Kako izračunate cos (tan ^ -1 (3/4))?

Odgovor:

# cos (tan ^ -1 (3/4)) = 0,8

Pojasnilo:

# cos (tan ^ -1 (3/4)) =? # Let # tan ^ -1 (3/4) = theta #

#:. tan theta = 3/4 = P / B, P in B # so pravokotne in osnove

nato pravokotnega trikotnika # H ^ 2 = P ^ 2 + B ^ 2 = 3 ^ 2 + 4 ^ 2 = 25 #

#: H = 5;:. cos theta = B / H = 4/5 = 0,8

# cos (tan ^ -1 (3/4)) = cos theta = 0,8 #

#:. cos (tan ^ -1 (3/4)) = 0,8 Ans

Odgovor:

#4/5#

Pojasnilo:

#tan (tan ^ -1 (3/4)) = 3/4 #

# "Ime" y = tan ^ -1 (3/4) #

# "Potem imamo" #

#tan (y) = 3/4 #

# "Zdaj uporabite" sec² (x) = 1 + tan² (x) #

# => sec² (y) = 1 + tan² (y) = 1 + 9/16 = 25/16 #

# => sec (y) = 1 / cos (y) = pm 5/4 #

# => cos (y) = pm 4/5 #

# => cos (tan ^ -1 (3/4)) = pm 4/5

# "Moramo vzeti rešitev z znakom + kot" #

# -pi / 2 <= arctan (x) <= pi / 2 #

# "in" #

#cos (x)> 0, če je -pi / 2 <= x <= pi / 2 #

# => cos (tan ^ -1 (3/4)) = 4/5

# "Upoštevajte, da smo lahko uporabili tudi" #

#tan (y) = sin (y) / cos (y) #

# "in" #

# sin ^ 2 (y) + cos ^ 2 (y) = 1 #

#tan (y) = sin (y) / cos (y) = 3/4 #

# => pm sqrt (1-cos ^ 2 (y)) / cos (y) = 3/4 #

# => 1-cos ^ 2 (y) = ((3/4) cos (y)) ^ 2 #

# => (1 + 9/16) cos ^ 2 (y) = 1 #

# => cos ^ 2 (y) = 16/25 #

# => cos (y) = 4/5 #

Pokažite, da cos²π / 10 + cos²4π / 10 + cos² 6π / 10 + cos²9π / 10 = 2. Malo sem zmeden, če naredim Cos²4π / 10 = cos² (π-6π / 10) & cos²9π / 10 = cos² (π-π / 10), bo postal negativen kot cos (180 ° - theta) = - costheta v drugi kvadrant. Kako naj dokazujem vprašanje?

Glej spodaj. LHS = cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) + cos ^ 2 ((6pi) / 10) + cos ^ 2 ((9pi) / 10) = cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) + cos ^ 2 (pi- (4pi) / 10) + cos ^ 2 (pi- (pi) / 10) = cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) + cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) = 2 * [cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10)] = 2 * [cos ^ 2 (pi / 2- (4pi) / 10) + cos ^ 2 ((4pi) / 10)] = 2 * [sin ^ 2 ((4pi) / 10) + cos ^ 2 ((4pi) / 10)] = 2 * 1 = 2 = RHS

Kako izračunate cos (tan-3/4)?

Predvidevam, da misliš cos (arctan (3/4)), kjer je arctan (x) inverzna funkcija tan (x). (Včasih arctan (x), kot je napisano kot tan ^ -1 (x), vendar osebno se mi zdi to zmedeno, saj bi bilo morda morda napačno razumljeno kot 1 / tan (x).) Uporabiti moramo naslednje identitete: cos (x ) = 1 / sec (x) {Identity 1} tan ^ 2 (x) + 1 = sec ^ 2 (x), ali sec (x) = sqrt (tan ^ 2 (x) +1) {Identity 2} v mislih lahko enostavno najdemo cos (arctan (3/4)). cos (arctan (3/4)) = 1 / sec (arctan (3/4)) {Uporaba identitete 1} = 1 / sqrt (tan (arctan (3/4)) ^ 2+ 1) {Uporaba identitete 2} = 1 / sqrt ((3/4) ^ 2 + 1) {Po definiciji arctan (x)}

Kako izračunate greh (cos ^ -1 (5/13) + tan ^ -1 (3/4))?

Sin (cos ^ (- 1) (5/13) + tan ^ (- 1) (3/4)) = 63/65 Naj cos ^ (- 1) (5/13) = x potem rarrcosx = 5/13 rarrsinx = sqrt (1-cos ^ 2x) = sqrt (1- (5/13) ^ 2) = 12/13 rarrx = sin ^ (- 1) (12/13) = cos ^ (- 1) (5 / 13) Tudi, naj bo tan ^ (- 1) (3/4) = y potem rarrtany = 3/4 rarrsiny = 1 / cscy = 1 / sqrt (1 + cot ^ 2y) = 1 / sqrt (1+ (4 / 3) ^ 2) = 3/5 rarry = tan ^ (- 1) (3/4) = sin ^ (- 1) (3/5) rarrcos ^ (- 1) (5/13) + tan ^ (- 1) (3/4) = sin ^ (- 1) (12/13) + sin ^ (- 1) (3/5) = sin ^ (- 1) (12/13 * sqrt (1- (3 / 5) ^ 2) + 3/5 * sqrt (1- (12/13) ^ 2)) = sin ^ (- 1) (12/13 * 4/5 + 3/5 * 5/13) = 63 / 65 Zdaj, greh (cos ^ (- 1)