Question

7cos a-6sin a ——————— если tg a=1 3sin a-5cos a

Answer 1

Пошаговое объяснение:

сомневаюсь но вроде бы ответ верный :)

Answer 2

To solve the given equation, we'll first find the values of cosine and sine using the given value of tangent:

Given: tan(a) = 1

We know that tangent is the ratio of sine to cosine:

tan(a) = sin(a) / cos(a)

Substituting the given value:

1 = sin(a) / cos(a)

We can rewrite this equation as:

sin(a) = cos(a)

Now, we have two equations:

1. 7cos(a) - 6sin(a)
2. 3sin(a) - 5cos(a)

Substituting sin(a) = cos(a) in both equations:

1. 7cos(a) - 6cos(a) = cos(a) * (7 - 6) = cos(a)
2. 3cos(a) - 5cos(a) = cos(a) * (3 - 5) = -2cos(a)

Simplifying the equations, we have:

1. cos(a) = cos(a)
2. -2cos(a) = cos(a)

For equation 1, cos(a) can take any value since both sides are equal.

For equation 2, we can divide by cos(a) if it is not equal to zero:

-2cos(a) = cos(a) [Dividing by cos(a)]

-2 = 1

Since -2 is not equal to 1, this equation has no valid solution.

Therefore, the equation 7cos(a) - 6sin(a) is valid for all values of cos(a), and the equation 3sin(a) - 5cos(a) has no valid solution.