Alguns valores precisos de senos e cossenos.

$\sin \frac{\pi}{2}=1$

$\cos \frac{\pi}{2}=0$

$\sin \frac{\pi}{4}={{1}\over{\sqrt{2}}}$

$\cos \frac{\pi}{4}={{1}\over{\sqrt{2}}}$

$\sin \frac{\pi}{8}={{\sqrt{\sqrt{2}-1}}\over{2^{{{3}\over{4}}}}}$

$\cos \frac{\pi}{8}={{\sqrt{\sqrt{2}+1}}\over{2^{{{3}\over{4}}}}}$

$\sin \frac{\pi}{16}={{\sqrt{2^{{{3}\over{4}}}-\sqrt{\sqrt{2}+1}}}\over{2^{{{7}\over{8}}}}}$

$\cos \frac{\pi}{16}={{\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}\over{2^{{{7}\over{8}}}}}$

$\sin \frac{\pi}{32}={{\sqrt{2^{{{7}\over{8}}}-\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}\over{2^{{{15}\over{16}}}}}$

$\cos \frac{\pi}{32}={{\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}\over{2^{{{15}\over{16}}}}}$

$\sin \frac{\pi}{64}={{\sqrt{2^{{{15}\over{16}}}-\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}\over{2^{{{31}\over{32}}}}}$

$\cos \frac{\pi}{64}={{\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}\over{2^{{{31}\over{32}}}}}$

$\sin \frac{\pi}{128}={{\sqrt{2^{{{31}\over{32}}}-\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}\over{2^{{{63}\over{64}}}}}$

$\cos \frac{\pi}{128}={{\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}\over{2^{{{63}\over{64}}}}}$

$\sin \frac{\pi}{256}={{\sqrt{2^{{{63}\over{64}}}-\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}\over{2^{{{127}\over{128}}}}}$

$\cos \frac{\pi}{256}={{\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}\over{2^{{{127}\over{128}}}}}$

$\sin \frac{\pi}{512}={{\sqrt{2^{{{127}\over{128}}}-\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}\over{2^{{{255}\over{256}}}}}$

$\cos \frac{\pi}{512}={{\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}\over{2^{{{255}\over{256}}}}}$

$\sin \frac{\pi}{1024}={{\sqrt{2^{{{255}\over{256}}}-\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}\over{2^{{{511}\over{512}}}}}$

$\cos \frac{\pi}{1024}={{\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}\over{2^{{{511}\over{512}}}}}$

$\sin \frac{\pi}{2048}={{\sqrt{2^{{{511}\over{512}}}-\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}\over{2^{{{1023}\over{1024}}}}}$

$\cos \frac{\pi}{2048}={{\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}\over{2^{{{1023}\over{1024}}}}}$

$\sin \frac{\pi}{4096}={{\sqrt{2^{{{1023}\over{1024}}}-\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}\over{2^{{{2047}\over{2048}}}}}$

$\cos \frac{\pi}{4096}={{\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}\over{2^{{{2047}\over{2048}}}}}$

$\sin \frac{\pi}{8192}={{\sqrt{2^{{{2047}\over{2048}}}-\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}}\over{2^{{{4095}\over{4096}}}}}$

$\cos \frac{\pi}{8192}={{\sqrt{2^{{{2047}\over{2048}}}+\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}}\over{2^{{{4095}\over{4096}}}}}$

$\sin \frac{\pi}{16384}={{\sqrt{2^{{{4095}\over{4096}}}-\sqrt{2^{{{2047}\over{2048}}}+\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}}}\over{2^{{{8191}\over{8192}}}}}$

$\cos \frac{\pi}{16384}={{\sqrt{2^{{{4095}\over{4096}}}+\sqrt{2^{{{2047}\over{2048}}}+\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}}}\over{2^{{{8191}\over{8192}}}}}$

$\sin \frac{\pi}{32768}={{\sqrt{2^{{{8191}\over{8192}}}-\sqrt{2^{{{4095}\over{4096}}}+\sqrt{2^{{{2047}\over{2048}}}+\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}}}}\over{2^{{{16383}\over{16384}}}}}$

$\cos \frac{\pi}{32768}={{\sqrt{2^{{{8191}\over{8192}}}+\sqrt{2^{{{4095}\over{4096}}}+\sqrt{2^{{{2047}\over{2048}}}+\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}}}}\over{2^{{{16383}\over{16384}}}}}$

