IME / ITA ⇒ trigonometria ITA/IME Tópico resolvido

[gabrielmacc]

06) Prove a identidade


1 − 𝑠𝑒𝑛(8𝛼) = 2 cos2(45° + 4𝛼)

[petras]

gabrielmacc, O correto é:
[tex3]1 − 𝑠𝑒𝑛(8𝛼) = 2 cos^2(45° + 4𝛼)\\
cos(2\beta) = 2cos^2\beta - 1 \implies
2cos^2 β = 1 + cos 2β\\
\therefore 2 cos^2(45^o + 4α) = 1 + cos2(45^o + 4α) = 1 + cos(90^o + 8α)\\
cos(90^o + β) = − senβ\\
\therefore 2 cos2(45^o + 4α) = 1 + cos(90^o + 8α) = \boxed{1 − sen8α} c.q.d[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]1 − 𝑠𝑒𝑛(8𝛼) = 2 cos^2(45° + 4𝛼)\\
cos(2\beta) = 2cos^2\beta - 1 \implies
2cos^2 β = 1 + cos 2β\\
\therefore 2 cos^2(45^o + 4α) = 1 + cos2(45^o + 4α) = 1 + cos(90^o + 8α)\\
cos(90^o + β) = − senβ\\
\therefore 2 cos2(45^o + 4α) = 1 + cos(90^o + 8α) = \boxed{1 − sen8α} c.q.d[/tex3]