Finding for X given half-angle formulas

[bsmithysmith]

If \(\displaystyle csc(x)=4\), for \(\displaystyle 90º<x<180º\)
\(\displaystyle sin\left(\frac{x}{2}\right)=\)
\(\displaystyle cos\left(\frac{x}{2}\right)=\)
\(\displaystyle tan\left(\frac{x}{2}\right)=\)

I'm definitely stumped on this one. I know that this is the half-angle formulas. Luckily we all have sheets we can use for the exam. I know that:

\(\displaystyle csc(x)=4\) is the same as \(\displaystyle sin(x)=1/4\), am I correct?

From there, I don't know if I should do a sine inverse, or plug and chug for the half-angle formulas.

\(\displaystyle sin\left(\frac{x}{2}\right)=±√(\left(\frac{1}{2}\right)(1-Cos(2x)))\)

And I don't know if I have to plug for x on both sides, or if I have to find what cosine(2x) is, or If I have to plug in the double angle formula at the end there.

 

[MarkFL]

Yes, you are right:

\(\displaystyle \csc(x)=4\implies \sin(x)=\frac{1}{4}\)

Now, looking at the half-angle identities for sine, cosine, and tangent, we see we need to know $\cos(x)$ as well. So, we can use the Pythagorean identity:

\(\displaystyle \cos^2(x)=1-\sin^2(x)=1-\left(\frac{1}{4}\right)^2=\frac{15}{16}\)

Now, when we take the square root of both sides to get $\cos(x)$, which sign should we take on the right, given the quadrant in which $x$ is said to be?

 

[bsmithysmith]

Yes, you are right:

\(\displaystyle \csc(x)=4\implies \sin(x)=\frac{1}{4}\)

Now, looking at the half-angle identities for sine, cosine, and tangent, we see we need to know $\cos(x)$ as well. So, we can use the Pythagorean identity:

\(\displaystyle \cos^2(x)=1-\sin^2(x)=1-\left(\frac{1}{4}\right)^2=\frac{15}{16}\)

Now, when we take the square root of both sides to get $\cos(x)$, which sign should we take on the right, given the quadrant in which $x$ is said to be?

Since it's at quadrant 2, all cosine values will be negative, and sine values will be positive. I suspect it'll be the same thing for the sine value, but what about the tangent value? I have the cheat sheet available, but I'd still rather understand the process than plug and chug.

 

[MarkFL]

Yes, cosine is negative in the second quadrant, so since:

\(\displaystyle \cos^2(x)=\frac{15}{16}\)

we must therefore conclude that:

\(\displaystyle \cos(x)=-\frac{\sqrt{15}}{4}\)

So, now you have all you need to find the half-angled values of the primary trig. functions, using the various half-angle identities. If sine is positive, and cosine is negative, and given:

\(\displaystyle \tan(\theta)\equiv\frac{\sin(\theta)}{\cos(\theta)}\)

then what sign would you expect for the tangent function to have?

 

[CellCoree]

I'm sorry if I'm asking too many questions, but I'm really stuck here.

First of all, great job recognizing that csc(x)=4 is the same as sin(x)=1/4. This is a key step in solving this problem.

To use the half-angle formulas, we need to first find the value of cos(x). We can do this by using the Pythagorean identity, which states that sin^2(x) + cos^2(x) = 1. Since we know that sin(x) = 1/4, we can substitute this value in and solve for cos(x).

sin^2(x) + cos^2(x) = 1
(1/4)^2 + cos^2(x) = 1
1/16 + cos^2(x) = 1
cos^2(x) = 1 - 1/16
cos^2(x) = 15/16
cos(x) = ±√(15/16) = ±√15/4 = ±√15/2

Now, we can plug this value into the half-angle formulas to find the values of sin(x/2), cos(x/2), and tan(x/2).

sin\left(\frac{x}{2}\right)=±√\left(\frac{1-\cos(x)}{2}\right)
sin\left(\frac{x}{2}\right)=±√\left(\frac{1-\frac{\pm\sqrt{15}}{2}}{2}\right)
sin\left(\frac{x}{2}\right)=±√\left(\frac{2-\sqrt{15}}{4}\right)
sin\left(\frac{x}{2}\right)=±\frac{\sqrt{2-\sqrt{15}}}{2}

cos\left(\frac{x}{2}\right)=±√\left(\frac{1+\cos(x)}{2}\right)
cos\left(\frac{x}{2}\right)=±√\left(\frac{1+\frac{\pm\sqrt{15}}{2}}{2}\right)
cos\left(\frac{x}{2}\right)=±√\left(\frac{2+\sqrt{15}}{4}\right)
cos\left(\frac{x}{2}\right)=±\frac{\sqrt{2+\sqrt{15}}}{2}

tan\