Question

FInd the exact value of csc (21 degrees)

Answer 1

csc(21)

Answer 2

how do i input this into the calculator?

Answer 3

A very important trigonometric operation, being this, the inverse of sine is defined as the "cosecant" of any angle, writing in a mathematcal standard:
\[\frac{ 1 }{ \sin v }=\csc v\]

Answer 4

Is the problem really asking for "exact value"?
Because calculator probably won't help with that.

Answer 5

You have \(\csc \theta=\dfrac{1}{\sin \theta}\), so let's work with \(\sin\) for the moment. Fair warning: I prefer using radians over degrees - too many \({}^\circ\) `\circ` symbols to keep track of with the latter. Hopefully you know that \(21^\circ=\dfrac{\pi}{60}\text{ rad}\).

Notice that\[\frac{\pi}{60}=\frac{\pi}{10}-\frac{\pi}{12}\]which means
\[\sin\frac{\pi}{60}=\sin\left(\frac{\pi}{10}-\frac{\pi}{12}\right)=\sin\frac{\pi}{10}\cos\frac{\pi}{12}-\sin\frac{\pi}{12}\cos\frac{\pi}{10}\]Easy enough, but not sufficient if you don't know the trig ratio values of \(\dfrac{\pi}{10}\) and \(\dfrac{\pi}{12}\). Let's derive them.

\(\dfrac{\pi}{12}\) is much simpler to work with. Recalling the half-angle identity, you have that
\[\sin^2\frac{\pi}{12}=\frac{1}{2}\left(1-\cos\frac{\pi}{6}\right)\implies\sin\frac{\pi}{12}=\frac{\displaystyle \sqrt{2-\sqrt{3}}}{2}\\
\cos^2\frac{\pi}{12}=\frac{1}{2}\left(1+\cos\frac{\pi}{6}\right)\implies\cos\frac{\pi}{12}=\frac{\displaystyle \sqrt{2+\sqrt{3}}}{2}\]Not too challenging, I hope?

The next part is a bit more involved. Again making use of the half-angle identity, you get
\[\begin{cases}x=\sin\dfrac{\pi}{10}\\[1ex]
y=\cos\dfrac{\pi}{10}\end{cases}
\implies
\cos\frac{\pi}{5}=\begin{cases}1-2x^2\\[1ex]
2y^2-1\end{cases}\]Recall Euler's formula,
\[e^{it}=\cos t+i\sin t\]and DeMoivre's theorem, which states that for non-negative integers \(n\),
\[(\cos t+i\sin t)^n=\left(e^{it}\right)^n=e^{int}=\cos nt+i\sin nt\]Let \(n=5\) (why will be apparent in just a moment), then
\[\cos5t+i\sin 5t=(\cos t+i\sin t)^5\]Expanding and equating the real parts, you have
\[\cos5t=\cos^5t-10\cos^3t\sin^2t+5\cos t\sin^4t\]Via the Pythagorean identity, we can almost make this equation a polynomial in \(\cos\):
\[\cos5t=\cos^5t-10\cos^3t\left(1-\cos^2t\right)+5\left(1-\cos^2t\right)^2\cos t\]Notice that if we set \(t=\dfrac{\pi}{5}\), then we end up with an equation quintic in \(\cos\dfrac{\pi}{5}\). Doing so, expanding, and setting \(z=\cos\dfrac{\pi}{5}\), yields
\[-1=16z^5-20z^3+5z\]This equation is actually quite easy to solve. Observe that it has a root at \(z=-1\), so \(z+1\) is a factor.
\[\begin{align*}
0&=16z^5-20z^3+5z+1\\[1ex]
&=(z+1)(4z^2-2z-1)^2\\[1ex]
z&=-1,\,\frac{1\pm\sqrt5}{4}
\end{align*}\]Since \(\dfrac{\pi}{5}\) is an angle in the first quadrant (ie. \(0<\dfrac{\pi}{5}<\dfrac{\pi}{2}\)), it's obvious that \(z=\cos\dfrac{\pi}{5}\) is positive, which means \(z=\dfrac{1+\sqrt5}{4}\).

So, you find that
\[\cos\frac{\pi}{5}=\frac{1+\sqrt5}{4}=\begin{cases}1-2x^2\\[1ex]2y^2-1\end{cases}
\implies
\begin{cases}
x=\sin\dfrac{\pi}{10}=\displaystyle\sqrt{\frac{1-\dfrac{1+\sqrt5}{4}}{2}}\\[1ex]
y=\cos\dfrac{\pi}{10}=\displaystyle\sqrt{\frac{\dfrac{1+\sqrt5}{4}+1}{2}}
\end{cases}\]
Finally (!), you arrive at an exact form:
\[\begin{align*}\sin\frac{\pi}{60}&=\left(\displaystyle\sqrt{\frac{1-\dfrac{1+\sqrt5}{4}}{2}}\right)\left(\frac{\displaystyle \sqrt{2+\sqrt{3}}}{2}\right)-\left(\frac{\displaystyle \sqrt{2-\sqrt{3}}}{2}\right)\left(\displaystyle\sqrt{\frac{\dfrac{1+\sqrt5}{4}+1}{2}}\right)\\[2ex]
&=\frac{\sqrt{2(2+\sqrt3)(3-\sqrt5)}-\sqrt{2(2-\sqrt3)(5+\sqrt5)}}{8}
\end{align*}\]which in turn yields
\[\csc21^\circ=\csc\frac{\pi}{60}=\frac{8}{\sqrt{2(2+\sqrt3)(3-\sqrt5)}-\sqrt{2(2-\sqrt3)(5+\sqrt5)}}\]