Class 14
| \(t\) | \(\sin t\) | \(\cos t\) | \(\tan t\) |
|---|---|---|---|
| \(0\) | \(0\) | \(1\) | \(0\) |
| \(\frac{\pi}{6}\) | \(\frac{1}{2}\) | \(\frac{\sqrt{3}}{2}\) | \(\frac{\sqrt{3}}{3}\) |
| \(\frac{\pi}{4}\) | \(\frac{\sqrt{2}}{2}\) | \(\frac{\sqrt{2}}{2}\) | \(1\) |
| \(\frac{\pi}{3}\) | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{2}\) | \(\sqrt{3}\) |
| \(\frac{\pi}{2}\) | \(1\) | \(0\) | undefined |
\[\begin{align*} \cos(s+t) &= \cos s \cos t - \sin s\sin t\\[1.2em] \sin(s+t) &= \cos s\sin t + \sin s\cos t\\[1.2em] \tan(s+t) &= \frac{\tan s + \tan t}{1 - \tan s \tan t} \end{align*}\]
\[\begin{align*} \cos(s-t) &= \cos s \cos t + \sin s\sin t\\[1.2em] \sin(s-t) &= -\cos s\sin t + \sin s\cos t\\[1.2em] \tan(s-t) &= \frac{\tan s - \tan t}{1 + \tan s \tan t} \end{align*}\]
\[\begin{align*} \cos(2t) &= \cos^2 t - \sin^2 t\\[1.2em] &= 2\cos^2 t - 1\\[1.2em] &= 1 - 2\sin^2 t\\[1.2em] \end{align*}\]
\[\begin{align*} \sin(2t) &= 2\cos t\sin t\\[1.2em] \tan(2t) &= \frac{2\tan t}{1 - \tan^2 t} \end{align*}\]
\[\cos(2t) = 2\cos^2 t - 1\]
\[2\cos^2 t = 1 + \cos(2t)\]
\[\cos^2 t = \frac{1 + \cos(2 t)}{2}\]
\[\cos(2t) = 1 - 2\sin^2 t\]
\[2\sin^2 t = 1 - \cos(2t)\]
\[\sin^2 t = \frac{1 - \cos(2 t)}{2}\]
\[\cos^2 t = \frac{1 + \cos(2 t)}{2}\]
\[\sqrt{\cos^2 t} = \sqrt{\frac{1 + \cos(2 t)}{2}}\]
\[\left\lvert\cos t\right\rvert = \sqrt{\frac{1 + \cos(2 t)}{2}}\]
\[\left\lvert\cos \frac{s}{2}\right\rvert = \sqrt{\frac{1 + \cos s}{2}}\]
\[\sin^2 t = \frac{1 - \cos(2 t)}{2}\]
\[\sqrt{\sin^2 t} = \sqrt{\frac{1 - \cos(2 t)}{2}}\]
\[\left\lvert\sin t\right\rvert = \sqrt{\frac{1 - \cos(2 t)}{2}}\]
\[\left\lvert\sin \frac{s}{2}\right\rvert = \sqrt{\frac{1 - \cos s}{2}}\]
\(\displaystyle\cos\frac{7\pi}{12}\)
\(\displaystyle\sin\frac{7\pi}{8}\)
\[\begin{align*} \left\lvert\tan\frac{s}{2}\right\rvert &\class{fragment}{{}= \frac{\left\lvert \sin \frac{s}{2}\right\rvert}{\left\lvert\cos\frac{s}{2}\right\rvert}} \class{fragment}{{}= \frac{\sqrt{\frac{1 - \cos s}{2}}}{\sqrt{\frac{1 + \cos s}{2}}}} \class{fragment}{{}= \sqrt{\frac{1 - \cos s}{1 + \cos s}}} \class{fragment}{{}= \sqrt{\frac{(1 - \cos s)}{(1 + \cos s)}\frac{(1 + \cos s)}{(1 + \cos s)}}}\\ &\class{fragment}{{}= \sqrt{\frac{1 - \cos^2 s}{(1 + \cos s)^2}}} \class{fragment}{{}= \sqrt{\frac{\sin^2 s}{(1 + \cos s)^2}}} \class{fragment}{{}= \left\lvert\frac{\sin s}{1 + \cos s}\right\rvert} \class{fragment}{{}= \frac{\left\lvert\sin s\right\rvert}{1 + \cos s}}\\[1.2em] &\class{fragment}{\sin s < 0 \text{ if } (2k - 1)\pi < s < 2k\pi.}\\ &\class{fragment}{\text{Then } (2k - 1)\frac{\pi}{2} < \frac{s}{2} < k\pi \text{ and } \tan\frac{s}{2} < 0.}\\[1.2em] &\class{fragment}{\sin s > 0 \text{ if } 2k\pi < s < (2k + 1)\pi.}\\ &\class{fragment}{\text{Then } k\pi < \frac{s}{2} < (2k+1)\frac{\pi}{2} \text{ and } \tan\frac{s}{2} > 0.} \end{align*}\]
\[\tan \frac{s}{2} = \frac{\sin s}{1 + \cos s} = \frac{1 - \cos s}{\sin s}\]
Calculate \(\tan \frac{7\pi}{8}\)
\[\begin{align*} \cos(s+t) &= \cos s \cos t - \sin s\sin t\\[1.2em] \cos(s-t) &= \cos s \cos t + \sin s\sin t\\[1.2em] \end{align*}\]
\[\begin{align*} \cos(s+t) &= \cos s \cos t - \sin s\sin t\\[1.2em] -\cos(s-t) &= -\cos s \cos t - \sin s\sin t\\[1.2em] \end{align*}\]
\[\cos s\cos t = \frac{1}{2}\left(\cos(s+t) + \cos(s-t)\right)\]
\[\sin s\sin t = -\frac{1}{2}\left(\cos(s+t) - \cos(s-t)\right)\]
\[\begin{align*} \sin(s+t) &= \cos s\sin t + \sin s\cos t\\[1.2em] \sin(s-t) &= -\cos s\sin t + \sin s\cos t\\[1.2em] \end{align*}\]
\[\begin{align*} \sin(s+t) &= \cos s\sin t + \sin s\cos t\\[1.2em] -\sin(s-t) &= \cos s\sin t - \sin s\cos t\\[1.2em] \end{align*}\]
\[\sin s\cos t = \frac{1}{2}\left(\sin(s+t) + \sin(s-t)\right)\]
\[\cos s\sin t = \frac{1}{2}\left(\sin(s+t) - \sin(s-t)\right)\]
Simplify \(\cos(3t)\sin(5t)\)
Simplify \(\cos(4t)\cos(2x)\)
Product to sum:
\[\begin{align*} 2 \cos s\cos t &= \cos(s+t) + \cos(s-t)\\ -2\sin s\sin t &= \cos(s+t) - \cos(s-t)\\ 2\sin s\cos t &= \sin(s+t) + \sin(s-t)\\ 2\cos s\sin t &= \sin(s+t) - \sin(s-t) \end{align*}\]
Suppose \(u = s+t\) and \(v = s - t\). Can you express \(s\) and \(t\) in terms of \(u\) and \(v\)?