{"history":[],"article":{"id":"8d9111d9-d326-42a5-acf5-024a55ddf4ea","title":"Differentiation of trigonometric functions","slug":"differentiation-of-trigonometric-functions","content":":::summary\n:::\n\n# \nThe **differentiation of [trigonometric function](https://bliptext.com/articles/trigonometric-function)s** is the mathematical process of finding the [derivative](https://bliptext.com/articles/derivative) of a [trigonometric function](https://bliptext.com/articles/trigonometric-function), or its rate of change with respect to a variable.\nFor example, the derivative of the sine function is written (a) = cos(a), meaning that the rate of change of sin(x) at a particular angle x = a is given by the cosine of that angle.\nAll derivatives of circular trigonometric functions can be found from those of sin(x) and cos(x) by means of the [quotient rule](https://bliptext.com/articles/quotient-rule) applied to functions such as tan(x) = sin(x)/cos(x).\nKnowing these derivatives, the derivatives of the [inverse trigonometric functions](https://bliptext.com/articles/inverse-trigonometric-functions) are found using [implicit differentiation](https://bliptext.com/articles/implicit-differentiation).\n\n# Proofs of derivatives of trigonometric functions\n\n## Limit of sin(θ)/θ as θ tends to 0\nThe diagram at right shows a circle with centre O and radius r = 1.\nLet two radii OA and OB make an arc of θ radians.\nSince we are considering the limit as θ tends to zero, we may assume θ is a small positive number, say 0 < θ < 1⁄2 π in the first quadrant.\nIn the diagram, let R1 be the triangle OAB, R2 the [circular sector](https://bliptext.com/articles/circular-sector) OAB, and R3 the triangle OAC.\nThe [area of triangle](https://bliptext.com/articles/area-of-a-triangle) OAB is:\nThe [area of the circular sector](https://bliptext.com/articles/circular-sector) OAB is:\nThe area of the triangle OAC is given by:\nSince each region is contained in the next, one has:\nMoreover, since sin θ > 0 in the first quadrant, we may divide through by 1⁄2 sin θ, giving:\nIn the last step we took the reciprocals of the three positive terms, reversing the inequities.\nWe conclude that for 0 < θ < 1⁄2 π, the quantity sin(θ)/θ is always less than 1 and always greater than cos(θ).\nThus, as θ gets closer to 0, sin(θ)/θ is \"[squeezed](https://bliptext.com/articles/squeeze-theorem)\" between a ceiling at height 1 and a floor at height cos θ, which rises towards 1; hence sin(θ)/θ must tend to 1 as θ tends to 0 from the positive side:\"\"For the case where θ is a small negative number –1⁄2 π < θ < 0, we use the fact that sine is an [odd function](https://bliptext.com/articles/odd-function):\n# [Circle, centre O, radius 1 | upload.wikimedia.org/wikipedia/commons/1/15/limit///circle///FbN.jpeg]\n# [Squeeze: The curves y = 1 and y = cos θ shown in red, the curve y = sin(θ)/θ shown in blue. | upload.wikimedia.org/wikipedia/commons/f/f9/Squeeze///FbN.png]\n\n## Limit of (cos(θ)-1)/θ as θ tends to 0\nThe last section enables us to calculate this new limit relatively easily.\nThis is done by employing a simple trick.\nIn this calculation, the sign of θ is unimportant.\nUsing cos2θ – 1 = –sin2θ,\nthe fact that the limit of a product is the product of limits, and the limit result from the previous section, we find that:\n\n## Limit of tan(θ)/θ as θ tends to 0\nUsing the limit for the [sine](https://bliptext.com/articles/) function, the fact that the tangent function is odd, and the fact that the limit of a product is the product of limits, we find:\n\n## Derivative of the sine function\nWe calculate the derivative of the [sine function](https://bliptext.com/articles/sine-function) from the [limit definition](https://bliptext.com/articles/derivative):\nUsing the [angle addition formula](https://bliptext.com/articles/list-of-trigonometric-identities) sin(α+β) = sin α cos β + sin β cos α, we have:\nUsing the limits for the [sine](https://bliptext.com/articles/) and co[sine](https://bliptext.com/articles/) functions:\n\n## Derivative of the cosine function\n\n### From the definition of derivative\nWe again calculate the derivative of the [cosine function](https://bliptext.com/articles/cosine-function) from the limit definition:\nUsing the angle addition formula cos(α+β) = cos α cos β – sin α sin β, we have:\nUsing the limits for the [sine](https://bliptext.com/articles/) and co[sine](https://bliptext.com/articles/) functions:\n\n### From the chain rule\nTo compute the derivative of the cosine function from the chain rule, first observe the following three facts:\nThe first and the second are [trigonometric identities](https://bliptext.com/articles/list-of-trigonometric-identities), and the third is proven above.\nUsing these three facts, we can write the following,\nWe can differentiate this using the [chain rule](https://bliptext.com/articles/chain-rule).\nLetting, we have:\nTherefore, we have proven that\n\n## Derivative of the tangent function\n\n### From the definition of derivative\nTo calculate the derivative of the [tangent function](https://bliptext.com/articles/tangent-function) tan θ, we use [first principles](https://bliptext.com/articles/derivative).\nBy definition:\nUsing the well-known angle formula tan(α+β) = (tan α + tan β) / (1 - tan α tan β), we have:\nUsing the fact that the limit of a product is the product of the limits:\nUsing the limit for the [tangent](https://bliptext.com/articles/) function, and the fact that tan δ tends to 0 as δ tends to 0:\nWe see immediately that:\n\n### From the quotient rule\nOne can also compute the derivative of the tangent function using the [quotient rule](https://bliptext.com/articles/quotient-rule).\nThe numerator can be simplified to 1 by the [Pythagorean identity](https://bliptext.com/articles/list-of-trigonometric-identities), giving us,\nTherefore,\n\n# Proofs of derivatives of inverse trigonometric functions\nThe following derivatives are found by setting a [variable](https://bliptext.com/articles/variable-mathematics) y equal to the [inverse trigonometric function](https://bliptext.com/articles/inverse-trigonometric-function) that we wish to take the derivative of.\nUsing [implicit differentiation](https://bliptext.com/articles/implicit-differentiation) and then solving for dy/dx, the derivative of the inverse function is found in terms of y.\nTo convert dy/dx back into being in terms of x, we can draw a reference triangle on the unit circle, letting θ be y. Using the [Pythagorean theorem](https://bliptext.com/articles/pythagorean-theorem) and the definition of the regular trigonometric functions, we can finally express dy/dx in terms of x.\n\n## Differentiating the inverse sine function\nWe let\nWhere\nThen\nTaking the derivative with respect to x on both sides and solving for dy/dx:\nSubstituting in from above,\nSubstituting x=\\sin y in from above,\n\n## Differentiating the inverse cosine function\nWe let\nWhere\nThen\nTaking the derivative with respect to x on both sides and solving for dy/dx:\nSubstituting in from above, we get\nSubstituting in from above, we get\nAlternatively, once the derivative of \\arcsin x is established, the derivative of \\arccos x follows immediately by differentiating the identity so that.\n\n## Differentiating the inverse tangent function\nWe let\nWhere\nThen\nTaking the derivative with respect to x on both sides and solving for dy/dx:\nLeft side:\nRight side:\nTherefore,\nSubstituting in from above, we get\n\n## Differentiating the inverse cotangent function\nWe let\nwhere 0

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