Vis bevise nedenstående identitet? 1 / cos290 + 1 / (sqrt3sin250) = 4 / sqrt3

# LHS = 1 / (cos290 ^ @) + 1 / (sqrt3sin250 ^ @) #

# = 1 / (cos (360-70) ^ @) + 1 / (sqrt3sin (180 + 70) ^ @) #

# = 1 / (cos70 ^ @) - 1 / (sqrt3sin70 ^ @) #

# = (Sqrt3sin70 ^ @ - cos70 ^ @) / (sqrt3sin70 ^ @ cos70 ^ @) #

# = 1 / sqrt3 (2 {sqrt3sin70 ^ @ - cos70 ^ @}) / (2sin70 ^ @ cos70 ^ @) #

# = 1 / sqrt3 (2 * 2 {sin70 ^ @ * (sqrt3 / 2) -cos70 ^ @ * (1/2)}) / (sin140 ^ @) #

# = 1 / sqrt3 (4 {sin70 ^ @ * cos30 ^ @ - cos70 ^ @ * sin30 ^ @}) / (sin (180-40) ^ @) #

# = 1 / sqrt3 (4 {sin (70-30) ^ @}) / (sin40 ^ @) = 1 / sqrt3 (4 {annullere (sin40 ^ @)}) / annullere ((sin40 ^ @)) = 4 / sqrt3 = RHS #

Noter det #cos (360-A) ^ @ = cosA og sin (180 + A) ^ @ = - sinA #

# 1 / cos290 + 1 / (sqrt3sin250) #

# = 1 / cos (270 + 20) + 1 / (sqrt3sin (270-20)) #

# = 1 / sin20 - 1 / (sqrt3cos20) #

# = (sqrt3cos20-sin20) / (sqrt3sin20cos20) #

# = 2 / sqrt3 (sqrt3 / 2cos20-1 / 2sin20) / (sin20cos20) #

# = 4 / sqrt3 (sin60cos20-cos60sin20) / (2sin20cos20) #

# = 4 / sqrt3 sin (60-20) / (2sin20cos20) #

# = 4 / sqrt3 sin40 / sin40 #

# = 4 / sqrt3 #

Vis det tan (52,5 °) = sqrt6 - sqrt3 - sqrt2 + 2?

Rarrtan75 ° = tan (45 + 30) = (tan45 + tan30) / (1-tan45 * tan30) = (1+ (1 / sqrt (3))) / (1- (1 / sqrt (3)) = sqrt (3) +1) / (sqrt (3) -1) = 2 + sqrt (3) rarrtan52.5 = barneseng (90-37,5) = cot37.5 rarrcot37.5 = 1 / (tan (75/2) ) rarrtanx = (2tan (x / 2)) / (1-tan ^ 2 (x / 2)) rarrtanx-tanx * tan ^ 2 (x / 2) = 2tan (x / 2) rarrtanx * tan ^ 2 / 2) + 2tan (x / 2) -tanx = 0 Det er kvadratisk i tan (x / 2) Så rarrtan (x / 2) = (- 2 + sqrt (2 ^ 2-4 * tanx * )) / (2 * tanx) rarrtan (x / 2) = (- 2 + sqrt (4 (1 + tan ^ 2x))) / (2 * tanx) rarrtan (x / 2) = (- 1 + sqrt (1 + tan ^ 2x)) / tanx At sætte x = 75 vi få

Hvad er sqrt {-sqrt3 + sqrt (3 + 8 sqrt (7 + 4 sqrt3?

Hvis man kan bruge en kalkulator, er dens 2 Hvis ingen kalkulator er tilladt, skal man leve med loven af surds og bruge algebraisk manipulation for at forenkle det. Går denne vej: sqrt (7 + 4sqrt (3)) = sqrt (4 + 2 * 2sqrt (3) +3) = sqrt (2 ^ 2 + 2 * 2sqrt (3) + sqrt3 ^ 2) = sqrt + sqrt3) ^ 2) = 2 + sqrt3 {Dette bruger identiteten (a + b) ^ 2 = a ^ 2 + b ^ 2 + 2ab} sqrt (3 + 8sqrt (7 + 4sqrt3)) = sqrt 8 * (2 + sqrt3)) = sqrt (3 + 16 + 8sqrt3) = sqrt (16 + 2 * 4sqrt3 + 3) = sqrt ((4 + sqrt3) ^ 2) = 4 + sqrt3 {Dette bruger identiteten a + b) ^ 2 = a ^ 2 + b ^ 2 + 2ab} sqrt (-sqrt3 + sqrt (3 + 8sqrt (7 + 4sqrt3))) = sqr

Skriv det komplekse tal (sqrt3 + i) / (sqrt3-i) i standardformularen?

Farve (maroon) (=> (sqrt3 + i) / 2) ^ 2 Ved at rationalisere nævneren får vi standardformularen. (sqrt 3 + i) / (sqrt3 - i) Multiplicér og divider med (sqrt3 + i) => (sqrt3 + i) ^ 2 / ((sqrt3-i) * (sqrt3 + i)) => (sqrt3 + i) ^ 2 / (3 + 1) farve (indigo) ) / 2) ^ 2