${\small \sin \frac{\pi}{65536}={{\sqrt{2^{{{16383}\over{16384}}}-\sqrt{2^{{{8191}\over{8192}}}+\sqrt{2^{{{4095}\over{4096}}}+\sqrt{2^{{{2047}\over{2048}}}+\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}}}}}\over{2^{{{32767}\over{32768}}}}}}$

${\small \cos \frac{\pi}{65536}={{\sqrt{2^{{{16383}\over{16384}}}+\sqrt{2^{{{8191}\over{8192}}}+\sqrt{2^{{{4095}\over{4096}}}+\sqrt{2^{{{2047}\over{2048}}}+\sqrt{2^{{{1023}\over{1024}}}+\sqrt{2^{{{511}\over{512}}}+\sqrt{2^{{{255}\over{256}}}+\sqrt{2^{{{127}\over{128}}}+\sqrt{2^{{{63}\over{64}}}+\sqrt{2^{{{31}\over{32}}}+\sqrt{2^{{{15}\over{16}}}+\sqrt{2^{{{7}\over{8}}}+\sqrt{2^{{{3}\over{4}}}+\sqrt{\sqrt{2}+1}}}}}}}}}}}}}}}\over{2^{{{32767}\over{32768}}}}}}$

$\sin \frac{\pi}{3}={{\sqrt{3}}\over{2}}$

$\cos \frac{\pi}{3}={{1}\over{2}}$

$\sin \frac{\pi}{6}={{1}\over{2}}$

$\cos \frac{\pi}{6}={{\sqrt{3}}\over{2}}$

$\sin \frac{\pi}{12}={{\sqrt{2-\sqrt{3}}}\over{2}}$

$\cos \frac{\pi}{12}={{\sqrt{\sqrt{3}+2}}\over{2}}$

$\sin \frac{\pi}{24}={{\sqrt{2-\sqrt{\sqrt{3}+2}}}\over{2}}$

$\cos \frac{\pi}{24}={{\sqrt{\sqrt{\sqrt{3}+2}+2}}\over{2}}$

$\sin \frac{\pi}{48}={{\sqrt{2-\sqrt{\sqrt{\sqrt{3}+2}+2}}}\over{2}}$

$\cos \frac{\pi}{48}={{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}}\over{2}}$

$\sin \frac{\pi}{96}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}}}\over{2}}$

$\cos \frac{\pi}{96}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}}\over{2}}$

$\sin \frac{\pi}{192}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}}}\over{2}}$

$\cos \frac{\pi}{192}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}}\over{2}}$

$\sin \frac{\pi}{384}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}}}\over{2}}$

$\cos \frac{\pi}{384}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}}\over{2}}$

$\sin \frac{\pi}{768}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}}}\over{2}}$

$\cos \frac{\pi}{768}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}}\over{2}}$

$\sin \frac{\pi}{1536}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}}}\over{2}}$

$\cos \frac{\pi}{1536}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}}\over{2}}$

$\sin \frac{\pi}{3072}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}}}\over{2}}$

$\cos \frac{\pi}{3072}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}}\over{2}}$

$\sin \frac{\pi}{6144}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}}}\over{2}}$

$\cos \frac{\pi}{6144}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}\over{2}}$

$\sin \frac{\pi}{12288}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}}\over{2}}$

$\cos \frac{\pi}{12288}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}\over{2}}$

$\sin \frac{\pi}{24576}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}}\over{2}}$

$\cos \frac{\pi}{24576}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}\over{2}}$

$\sin \frac{\pi}{49152}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}}\over{2}}$

$\cos \frac{\pi}{49152}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}\over{2}}$

$\sin \frac{\pi}{98304}={{\sqrt{2-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}}\over{2}}$

$\cos \frac{\pi}{98304}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{3}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}+2}}\over{2}}$

$\cos \frac{\pi}{5}={{\sqrt{5}+1}\over{4}}$

$\sin \frac{\pi}{10}={{\sqrt{3-\sqrt{5}}}\over{2^{{{3}\over{2}}}}}$

$\cos \frac{\pi}{10}={{\sqrt{\sqrt{5}+5}}\over{2^{{{3}\over{2}}}}}$

$\sin \frac{\pi}{20}={{\sqrt{2^{{{3}\over{2}}}-\sqrt{\sqrt{5}+5}}}\over{2^{{{5}\over{4}}}}}$

$\cos \frac{\pi}{20}={{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}}\over{2^{{{5}\over{4}}}}}$

$\sin \frac{\pi}{40}={{\sqrt{2^{{{5}\over{4}}}-\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}}}\over{2^{{{9}\over{8}}}}}$

