Help

Hey Everyone,
First time poster. I just have a couple questions that I was wondering if anyone could help me with (big test tomorrow). Thanks in advance.

1.
Given tan^-1 (x) = u :

a) State the domain of tan^-1 (x)

b) Find the exact value of sin (2tan^-1(5)).

c) Find an algebraic expression in x for f(x)=sin(2tan^-1(x))

2
Solve for x exactly:

a) 2cos^2(x) - sin(x) = 1

b) 2cos(x) + 1 < 0

3. A quadrilateral ABCD with AB=10, BC=10, CD=12 and DA=15, and angle A=50 degrees. Find the area of the quadrilateral.

Thanks again. I'll try to return the favor by answering some others questions at some point.

2. Originally Posted by bluejfr09
1.
Given tan^-1 (x) = u :

a) State the domain of tan^-1 (x)

b) Find the exact value of sin (2tan^-1(5)).

c) Find an algebraic expression in x for f(x)=sin(2tan^-1(x))
a) For the domain we usually look for where x is not defined. So do you know of any values of x where $\displaystyle tan^{-1}(x)$ does not exist?

b)
$\displaystyle sin(2~tan^{-1}(5)) = 2sin(tan^{-1}(5))~cos(tan^{-1}(5))$

Now, sketch a right triangle with hypotenuse of length 1. (We can actually choose any length we like for this. I choose 1 as a standard) and let one of the non-right angles be $\displaystyle \theta$. My typical orientation is that the hypotenuse slopes up and to the right, and my $\displaystyle \theta$ is in the lower left corner. Define $\displaystyle \theta = tan^{-1}(5)$. This means that $\displaystyle tan(\theta) = 5$. So we know that the legs of the right triangle (x and y in an obvious orientation) are related by
$\displaystyle tan(\theta) = \frac{y}{x} = 5 \implies y = 5x$

So the Pythagorean theorem says that
$\displaystyle x^2 + y^2 = 1$

$\displaystyle x^2 + (5x)^2 = 1$

$\displaystyle 26x^2 = 1$

$\displaystyle x = \frac{1}{\sqrt{26}} \implies y = \frac{5}{\sqrt{26}}$

So
$\displaystyle sin(tan^{-1}(5)) = sin(\theta) = \frac{5}{\sqrt{26}}$
and
$\displaystyle cos(tan^{-1}(5)) = cos(\theta) = \frac{1}{\sqrt{26}}$

So finally:
$\displaystyle sin(2~tan^{-1}(5)) = 2sin(tan^{-1}(5))~cos(tan^{-1}(5)) = 2 \cdot \frac{5}{\sqrt{26}} \cdot \frac{1}{\sqrt{26}}$

$\displaystyle = \frac{10}{26} = \frac{5}{13}$

c) This is virtually identical to b). In fact, the right triangle is exactly the same, it's just that instead of 5 you have an x. (You'll want to change the variables x and y that I used in the above derivation, of course.)

-Dan

3. Originally Posted by bluejfr09
2
Solve for x exactly:

a) 2cos^2(x) - sin(x) = 1
Probably the best way to tackle a) is to recall that
$\displaystyle cos^2(x) = 1 - sin^2(x)$

Sub this into your equation and you'll have a quadratic in sin(x) that you can solve.

Originally Posted by bluejfr09
b) 2cos(x) + 1 < 0
$\displaystyle cos(x) < -\frac{1}{2}$

The easiest thing to do here would be to graph the function $\displaystyle y = cos(x) + \frac{1}{2}$ and see where it is negative.

-Dan

4. Thank you!