Question

Sin2xsin5x+cos2xcos5x>корень 3/2

Answer 1

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Answer 2

We can start by using the product-to-sum identities to rewrite the left-hand side of the inequality:

sin(2x)sin(5x) + cos(2x)cos(5x) = (1/2)(cos(3x) - cos(7x)) + (1/2)(cos(3x) + cos(7x)) = cos(3x)

Now we need to find the range of cos(3x) in order to determine whether it is greater than the given value of √3/2. Since the range of the cosine function is [-1,1], we can find the range of cos(3x) by considering the interval [0,2π/3] (since cos(3x) has a period of 2π/3).

We know that cos(0) = 1 and cos(2π/3) = -1/2, so by the intermediate value theorem, cos(3x) takes on all values in the interval [-1,1] over the interval [0,2π/3]. Therefore, the inequality sin(2x)sin(5x) + cos(2x)cos(5x) > √3/2 is true for all values of x in the interval [0,2π/3].

In other words, if x is any angle in the interval [0,2π/3], then sin(2x)sin(5x) + cos(2x)cos(5x) is greater than √3/2. Outside of this interval, the inequality may or may not be true, depending on the value of cos(3x) in that range.

Note that we assumed x to be in radians, as is convention in most mathematical contexts.