$\cos \frac{\pi}{40}={{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}}\over{2^{{{9}\over{8}}}}}$

$\sin \frac{\pi}{80}={{\sqrt{2^{{{9}\over{8}}}-\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}}}\over{2^{{{17}\over{16}}}}}$

$\cos \frac{\pi}{80}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}}\over{2^{{{17}\over{16}}}}}$

$\sin \frac{\pi}{160}={{\sqrt{2^{{{17}\over{16}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}}}\over{2^{{{33}\over{32}}}}}$

$\cos \frac{\pi}{160}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}}\over{2^{{{33}\over{32}}}}}$

$\sin \frac{\pi}{320}={{\sqrt{2^{{{33}\over{32}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}}}\over{2^{{{65}\over{64}}}}}$

$\cos \frac{\pi}{320}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}}\over{2^{{{65}\over{64}}}}}$

$\sin \frac{\pi}{640}={{\sqrt{2^{{{65}\over{64}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}}}\over{2^{{{129}\over{128}}}}}$

$\cos \frac{\pi}{640}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}}\over{2^{{{129}\over{128}}}}}$

$\sin \frac{\pi}{1280}={{\sqrt{2^{{{129}\over{128}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}}}\over{2^{{{257}\over{256}}}}}$

$\cos \frac{\pi}{1280}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}}\over{2^{{{257}\over{256}}}}}$

$\sin \frac{\pi}{2560}={{\sqrt{2^{{{257}\over{256}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}}}\over{2^{{{513}\over{512}}}}}$

$\cos \frac{\pi}{2560}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}}\over{2^{{{513}\over{512}}}}}$

$\sin \frac{\pi}{5120}={{\sqrt{2^{{{513}\over{512}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}}}\over{2^{{{1025}\over{1024}}}}}$

$\cos \frac{\pi}{5120}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}}\over{2^{{{1025}\over{1024}}}}}$

$\sin \frac{\pi}{10240}={{\sqrt{2^{{{1025}\over{1024}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}}}\over{2^{{{2049}\over{2048}}}}}$

$\cos \frac{\pi}{10240}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}}\over{2^{{{2049}\over{2048}}}}}$

$\sin \frac{\pi}{20480}={{\sqrt{2^{{{2049}\over{2048}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}}}\over{2^{{{4097}\over{4096}}}}}$

$\cos \frac{\pi}{20480}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}+2^{{{2049}\over{2048}}}}}\over{2^{{{4097}\over{4096}}}}}$

$\sin \frac{\pi}{40960}={{\sqrt{2^{{{4097}\over{4096}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}+2^{{{2049}\over{2048}}}}}}\over{2^{{{8193}\over{8192}}}}}$

$\cos \frac{\pi}{40960}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}+2^{{{2049}\over{2048}}}}+2^{{{4097}\over{4096}}}}}\over{2^{{{8193}\over{8192}}}}}$

${\small \sin \frac{\pi}{81920}={{\sqrt{2^{{{8193}\over{8192}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}+2^{{{2049}\over{2048}}}}+2^{{{4097}\over{4096}}}}}}\over{2^{{{16385}\over{16384}}}}}}$

${\small \cos \frac{\pi}{81920}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}+2^{{{2049}\over{2048}}}}+2^{{{4097}\over{4096}}}}+2^{{{8193}\over{8192}}}}}\over{2^{{{16385}\over{16384}}}}}}$

${\small \sin \frac{\pi}{163840}={{\sqrt{2^{{{16385}\over{16384}}}-\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}+2^{{{2049}\over{2048}}}}+2^{{{4097}\over{4096}}}}+2^{{{8193}\over{8192}}}}}}\over{2^{{{32769}\over{32768}}}}}}$

${\small \cos \frac{\pi}{163840}={{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{5}+5}+2^{{{3}\over{2}}}}+2^{{{5}\over{4}}}}+2^{{{9}\over{8}}}}+2^{{{17}\over{16}}}}+2^{{{33}\over{32}}}}+2^{{{65}\over{64}}}}+2^{{{129}\over{128}}}}+2^{{{257}\over{256}}}}+2^{{{513}\over{512}}}}+2^{{{1025}\over{1024}}}}+2^{{{2049}\over{2048}}}}+2^{{{4097}\over{4096}}}}+2^{{{8193}\over{8192}}}}+2^{{{16385}\over{16384}}}}}\over{2^{{{32769}\over{32768}}}}}}$

Integral da cotangente.

