Find the exact value of tan 13π/8 using half-angle identities. sqrt(2)-sqrt(3)/2 -sqrt(2)-1 1-sqrt(2) sqrt(2)-sqrt(2)/2 possible answers
i guess \(\frac{13\pi}{8}\) is half of \(\frac{13\pi}{4}\)
sqrt(2)-sqrt(3)/2 -sqrt(2)-1 1-sqrt(2) sqrt(2)-sqrt(2)/2 possible answers
oh ok so that half 13x/4?
perhaps easiest to use \[\tan(\frac{13\pi}{8})=\frac{\sin(\frac{13\pi}{4})}{1+\cos(\frac{13\pi}{4})}\]
yes \(\frac{13\pi}{8}\) is half of \(\frac{13\pi}{4}\)
so what the next step to getting the possible answer i listed
sqrt(2)-sqrt(3)/2 -sqrt(2)-1 1-sqrt(2) sqrt(2)-sqrt(2)/2
the last step is to compute \(\frac{\sin(\frac{13\pi}{4})}{1+\cos(\frac{13\pi}{4})}\) and see what you get
i got 1.16 rounded but it doesnt match the other answes so is there something im missing
since \(\frac{13\pi}{4}\) is coterminal with \(\frac{3\pi}{4}\) you can find \(\sin(\frac{3\pi}{4})\) and \(\cos(\frac{3\pi}{4})\) instead don't use a calculator, write down the exact answer
oh ok soo use sin=135 and cos 135 too so -sqrt(2)/2 sqrt(2)
oh ok soo use sin=135 and cos 135 too so -sqrt(2)/2 sqrt(2)/2
13pi/4 - 2pi = 5pi/4 i think u can use 5pi/4 or 225 degrees not 135 d...
oh so use 225 so what do i do next?
oh damn you are right sorry
it is coterminal with \(\frac{5\pi}{4}\)
forget about degrees, you are an adult and so you should only use them if you are "solving" a triangle you need to know that \(\cos(\frac{5\pi}{4})=-\frac{\sqrt{2}}{2}\) and also \(\sin(\frac{5\pi}{4})=-\frac{\sqrt{2}}{2}\)
so they answer but it doesnt match any of my answers ?? so idk
\[-\frac{\frac{\sqrt{2}}{2}}{1-\frac{\sqrt{2}}{2}}\]
simplify by multiplying top and bottom by 2
\[\frac{-\sqrt{2}}{2-\sqrt{2}}\]
the answer 2? if u simplfy
multiply by conyugate of denominator...
huh so multpy by sqrt 2
no, but multiply (2+sqrt(2))/(2+sqrt(2)
1 right
not yet
???
can u simplify it : |dw:1353211817999:dw|
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