Question

Алгебра,умоляю,сейчас уже к/р срочно пожалуйста 1)sin765градусов Cos19П:6 2)cos(a-b)-cos(a+b) a-альфа , b-бетта 3)решите уравнение 2sin * x/2=1-cos * x

Answer 1

1)sin(765)=sin(720+45)=-sin45=-√2/2
cos19п/6=cos570=cos(540+30)=-sin30=-1/2

2)cos(a-b)-cos(a+b)=cosacosb+sinasinb-cosacosb+sinasinb=2sinasinb

3) 2sinx/2=1-cosx тут если смогу решить в скором времени в комментарии добавлю

Answer 2

1) sin 765 градусов, cos 19π/6

To find the values of sin 765 degrees and cos 19π/6, we can use the properties of the unit circle and the periodicity of trigonometric functions.

1) sin 765 degrees:

Since the sine function has a period of 360 degrees (or 2π radians), we can find an equivalent angle within a single period.

765 degrees is equivalent to 765 - 360 = 405 degrees.

Now, we can use the fact that sine is an odd function, which means sin(-x) = -sin(x).

sin 405 degrees = -sin(-405 degrees)

Using a reference angle of 45 degrees, we can find the values of sine:

sin 405 degrees = -sin(405 - 360) = -sin 45 degrees = -√2/2.

Therefore, sin 765 degrees = -√2/2.

2) cos(a - b) - cos(a + b) + a - α, b - β:

To simplify the expression cos(a - b) - cos(a + b) + a - α, b - β, we can use the trigonometric identities.

cos(a - b) = cos a * cos b + sin a * sin b cos(a + b) = cos a * cos b - sin a * sin b

Substituting these values into the expression:

cos(a - b) - cos(a + b) + a - α, b - β = (cos a * cos b + sin a * sin b) - (cos a * cos b - sin a * sin b) + a - α, b - β = 2 * sin a * sin b + a - α, b - β

Please provide the values of a, α, b, and β so that we can further simplify the expression.

3) Solve the equation 2sin(x/2) = 1 - cos(x):

To solve the equation 2sin(x/2) = 1 - cos(x), we can use the double-angle identity for sine and the identity sin^2(x) + cos^2(x) = 1.

First, let's rewrite the equation using the double-angle identity:

2sin(x/2) = 1 - cos(x) 2 * 2sin^2(x/2) = 1 - (1 - 2sin^2(x/2)) 4sin^2(x/2) = 2sin^2(x/2) 2sin^2(x/2) = 0

Now, we have a quadratic equation. Let's solve it:

sin^2(x/2) = 0

Taking the square root of both sides:

sin(x/2) = 0

Since sin(x/2) = 0, we know that x/2 must be an integer multiple of π.

Therefore, x/2 = nπ, where n is an integer.

Solving for x:

x = 2nπ, where n is an integer.

So the general solution to the equation 2sin(x/2) = 1 - cos(x) is x = 2nπ, where n is an integer.