MHB Finding for X given half-angle formulas

[bsmithysmith]

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

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:

$$csc(x)=4$$ is the same as $$sin(x)=1/4$$, 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.

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

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:

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

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:

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

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:

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

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:

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

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:

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

we must therefore conclude that:

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

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:

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

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