hectictar

At first I was thinking like asinus that these equations were true for all values of beta, but then I realized that can't be because if beta were 90° that would make it 1 - 0 = 0, which is not true. So then I thought about this:

\(\sin \beta - \cos^2 \beta = 0 \\ - \cos^2 \beta = - \sin \beta \\ - \cos^2 \beta + 1 = - \sin \beta + 1 \\ 1 - \cos^2 \beta = - \sin \beta + 1 \\ \sin^2 \beta = - \sin \beta + 1 \\ 0 = - \sin^2 \beta - \sin \beta + 1 \\ 0 = \sin^2 \beta + \sin \beta - 1\)

Then use the quadratic formula or complete the square to find out that

\(\sin \beta = \frac{\sqrt{5}-1}{2}\)

and

\(\sin \beta = \frac{-\sqrt{5}-1}{2}\)

So

\( \beta =\arcsin( \frac{\sqrt{5}-1}{2} )\)

and

\(\beta = \arcsin (\frac{-\sqrt{5}-1}{2})\)

I don't really know if that helps figure out the rest of the problem or not.

\((\csc (x) - \cot (x))(\sec (x)+1) \\~\\ = (\frac{1}{\sin (x)}-\frac{\cos (x)}{\sin (x)})(\frac{1}{\cos (x)}+1) \\~\\ = (\frac{1-\cos (x)}{\sin (x)})(\frac{1}{\cos (x)}+\frac{\cos (x)}{\cos (x)}) \\~\\ = (\frac{1-\cos (x)}{\sin (x)})(\frac{1+\cos (x)}{\cos (x)}) \\~\\ = \frac{1 + \cos (x) - \cos (x) - \cos^2 (x)}{\sin (x) \cos (x)} \\~\\ = \frac{1 - \cos^2 (x)}{\sin (x) \cos (x)} \\~\\ = \frac{\sin^2 (x)}{\sin (x) \cos (x)} \\~\\ = \frac{\sin (x)}{ \cos (x)} \\~\\ = \tan (x)\)

It actually is possible.

Sometimes if I expand a post by clicking the down arrow from the view all questions page, then click "load next questions" it displays a post with each letter on a new line.

L

i

k

e

T

h

i

s

.

Sometimes it's the post that I expanded and sometimes it's not. I can't figure out what exactly causes it. Even if I collapse the post before clicking "load next questions" it still can do it.

It's not a major problem, just a minor annoyance. I thought that I'd add it here just in case.

One other thing I noticed is it seems like posts get "hidden" under the "_ hours ago" headlines. Sometimes a post will disappear, but if you wait for the next post to come in it will "push" the one that was previously hidden out and the one behind it disappears.

0.490 = 0.490/x

Multiply both sides by x.

0.490x = 0.490

Divide both sides by 0.490.

x = 0.490/0.490

Anything divided by itself is 1.

x = 1

I actually drew the second picture wrong. I think the confusion is over whether the 65º angle is from vertical or horizontal. This is how the second one I did should look like:

I was making it really fast so I messed up.

But when it's like this the height actually is 80 - 45tan25º ≈ 59.016 inches

The shooters height will be (80 + x) inches

The third angle in the triangle is 180 - 90 - 65 = 25º

tan25º = x/45

45tan25º = x

Shooter's height = 80 + 45tan25 ≈ 100.984 inches

That is over 8 feet tall! 

Okay I just figured out that it probably means this:

In which case the height is 80 - 45tan25 ≈ 59.016 inches

About 4.9 feet, which makes more sense.

1 × 41

That's it. 41 is prime. There are only 2 factors.

I tried goat cheese for the first time today. I'm proud of that because I normally am afraid to try new foods.

To find θ, just take the arcsin of both sides. Arcsin can also be written asin or sin^-1 .

sin θ = 10/11

arcsin(sin θ) = arcsin(10/11)

θ = arcsin(10/11) ≈ 65.380º