Seja $u = \sin x$, $du\ =\ \cos x\ dx$.

$\displaystyle\int \cot x\ dx = \displaystyle\int \dfrac{\cos x}{\sin x}\ dx\ = \displaystyle\int \dfrac{du}{u} = \fbox{$\log |\sin x| + c$}$

Integral da tangente.

Seja $u = \cos x$, $du\ =\ -\sin x\ dx$.

$\displaystyle\int \tan x\ dx = \displaystyle\int \dfrac{\sin x}{\cos x}\ dx\ =\ -\displaystyle\int \dfrac{du}{u}\ = \fbox{$-\log |\cos x| + c$}$

Trigonometria: transformação de soma em produto.

Sabemos que:

$\sin (a + b) = (\sin a)(\cos b) + (\sin b)(\cos a)$ (I)

$\sin (a - b) = (\sin a)(\cos b) - (\sin b)(\cos a)$ (II)

$\cos (a + b) = (\cos a)(\cos b) - (\sin a)(\sin b)$ (III)

$\cos (a - b) = (\cos a)(\cos b) + (\sin a)(\sin b)$ (IV)

Somando (I) e (II): $2(\sin a)(\cos b) = \sin (a + b) + \sin (a - b)$.

Subtraindo (II) de (I): $2(\sin b)(\cos a) = \sin (a + b) - \sin (a - b)$.

Somando (III) e (IV): $2(\cos a)(\cos b) = \cos (a + b) + \cos (a - b)$.

Subtraindo (IV) de (III): $-2(\sin a)(\sin b) = \cos (a + b) - \cos (a - b)$.

Fazendo $p = a + b$ e $q = a - b$, teremos que $a = \dfrac{p + q}{2}$ e $b = \dfrac{p - q}{2}$. Substituindo:

$\sin p + \sin q = 2\left(\sin \dfrac{p + q}{2}\right)\left(\sin \dfrac{p - q}{2}\right)$

$\sin p - \sin q = 2\left(\cos \dfrac{p + q}{2}\right)\left(\sin \dfrac{p - q}{2}\right)$

$\cos p + \cos q = 2\left(\cos \dfrac{p + q}{2}\right)\left(\cos \dfrac{p - q}{2}\right)$

$\cos p - \cos q = -2\left(\sin \dfrac{p + q}{2}\right)\left(\sin \dfrac{p - q}{2}\right)$

Resolver a inequação $\tan \alpha > \sqrt{3}$.

$\arctan \sqrt{3} = \dfrac{\pi}{3}$

$S = \displaystyle\bigcup_{k \in \mathbb{Z}}\left]\dfrac{\pi}{3} + k\pi, \dfrac{\pi}{2} + k\pi\right[$

Em $\mathbb{U} = \left]\dfrac{\pi}{2}, \pi\right[$, resolver $\sin x + \sin 2x + \sin 3x = 0$.

$\sin x + \sin 2x + \sin 3x = 2(\sin 2x)(\cos x) + \sin 2x = (\sin 2x)(2\cos x + 1)$

$\sin 2x = 0\ \vee\ 2\cos x + 1 = 0\ \Rightarrow\ x = \dfrac{2\pi}{3}$

$S = \left\{\dfrac{2\pi}{3}\right\}$

Se $\alpha$ e $\beta$ são os ângulos opostos aos catetos de um triângulo retângulo, quanto é $\delta = (\cos \alpha - \cos \beta)^2 + (\sin \alpha + \sin \beta)^2$?

$\alpha + \beta = \dfrac{\pi}{2}$

$\delta = \cos^2 \alpha + \cos^2 \beta - 2(\cos \alpha)(\cos \beta) + \sin^2 \alpha + \sin^2 \beta + 2(\sin \alpha)(\sin \beta)$

$\delta = 2 - 2\cos (\alpha + \beta) = \fbox{$2$}$

Sabendo-se que $x = 4r \cdot \cos a \cdot \sin b$, $y = 6r \cdot \sin a \cdot \sin b$ e $z = 8r \cdot \cos b$, calcular $\alpha = \dfrac{x^2}{4} + \dfrac{y^2}{9} + \dfrac{z^2}{16}$.

$\alpha = 4r^2[(\cos^2 a)(\sin^2 b) + (\sin^2 a)(\sin^2 b) + cos^2 b] = 4r^2(\sin^2 b + \cos^2 b) = \fbox{$4r^2$}$

Se $\cos 2a = \dfrac{1}{2}$, quanto é $\tan^2 a + \sec^2 a$?

