Find the equation of tangents to the curve y = cos(x + y), -2pi<=x<=2pi that are parallel to the line x+2y=0
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y = cos(x + y)
differentiate with respect to x
dy/dx = -sin(x + y).{ 1 + dy/dx}
dy/dx = -sin(x + y) - sin(x +y).dy/dx
dy/dx { 1+ sin(x + y)} = - sin(x + y)
dy/dx = -sin(x + y)/{1 + sin(x + y)}
because equation of tangent is parallel to x + 2y = 0 so,
dy/dx = -1/2 = -sin(x + y)/{1 + sin(x + y)}
2sin(x + y) = 1 + sin(x + y)
sin(x + y) = 1
x + y = π/2 and -3π/2
so, y = cos(x +y) = 0
hence, (π/2,0) and ( -3π/2, 0) are two points form which two tangents are passing and parallel to x + 2y = 0
now , equation of tangents:
(y - 0) = -1/2(x -π/2) => x + 2y = π/2
(y -0) = -1/2(x +3π/2) => x + 2y = -3π/2
differentiate with respect to x
dy/dx = -sin(x + y).{ 1 + dy/dx}
dy/dx = -sin(x + y) - sin(x +y).dy/dx
dy/dx { 1+ sin(x + y)} = - sin(x + y)
dy/dx = -sin(x + y)/{1 + sin(x + y)}
because equation of tangent is parallel to x + 2y = 0 so,
dy/dx = -1/2 = -sin(x + y)/{1 + sin(x + y)}
2sin(x + y) = 1 + sin(x + y)
sin(x + y) = 1
x + y = π/2 and -3π/2
so, y = cos(x +y) = 0
hence, (π/2,0) and ( -3π/2, 0) are two points form which two tangents are passing and parallel to x + 2y = 0
now , equation of tangents:
(y - 0) = -1/2(x -π/2) => x + 2y = π/2
(y -0) = -1/2(x +3π/2) => x + 2y = -3π/2
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Mathematics textbook for class 12 part 1
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