Determine o valor de log (cos0) + 
- cos 
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Olá, Sara.
![\log \cos 0+2^{3\cos0,5 \pi }-\cos \frac{2 \pi }{3} =\log1+2^{3\cos\frac \pi 2}-(\cos^2\frac\pi 3-\sin^2\frac\pi 3)=\\\\=0+2^{3\cdot0}-[(\frac12)^2-(\frac{\sqrt3}2)^2]=1-[\frac14-\frac34]=1-(-\frac24)=1+\frac12=\frac32](https://tex.z-dn.net/?f=%5Clog+%5Ccos+0%2B2%5E%7B3%5Ccos0%2C5+%5Cpi+%7D-%5Ccos+%5Cfrac%7B2+%5Cpi+%7D%7B3%7D+%3D%5Clog1%2B2%5E%7B3%5Ccos%5Cfrac+%5Cpi+2%7D-%28%5Ccos%5E2%5Cfrac%5Cpi+3-%5Csin%5E2%5Cfrac%5Cpi+3%29%3D%5C%5C%5C%5C%3D0%2B2%5E%7B3%5Ccdot0%7D-%5B%28%5Cfrac12%29%5E2-%28%5Cfrac%7B%5Csqrt3%7D2%29%5E2%5D%3D1-%5B%5Cfrac14-%5Cfrac34%5D%3D1-%28-%5Cfrac24%29%3D1%2B%5Cfrac12%3D%5Cfrac32 "\log \cos 0+2^{3\cos0,5 \pi }-\cos \frac{2 \pi }{3} =\log1+2^{3\cos\frac \pi 2}-(\cos^2\frac\pi 3-\sin^2\frac\pi 3)=\\\\=0+2^{3\cdot0}-[(\frac12)^2-(\frac{\sqrt3}2)^2]=1-[\frac14-\frac34]=1-(-\frac24)=1+\frac12=\frac32")
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