Întrebarea nr. 132a1

Întrebarea nr. 132a1
Anonim

Răspuns:

Vedeți mai jos

Explicaţie:

# LHS = 1-sin4x + patut ((3pi) / 4.2x) * cos4x #

# = 1-sin4x + (pat copii ((3pi) / 4) * cot2x + 1) / (cot2x-patut ((3pi) / 4)) * # cos4x

# = 1-sin4x + ((pat copii (pi-pi / 4) * cot2x + 1) / (cot2x-cot (pi-pi / 4))) * # cos4x

# = 1-sin4x + (- patut (pi / 4) * cot2x + 1) / (cot2x - (- patut (pi / 4))) * # cos4x

# = 1-sin4x + (1-cot2x) / (1 + cot2x) * # cos4x

# = 1-sin4x + (1- (cos2x) / (sin2x)) / (1+ (cos2x) / (sin2x)) * # cos4x

# = 1-sin4x + (sin2x-cos2x) / (sin2x + cos2x) * cos4x #

# = 1 + (2 (sin2x * cos4x-cos4x * cos2x-sin4x * sin2x-sin4x * cos2x)) / (2 (sin2x + cos2x)) #

# = 1 + (sin (4x + 2x) -sin (4x-2x) -cos (4x + 2x) -cos (4x-2x) -cos (4x-2x) + cos (4x + 2x) -sin (4x + 2x) -sin (4x-2x)) / (2 (sin2x + cos2x) #

# = 1 + (sin6x-sin2x-cos6x-cos2x-cos2x + cos6x-sin6x-sin (2x)) / (2 (sin2x + cos2x) #

# = 1-a anula ((2 (sin2x + cos2x)) / (2 (sin2x + cos2x))) #

# = 1-1 = 0 = # RHS