Sagot:
Ang pagkakakilanlan ay dapat totoo para sa anumang numero
Paliwanag:
Verify secx • cscx + cotx = tanx + 2cosx • cscx?
RHS = tanx + 2cosx * cscx = sinx / cosx + (2cosx) / sinx = (sin ^ 2x + 2cos ^ 2x) / (sinx * cosx) = (sin ^ 2x + cos ^ 2x + cos ^ 2x) cosx) = (1 + cos ^ 2x) / (sinx * cosx) = 1 / (sinx * cosx) + (cos ^ 2x) / (sinx * cosx) = cscx * secx + cotx = LHS
Paano ako magpapatunay na ito ay isang pagkakakilanlan? Salamat. (1-sin ^ 2 (x / 2)) / (1 + sin ^ 2 (x / 2)) = (1 + cosx) / (3-cosx)
LHS = (1-sin ^ 2 (x / 2)) / (1 + sin ^ 2 (x / 2) = (cos ^ 2 (x / 2)) / (1 + (2cos ^ 2 (x / 2)) / (2-2cos ^ 2 (x / 2)) = (1 + cosx) / (4 (1 + cosx)) = (1 + cosx) / ( 3-cosx) = RHS
Paano mo i-verify ang pagkakakilanlan segundo ^ 2 (x / 2) = (2secx + 2) / (secx + 2 + cosx)?
Kinakailangang patunayan: sec ^ 2 (x / 2) = (2secx + 2) / (secx + 2 + cosx) "Right Hand Side" = (2secx + 2) / (secx + 2 + cosx) / cosx => (2 * 1 / cosx + 2) / (1 / cosx + 2 + cosx) Ngayon, i-multiply ang tuktok at ibaba ng cosx => (cosx xx (2 * 1 / cosx + 2) (1 / cosx + 2 + cosx)) => (2 + 2cosx) / (1 + 2cosx + cos ^ 2x) Factorize sa ibaba, => (2 (1 + cosx) > 2 / (1 + cosx) Alalahanin ang pagkakakilanlan: cos2x = 2cos ^ 2x-1 => 1 + cos2x = 2cos ^ 2x Katulad: 1 + cosx = 2cos ^ 2 (x / 2) => "Right Hand Side" 2 / (2cos ^ 2 (x / 2)) = 1 / cos ^ 2 (x / 2) = kulay (asul) (sec ^ 2 (x / 2))