Question

1. (1-cos a)(1+cos a) 2. sin(n+a)cos(3n/2-a)+ctg(n+a)ctg(n/2+a) 3. cos 240; ctg(-120)

Answer 1

Пошаговое объяснение:

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Answer 2

Let's break down each of the expressions you provided:

1. (1 - cos a)(1 + cos a): This is a simple trigonometric expression. You can use the identity for the difference of squares, which is: a^2 - b^2 = (a + b)(a - b)

   In this case, a = 1 and b = cos(a). So: (1 - cos a)(1 + cos a) = 1^2 - cos^2(a) = 1 - cos^2(a)

2. sin(n + a)cos(3n/2 - a) + ctg(n + a)ctg(n/2 + a): It seems like you are using the abbreviation "ctg" for the cotangent function. Let's work with this expression:

   sin(n + a)cos(3n/2 - a) + cot(n + a)cot(n/2 + a)

   You can use trigonometric identities to simplify this expression further. The identities you'll need are:

   • sin(3n/2 - a) = -cos(a)
   • cot(x) = 1/tan(x)

   So, the expression becomes:

   -sin(a)cos(a) + 1/tan(n + a) * 1/tan(n/2 + a)

   You can simplify the first part by using the identity sin(2x) = 2sin(x)cos(x):

   -2sin(a)cos(a)

   And for the second part, you have the product of two cotangents, so it becomes:

   1/(tan(n + a)tan(n/2 + a))

   Now, you've simplified the expression as much as possible.

3. cos 240; ctg(-120): These are the values of the trigonometric functions at the given angles. Let's calculate them:

   a. cos(240 degrees): To find the cosine of 240 degrees, you can use the fact that cos is negative in the third quadrant. So, cos(240 degrees) = -cos(60 degrees), and you can use the known value for cos(60 degrees), which is 0.5. Therefore, cos(240 degrees) = -0.5.

   b. ctg(-120 degrees): "ctg" usually refers to the cotangent function. The cotangent of an angle is the reciprocal of the tangent. The tangent of -120 degrees is the same as the tangent of 120 degrees because the tangent function has a period of 180 degrees. So, first, you need to find the tangent of 120 degrees:

   tan(120 degrees) = √3

   Now, find the cotangent:

   ctg(-120 degrees) = 1 / tan(120 degrees) = 1 / √3

   So, ctg(-120) = 1/√3.