Cari nilai theta, jika, Cos (theta) / 1 - sin (theta) + cos (theta) / 1 + sin (theta) = 4?

Jawapan:

# theta = pi / 3 # atau #60^@#

Penjelasan:

Baik. Kami mempunyai:

# costheta / (1-sintheta) + costheta / (1 + sintheta) = 4 #

Mari kita abaikan # RHS # untuk sekarang.

# costheta / (1-sintheta) + costheta / (1 + sintheta) #

# (costheta (1 + sintheta) + costheta (1-sintheta)) / ((1-sintheta) (1 + sintheta)) #

# (costheta ((1-sintheta) + (1 + sintheta))) / (1-sin ^ 2theta) #

# (costheta (1-sintheta + 1 + sintheta)) / (1-sin ^ 2theta) #

# (2costheta) / (1-sin ^ 2theta) #

Menurut Identiti Pythagorean, # sin ^ 2theta + cos ^ 2theta = 1 #. Jadi:

# cos ^ 2theta = 1-sin ^ 2theta #

Sekarang kita tahu bahawa, kita boleh menulis:

# (2costheta) / cos ^ 2theta #

# 2 / costheta = 4 #

# costheta / 2 = 1/4 #

# costheta = 1/2 #

# theta = cos ^ -1 (1/2) #

# theta = pi / 3 #, bila # 0 <= theta <= pi #.

Dalam darjah, # theta = 60 ^ @ # bila # 0 ^ @ <= theta <= 180 ^ @ #

Jawapan:

# rarrcosx = 1/2 #

Penjelasan:

Diberikan, # rarrcosx / (1-sinx) + cosx / (1 + sinx) = 4 #

#rarrcosx 1 / (1-sinx) + 1 / (1 + sinx) = 4 #

#rarrcosx (1 + cancel (sinx) + 1cancel (-sinx)) / ((1-sinx) * (1 + sinx) = 4 #

#rar (2cosx) / (1-sin ^ 2x) = 4 #

# rarrcosx / cos ^ 2x = 2 #

# rarrcosx = 1/2 #

Memudahkan (1- cos theta + sin theta) / (1+ cos theta + sin theta)?

= sin (theta) / (1 + cos (theta)) (1-cos (theta) + sin (theta)) / (1 + cos (theta) (1 + cos (theta) + sin (theta)) / (1 + cos (theta) + sin (theta)) ^ 2 = 2 (theta)) / (1 + cos ^ 2 (theta) + sin ^ 2 (theta) +2 sin (theta) +2 cos (theta) sin (theta)) ^ 2-cos ^ 2 (theta)) / (2 + 2 sin (theta) +2 cos (theta) (1 + cos (theta)) + 2 sin (theta) (1 + cos (theta)) = (1/2) ((1 + sin (theta) ) (1 + sin (theta)) = (1/2) (1 + sin (theta)) / (1 + cos (theta) (1 + sin (theta)) (1 + sin (theta))) = (1/2) (1 + sin (theta)) / (1 + cos (theta)) - (1/2) (1-sin ^ 2 (theta)) / ((1 + cos (theta)) (1 + sin (theta) 1 + sin (theta)) / (1 + cos (th

Sin theta / x = cos theta / y maka sin theta - cos theta =?

Jika frac { sin theta} {x} = frac {cos theta] {y} maka sin theta - cos theta = pm frac {x - y} {sqrt {x ^ 2 + y ^ sin theta} {x} = frac {cos theta] {y} frac { sin theta} { cos theta} = frac {x} {y} tan theta = x / dan bersebelahan y jadi cos theta = frac { pm y} {sqrt {x ^ 2 + y ^ 2} sin theta = tan theta cos theta sin theta - cos theta = tan theta cos theta - cos theta = cos theta ( tan theta - 1) = frac { pm y} {sqrt {x ^ 2 + y ^ 2}} (x / y -1) sin theta - cos theta = pm frac {x - y } {sqrt {x ^ 2 + y ^ 2}}

Tunjukkan bahawa, (1 + cos theta + i * sin theta) ^ n + (1 + cos theta - i * sin theta) ^ n = 2 ^ (n + 1) * (cos theta / n * theta / 2)?

Sila lihat di bawah. Letakkan 1 + costheta + isintheta = r (cosalpha + isinalpha), di sini r = sqrt ((1 + costheta) ^ 2 + sin ^ 2theta) = sqrt (2 + 2costheta) = sqrt (2 + 4cos ^ 2 (theta / ) -2) = 2cos (theta / 2) dan tanalpha = sintheta / (1 + costheta) == (2sin (theta / 2) cos (theta / (theta / 2) atau alpha = theta / 2 maka 1 + costheta-isintheta = r (cos (-alpha) + isin (-alpha)) = r (cosalpha-isinalpha) dan kita boleh menulis (1 + ^ n + (1 + costheta-isintheta) ^ menggunakan teorem DE MOivre sebagai cos ^ / 2) = 2 ^ (n + 1) cos ^ n (theta / 2) cos ((ntheta) / 2)