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

Odgovor:

#sin (cos ^ (- 1) (5/13) + tan ^ (- 1) (3/4)) = 63/65 #

Pojasnilo:

Let #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, pustite #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, #sin (cos ^ (- 1) (5/13) + tan ^ (- 1) (3/4)) #

# = sin (sin ^ (- 1) (63/65)) = 63/65 #

Kako ocenjujete greh (cos ^ -1 (1/2)) brez kalkulatorja?

Sin (cos ^ (- 1) (1/2)) = sqrt (3) / 2 Naj bo cos ^ (- 1) (1/2) = x, nato cosx = 1/2 rarrsinx = sqrt (1-cos ^ 2x ) = sqrt (1- (1/2) ^ 2) = sqrt (3) / 2 rarrx = sin ^ (- 1) (sqrt (3) / 2) = cos ^ (- 1) (1/2) , sin (cos ^ (- 1) (1/2)) = sin (sin ^ (- 1) (sqrt (3) / 2)) = sqrt (3) / 2

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 cos (tan ^ -1 (3/4))?

Cos (tan ^ -1 (3/4)) = 0,8 cos (tan ^ -1 (3/4)) =? Naj bo tan ^ -1 (3/4) = theta:. tan theta = 3/4 = P / B, P in B sta pravokotni in osnovni pravokotni trikotnik, nato 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]