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The expression (\cos(x + y) \cos x + \cos y) does not simplify to a standard identity. Instead, it can be rewritten using the angle addition formula for cosine: (\cos(x + y) = \cos x \cos y - \sin x \sin y). Therefore, the original expression is not generally true, and its simplification would depend on specific values of (x) and (y).
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What is the implicit differentiation of y equals sin x plus y?
y = sin(x+y) cos( x + y )[(1 + y')] = y' cos(x + y ) + y'cos(x + y ) = y' y'-y'cos( x+ y) = cos( x + y ) y'[1-cos(x+y)]= cos(x+y) y'= [cos(x+y)]/ [1-cos(x+y)]
How do you solve cos squared x - cosx equals 2 for 0x2pi?
cos2(x) - cos(x) = 2 Let y = cos(x) then y2 - y = 2 or y2 - y - 2 = 0 factorising, (y - 2)(y + 1) = 0 that is y = 2 or y = -1 Substitutng back, this would require cos(x) = 2 or cos(x) = -1 But cos(x) cannot be 2 so cos(x) = -1 Then x = cos-1(-1) => x = pi radians.
What is the derivative of cos pi x plus sin pi y all to the 8th power equals 44?
(cos(pi x) + sin(pi y) )^8 = 44 differentiate both sides with respect to x 8 ( cos(pi x) + sin (pi y ) )^7 d/dx ( cos(pi x) + sin (pi y) = 0 8 ( cos(pi x) + sin (pi y ) )^7 (-sin (pi x) pi + cos (pi y) pi dy/dx ) = 0 8 ( cos(pi x) + sin (pi y ) )^7 (pi cos(pi y) dy/dx - pi sin (pi x) ) = 0 cos(pi y) dy/dx - pi sin(pi x) = 0 cos(pi y) dy/dx = sin(pi x) dy/dx = sin (pi x) / cos(pi y)
Cos x equals -cos x plus 1?
No, but cos(-x) = cos(x), because the cosine function is an even function.
Find the second derivative of y equals sin sqaured x?
y = sin2(x) y' = 2sin(x)cos(x) y'' = 2 [ cos(x)cos(x) + sin(x)(-sin(x)) ] = 2 [ cos2(x) - sin2(x) ] = 2 [ 1 - sin2(x) - sin2(x) ] = 2 [ 1 - 2sin2(x) ]
What is the implicit differentiation of y equals sin x plus y?
y = sin(x+y) cos( x + y )[(1 + y')] = y' cos(x + y ) + y'cos(x + y ) = y' y'-y'cos( x+ y) = cos( x + y ) y'[1-cos(x+y)]= cos(x+y) y'= [cos(x+y)]/ [1-cos(x+y)]
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) ]
What are the sum and difference identities for the sine cosine and tangent functions?
Sine sum identity: sin (x + y) = (sin x)(cos y) + (cos x)(sin y)Sine difference identity: sin (x - y) = (sin x)(cos y) - (cos x)(sin y)Cosine sum identity: cos (x + y) = (cos x)(cos y) - (sin x)(sin y)Cosine difference identity: cos (x - y) = (cos x)(cos y) + (sin x)(sin y)Tangent sum identity: tan (x + y) = [(tan x) + (tan y)]/[1 - (tan x)(tan y)]Tangent difference identity: tan (x - y) = [(tan x) - (tan y)]/[1 + (tan x)(tan y)]
How do you prove that 2 sin 3x divided by sin x plus 2 cos 3x divided by cos x equals 8 cos 2x?
You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.
How do you solve cos squared x - cosx equals 2 for 0x2pi?
cos2(x) - cos(x) = 2 Let y = cos(x) then y2 - y = 2 or y2 - y - 2 = 0 factorising, (y - 2)(y + 1) = 0 that is y = 2 or y = -1 Substitutng back, this would require cos(x) = 2 or cos(x) = -1 But cos(x) cannot be 2 so cos(x) = -1 Then x = cos-1(-1) => x = pi radians.
What is the derivative of cos pi x plus sin pi y all to the 8th power equals 44?
(cos(pi x) + sin(pi y) )^8 = 44 differentiate both sides with respect to x 8 ( cos(pi x) + sin (pi y ) )^7 d/dx ( cos(pi x) + sin (pi y) = 0 8 ( cos(pi x) + sin (pi y ) )^7 (-sin (pi x) pi + cos (pi y) pi dy/dx ) = 0 8 ( cos(pi x) + sin (pi y ) )^7 (pi cos(pi y) dy/dx - pi sin (pi x) ) = 0 cos(pi y) dy/dx - pi sin(pi x) = 0 cos(pi y) dy/dx = sin(pi x) dy/dx = sin (pi x) / cos(pi y)
Find y'' if y equals 6x sinx?
That means you must take the derivative of the derivative. In this case, you must use the product rule. y = 6x sin x y'= 6[x (sin x)' + (x)' sin x] = 6[x cos x + sin x] y'' = 6[x (cos x)' + (x)' cos x + cos x] = 6[x (-sin x) + cos x + cos x] = 6[-x sin x + 2 cos x]
How do you solve csc x sin x equals cos x cot x plus?
Suppose csc(x)*sin(x) = cos(x)*cot(x) + y then, ince csc(x) = 1/sin(x), and cot(x) = cos(x)/sin(x), 1 = cos(x)*cos(x)/sin(x) + y so y = 1 - cos2(x)/sin(x) = 1 - [1 - sin2(x)]/sin(x) = [sin2(x) + sin(x) - 1]/sin(x)
If Sin equals x and Cos equals y then x squared equals what function of y?
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
What is x equal y sq over cos x pi?
Can you please claify if you mean x=y^2/ pi*cos(x) , or x=y^2/cos(pi), since they are very different sums.
Cos x equals -cos x plus 1?
No, but cos(-x) = cos(x), because the cosine function is an even function.
Find the derivative of y x2 sin x 2xcos x - 2sin x?
y = (x^2)(sin x)(2x)(cos x) - 2sin xy' = [[(x^2)(sin x)][(2x)(cos x)]]' - (2sin x)'y' = [[(x^2)(sin x)]'[(2x)(cos x)] + [(2x)(cos x)]'[(x^2)(sin x)]]- (2sin x)'y' = [[(x^2)'(sin x) + (sin x)'(x^2)][(2x)(cos x)] + [(2x)'(cos x) + (cos x)'(2x)][(x^2)(sin x)] ] - 2(cos x)y' = [[(2x)(sin x )+ (cos x)(x^2)][(2x)(cos x)] + [2cos x - (sin x)(2x)][(x^2)(sin x)]] - 2(cos x)y' = (4x^2)(sin x cos x) + (2x^3)(cos x)^2 + (2x^2)(sin x cos x) - (2x^3)(sin x)^2 - 2cos xy' = (6x^2)(sin x cos x) + (2x^3)(cos x)^2 - (2x^3)(sin x)^2 - 2cos x (if you want, you can stop here, or you can continue)y' = (3x^2)(2sin x cos x) + (2x^3)[(cos x)^2 - (sin x)^2] - 2cos xy' = (3x^2)(sin 2x) + (2x^3)(cos 2x) - 2 cos xy' = (2x^3)(cos 2x) + (3x^2)(sin 2x) - 2 cos x
