The simplest trigonometry function y = Sin (x) is given; it is differentiable at each point from the entire domain of definition. It is necessary to prove that the derivative of the sine of any argument is equal to the cosine of the same angle, that is, y '= Cos (x).
The proof is based on the definition of a derivative function
We define x (arbitrary) in some small neighborhood Δx of a specific point x 0 . We show the value of the function in it and at the point x to find the increment of the given function. If Δx is an increment of the argument, then the new argument is x 0 + Δx = x, the value of this function for a given value of the argument y (x) is Sin (x 0 + Δx), the value of the function at a specific point y (x 0 ) is also known .
Now we have Δy = Sin (x 0 + Δx) -Sin (x 0 ) - the resulting increment of the function.
Using the sine formula, the sum of two unequal angles will transform the difference Δ.
Δy = Sin (x 0 ) Cos (Δx) + Cos (x 0 ) Sin (Δx) minus Sin (x 0 ) = (Cos (Δx) -1) Sin (x 0 ) + Cos (x 0 ) Sin (Δx).
We performed a permutation of the terms, grouped the first with the third Sin (x 0 ), took out the common factor - sine - out of the brackets. Received in the expression the difference Cos (Δx) -1. It remains to change the sign in front of the bracket and in brackets. Knowing what 1-Cos (Δx) is, make a replacement and get a simplified expression Δy, which we then divide by Δx.
Δ / Δ will have the form: Cos (x 0 ) · Sin (Δ) / Δ-2 · Sin 2 (0,5 · Δ) · Sin ( 0 ) / Δ. This is the ratio of the increment of the function to the allowed increment of the argument.
It remains to find the limit of the relation lim obtained for us when Δ tends to zero.
It is known that the limit Sin (Δ) / Δx is 1, under this condition. And the expression 2 · Sin 2 (0.5 · Δ) / Δ in the obtained quotient is summed up by transformations to the product containing the first remarkable limit as a factor: we divide the numerator and the negator of the fraction by 2, and replace the square of the sine by the product. Like this:
(Sin (0.5 · Δx) / (0.5 · Δx)) · Sin (Δx / 2).
The limit of this expression for Δx tending to zero will be equal to zero (1 times 0). It turns out that the limit of the ratio Δy / Δ is Cos (x 0 ) · 1-0, this is Cos (x 0 ), an expression that does not depend on Δ, tending to 0. Hence the conclusion follows: the derivative of the sine of any angle x is to the cosine of x, we write this: y '= Cos (x).
The resulting formula is entered in the well-known table of derivatives, where all elementary functions are collected
When solving problems where the sine derivative is found, one can use the differentiation rules and ready-made formulas from the table. For example: find the derivative of the simplest function y = 3 · Sin (x) -15. We use the elementary rules of differentiation, removing the numerical factor behind the sign of the derivative, and calculating the derivative of a constant number (it is zero). We apply the tabular value of the derivative of the sine of the angle x, equal to Cos (x). We get the answer: y '= 3 · Cos (x) -O. This derivative, in turn, is also an elementary function y = 3 · Cos (x).
The derivative of the sine squared of any argument
When calculating this expression (Sin 2 (x)) ', it is necessary to recall how a complex function differentiates. So, y = Sin 2 (x) - is a power function, since the sine is squared. Its argument is also a trigonometric function, difficult argument. The result in this case is equal to the product, the first factor of which is the derivative of the square of the given complex argument, and the second is the derivative of the sine. Here's what the rule of differentiating a function from a function looks like: (u (v (x))) 'is equal to (u (v (x)))' (v (x)) '. The expression v (x) is a complex argument (internal function). If given the function "the game is equal to the sine squared x", then the derivative of this complex function will be y '= 2 · Sin (x) · Cos (x). In the product, the first double factor is the derivative of the known power function, and Cos (x) is the derivative of the sine, the argument of a complex quadratic function. The final result can be transformed using the trigonometric double-angle sine formula. Answer: the derivative is equal to Sin (2 · x). This formula is easy to remember, it is often used as a table.