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Need help with working this Trig problem out. A sin alpha plus cos alpha equals 1 B sin alpha - cos alpha equals 1 Solution is AB equals 1?
A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.
How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
Differentiation of sin x 1 plus cosx?
The differentiation of sin x plus cosx is cos (x)-sin(x).
Prove each Indentity tanx mins sinx divided by tanxsinx equals tanxsinx divided by tanx plus sinx?
(tan x - sin x)/(tan x sin x) = (tan x sin x)/(tan x + sin x)[sin x/cos x) - sin x]/[(sin x/cos x)sin x] =? [(sin x/cos x)sin x]/[sin x/cos x) + sin x][(sin x - sin x cos x)/cos x]/(sin2 x/cos x) =? (sin2 x/cos x)/[(sin x + sin x cos x)/cos x)(sin x - sin x cos x)/sin2 x =? sin2 x/(sin x + sin x cos x)[sin x(1 - cos x)]/sin2 x =? sin2 x/[sin x(1 + cos x)(1 - cos x)/sin x =? sin x/(1 + cos x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[(1 + cos x)(1 - cos x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - cos2 x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - (1 - sin2 x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/sin2 x(1 - cos x)/sin x = (1 - cos x)/sin x True
How do you express sin x plus cos x divided by cos x in terms of tan x?
(sin x + cos x) / cosx = sin x / cos x + cosx / cos x = tan x + 1
When does cos x equal -sin x?
The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).
How do you show that 2 sin squared x minus 1 divided by sin x minus cos x equals sin x plus cos x?
(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x
Who formulated eix equals cos x plus you sin x?
leonhard euler
Differentiate y equals xsinx plus cosx?
y = x sin(x) + cos(x)Derivative of the first term = x cos(x) + sin(x)Derivative of the second term = -sin(x)y' = Sum of the derivatives = x cos(x) + sin(x) - sin(x)= [ x cos(x) ]
Need help with working this Trig problem out. A sin alpha plus cos alpha equals 1 B sin alpha - cos alpha equals 1 Solution is AB equals 1?
A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.
How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
Differentiation of sin x 1 plus cosx?
The differentiation of sin x plus cosx is cos (x)-sin(x).
Sin x plus cos x?
sin(x) + cos(x) = sqrt(2) · sin(45°+x)
Simplyfy cos x cot x plus tan x equals?
To simplify this sort of things, it helps if, first of all, you convert everything to sines and cosines.cos x cot x + tan x (original equation)= cos (cos x / sin x) + (sin x / cos x) (convert to sin and cos)= cos2x / sin x + sin x / cos x (multiplying in the first term)= (sin x cos2x + sin x cos x) / sin x cos x (converting common denominator)= (sin x cos x) (cos x + 1) / (sin x cos x) (factoring the numerator)= cos x + 1 (cancelling factors in numerator and denominator)
How do you prove sin x tan x equals cos x?
You can't. tan x = sin x/cos x So sin x tan x = sin x (sin x/cos x) = sin^2 x/cos x.
How do you factor sin squared times x plus cos2x -cosx equals 0?
2
If sin x - cos x equals 1 over 3 what is sin x?
Sin[x] = Cos[x] + (1/3)
