MHB Trigonometric Values Exact Values

[ardentmed]

Hey guys,

I've a few more questions this time around from my problem set:

(Ignore the first question , I only need help with 2a, b,and c.)

Question:

For the first one, I assessed the inside of the inverse trigonometric function first:

sin^-1 (2/3) = 0.785
Then tan-1(.785) ~ .66577,

Likewise, for b, I got 2.14173 and undefined for c.

However, I am having trouble finding these in exact values. How would I go about doing that? For example, is sin^-1 (2/3) a common trigonometric value I should be committing to memory? Thanks again guys.

 

[MarkFL]

Let's look at the first one, and then you can try the others...

$$\sin^{-1}\left(\frac{2}{3}\right)$$

can represent an angle in a right triangle whose opposite side is 2 and the hypotenuse is 3. So, we need to find the adjacent side $x$, and by Pythagoras we find:

$$x=\sqrt{3^2-2^2}=\sqrt{5}$$

Since the cotangent function is defined as adjacent over opposite, we mat then state:

$$\cot\left(\sin^{-1}\left(\frac{2}{3}\right)\right)=\frac{\sqrt{5}}{2}$$

Can you now use a similar method for the others? With the negative values, use known identities and symmetries.

 

[ardentmed]

Let's look at the first one, and then you can try the others...

$$\sin^{-1}\left(\frac{2}{3}\right)$$

can represent an angle in a right triangle whose opposite side is 2 and the hypotenuse is 3. So, we need to find the adjacent side $x$, and by Pythagoras we find:

$$x=\sqrt{3^2-2^2}=\sqrt{5}$$

Since the cotangent function is defined as adjacent over opposite, we mat then state:

$$\cot\left(\sin^{-1}\left(\frac{2}{3}\right)\right)=\frac{\sqrt{5}}{2}$$

Can you now use a similar method for the others? With the negative values, use known identities and symmetries.

Thanks, I got the same answer for the first one after drawing the triangles out. For the second and third one, I got 5/√ ̅34 and √ ̅15 / 4. Am I on the right track?

 

[MarkFL]

The second one is incorrect (2b)...but without seeing your work, I can't tell you where you went wrong.

 

[Dethrone]

I suspect he's misreading the triangle. $$sec\theta = \frac{H}{A}$$, h is the hypotenuse, and a is the adjacent. 1c) is also wrong for the same reason.

 

[MarkFL]

I suspect he's misreading the triangle. $$sec\theta = \frac{H}{A}$$, h is the hypotenuse, and a is the adjacent.

Yes, I suspect that as well, but I am trying to discourage the habit of simply posting an answer and saying "is that right?" I want to see work and reasoning, this way I can point to where the error is if the result is wrong. :D

 

[ardentmed]

I suspect he's misreading the triangle. $$sec\theta = \frac{H}{A}$$, h is the hypotenuse, and a is the adjacent. 1c) is also wrong for the same reason.

Oh, my bad. Alright, well since drawing out the triangles on a computer is such a hassle, I'll just post the new answers:

I got √ ̅34 /5and 4/√ ̅15

I'm hoping these are right. Thanks guys!

 

[Dethrone]

Those are correct, but if you really want to, you can rationalize the denominator for 1c).