$\dfrac{1}{2} = 2\cos^2 a - 1\ \Rightarrow\ \cos^2 a = \dfrac{3}{4}$

$\tan^2 a + \sec^2 a = 2\sec^2 a - 1 = \fbox{$\dfrac{5}{3}$}$

Se um arco $\theta$, $0 < \theta < \dfrac{\pi}{2}$ é tal que o dobro do seu seno é igual ao triplo do quadrado de sua tangente, qual o valor de $\cos \theta$?

$2\sin \theta = 3\tan^2 \theta$

$\dfrac{2}{3} = \dfrac{\sin \theta}{\cos^2 \theta} = \dfrac{\sin \theta}{1 - \sin^2 \theta}$

$2 - 2\sin^2 \theta = 3\sin \theta$

$2\sin^2 \theta + 3\sin \theta - 2 = 0$

$\sin \theta = \dfrac{1}{2}\ \Rightarrow\ \fbox{$\cos \theta = \dfrac{\sqrt{3}}{2}$}$

Cosseno da soma de arcos.

$\cos (a + b) = (\cos a)(\cos b) - (\sin a)(\sin b)$

No primeiro quadrante, tomemos $a = m(D\hat{A}B)$ e $b = m(B\hat{A}C)$.

$m(\overline{AH}) = \dfrac{\cos (a + b)}{\cos a}$

$m(\overline{HG}) = (\cos b) - m(\overline{AH}) = (\cos b) - \dfrac{\cos (a + b)}{\cos a}$

$(\sin a)(\sin b) = (\cos a)(\cos b) - \cos (a + b)$

Se, como um caso particular, $a$ está no segundo quadrante, podemos fazer a redução ao primeiro quadrante:

$\cos (a + b) = \cos (\pi - a´ + b) = [(\cos (\pi - a´)](\cos b) - [(\sin (\pi - a´)](\sin b) =$

$= - (\cos a´)(\cos b) - (\sin a´)(\sin b) = (\cos a)(\cos b) - (\sin a)(\sin b)$.

Analogamente, para $a$ ou $b$ em quaisquer dos quadrantes, verificando também quando $a$ ou $b$ pertencem aos eixos, teremos que a fórmula é válida para todos os valores.

Quod Erat Demonstrandum.

Ângulo de Antonio Vandré.

Sejam $P(a, b)$, $Q(c, d)$, o eixo $\overrightarrow{PQ}$, e uma função $f: I \rightarrow \mathbb{R}$. O ângulo $\theta$ de um ponto de $f$ com o eixo $\overrightarrow{PQ}$ é tal que $\cos \theta = \dfrac{(c - a)(x - a) + (d - b)[f(x) - b]}{\sqrt{[(c - a)^2 + (d - b)^2]\{(x - a)^2 + [f(x) - b]^2\}}}$.

Chamando tal ângulo de Ângulo de Antonio Vandré,

$\mathcal{\alpha_A}_{f(x)}^{[(a, b), (c, d)]} = \arccos \dfrac{(c - a)(x - a) + (d - b)[f(x) - b]}{\sqrt{[(c - a)^2 + (d - b)^2]\{(x - a)^2 + [f(x) - b]^2\}}}$.

Exemplo: $f(x) = 0$, $P(0, 1)$, $Q(0, 2)$:

$\mathcal{\alpha_A}_{0}^{[(0, 1), (0, 2)]} = \arccos \dfrac{-1}{\sqrt{x^2 + 1}}$.

Exercício: satélites estacionários; função corda.

Sejam dois satélites estacionários, um à longitude $80^o$ e outro à longitude $30^o$. Sabendo que satélites estacionários estão a aproximadamente $42000\ km$ do centro da Terra, qual a distância entre eles?

Resolução:

$cord\ \dfrac{5\pi}{18} = \sqrt{2(1 - \cos \dfrac{5\pi}{18})} \approx 0,85$

Logo distanciam-se de, aproximadamente, $42000 \cdot 0,85 \approx \fbox{$36000\ km$}$.

Derivada da inversa da função corda.

Sabendo que $arccord\ x = \arccos \left(1 - \dfrac{x^2}{2}\right)$, utilizando a regra da cadeia:

$(arccord\ x)' = \dfrac{x}{\sqrt{1 - \left(1 - \dfrac{x^2}{2}\right)^2}} = \fbox{$\dfrac{2}{\sqrt{4 - x^2}}$}$.

A inversa, a derivada, e a integral da função corda.

${\large cord\ \alpha = \sqrt{2(1 - \cos \alpha)}}$

Inversa: seja $arccord: \underset{x\ \mapsto\ arccord\ x}{[0, 2] \rightarrow [0, \pi]},\ \fbox{$arccord\ x = \arccos \left(1 - \dfrac{x^2}{2}\right)$}$.

Derivada: $\fbox{$(cord\ \alpha)' = \dfrac{\sin \alpha}{\sqrt{2 - 2\cos \alpha}}$}$.

Observemos que, para $0 \le \alpha \le 2\pi$, $cord\ \alpha = 2\sin \dfrac{\alpha}{2}$.

Logo,

$\fbox{$\displaystyle\int cord\ \alpha\ d\alpha\ =\ -4\cos \dfrac{\beta}{2} + c,\ \alpha = 2k\pi + \beta,\ k \in \mathbb{Z}, 0 \le \beta < 2\pi$}$.

Seja $V$ o espaço vetorial de dimensão infinita gerado por $\{\sin \alpha x\ :\ \alpha \in \mathbb{Z}\}$ e $\langle f, g \rangle = \displaystyle\int_{-\pi}^{\pi} f(x)g(x)\ dx$, mostre que $\sin mx$ e $\sin nx$, com $m, n \in \mathbb{Z},\ m \neq n$ são linearmente independentes.

Seja $V$ o espaço vetorial de dimensão infinita gerado por $\{\sin \alpha x\ :\ \alpha \in \mathbb{Z}\}$ e $\langle f, g \rangle = \displaystyle\int_{-\pi}^{\pi} f(x)g(x)\ dx$, mostre que $\sin mx$ e $\sin nx$, com $m, n \in \mathbb{Z},\ m \neq n$ são linearmente independentes.

Resolução:

Basta mostrar que $\sin mx$ e $\sin nx$, com $m, n \in \mathbb{Z},\ m \neq n$ são perpendiculares.

De fato, $\displaystyle\int_{-\pi}^{\pi} (\sin mx)(\sin nx)\ dx\ =\ \dfrac{1}{2}\displaystyle\int_{-\pi}^{\pi} \cos (m - n)x - \cos (m + n)x\ dx\ =\ 0$.

Quod Erat Demonstrandum.

Determinar o volume de $4$ nódulos resultantes da rotação da função $y = \sin x$ em torno do eixo $x$.

Determinar o volume de $4$ nódulos resultantes da rotação da função $y = \sin x$ em torno do eixo $x$.

Resolução:

$V = 4\pi \displaystyle\int_0^{\pi} \sin^2 x\ dx = 4 \displaystyle\int_0^{\pi} \dfrac{1 - \cos 2x}{2}\ dx = \left. [2x - \sin(2x)] \right|_0^{\pi} = \fbox{$2\pi$}$

Integral do arco-tangente.

$\displaystyle\int \arctan x\ dx\ \overset{\text{Por partes}}=\ x \cdot \arctan x - \underset{I}{\underbrace{\displaystyle\int \dfrac{x}{x^2 + 1} dx}}$

Seja $u = x^2 + 1$, $du = 2x\ dx$.

$I = \dfrac{1}{2} \displaystyle\int \dfrac{du}{u} = \dfrac{\log |u|}{2} + c = \dfrac{\log |x^2 + 1|}{2} + c = \dfrac{\log (x^2 + 1)}{2} + c$

Logo, $\fbox{$\displaystyle\int \arctan x\ dx\ =\ x \cdot \arctan x - \dfrac{\log (x^2 + 1)}{2} + C$}$.

Calcular $I = \displaystyle\int_0^{\dfrac{\pi}{6}} \tan^2 2x\ dx$.

$I = \displaystyle\int_0^{\dfrac{\pi}{6}} [(\sec^2 2x) - 1]\ dx = \left. (\dfrac{\tan 2x}{2} - x)\right|_0^{\dfrac{\pi}{6}} = \fbox{$\dfrac{\sqrt{3}}{2} - \dfrac{\pi}{6}$}$

Calcular $I\ =\ \displaystyle\int \cos^3 x\ dx$.

$I\ =\ \displaystyle\int (\cos x)(1 - \sin^2 x)\ dx$.

Seja $u = \sin x$, $du = \cos x\ dx$.

$I\ =\ \displaystyle\int (1 - u^2)\ du\ =\ u - \dfrac{u^3}{3} + c$.

Logo $\fbox{$\displaystyle\int \cos^3 x\ dx\ =\ \sin x - \dfrac{\sin^3 x}{3} + c$}$.