In △ ABC, B / C = cosa, then angle A=

In △ ABC, B / C = cosa, then angle A=

From the sine theorem: B / C = SINB / sinc, so, SINB / sinc = cosa, that is: SINB = cosasinc triangle, SINB = sin (a + C) so, sin (a + C) = cosasincsinacosc + cosasinc = cosasincsinacosc = 0sina ≠ 0, so, COSC = 0, we can get: C = 90 ° PS: we can only get the angle C

In the triangle ABC, given a = Π / 3, SINB = 5 / 13, find COSC
Is COSC = cos [Π - (a + b)] = cos (a + b) right?

No
cosC=cos[∏-(A+B)]=-cos(A+B)

In RT △ ABC, ∠ C = 90 °, ab = 13, AC = 5. Find the value of sin ^ A + cos ^ a

Because RT triangle, and angle c = 90 °, ab = 13, AC = 5
From ab ^ 2 = AC ^ 2 + BC ^ 2
BC=12
sinA=BC/AB=12/13
cosA=AC/AB=5/13
So cosa + Sina = 12 / 13 + 5 / 13 = 17 / 13
I hope it can help you
You can keep asking me if you don't know

Find the inverse function of y = ax + B / CX + D, and find out the domain and range of two functions. By comparing the domain and range, what conclusion can you draw?

Y = [C / (AC + b)] x - (ACD + BD) / C to find the inverse function of y = f (x), we only transform the original function into the form of x = f (y), and then let y = x, f (x ') = f (y), then y = f (x') is the inverse function of y = f (x)

Given the function y = X-1 ／ 3x-2, the inverse function is?; the range of the inverse function is?

The definition field of y = X-1 / 3x-2 is x ≠ 2 / 3, and the value field is r
From y = X-1 / 3x-2 solution
　　x=（2y-1）/（3y-1）
If x and y are exchanged, the inverse function of y = X-1 / 3x-2 is
　　y=（2x-1）/（3x-1）（x≠1/3 ）
The domain x ≠ 2 / 3 of y = X-1 / 3x-2 is the domain y ≠ 2 / 3 of the inverse function y = (2x-1) / (3x-1)

How to find the range of inverse function of function g (x) = sin (3x + 1)?

A:
The range of the inverse function is the domain of the original function
The domain of G (x) = sin (3x + 1) is the real number range R
So: the range of its inverse function is real number r

Given that the function f (x) = (2x-m) / (x ^ 2 + 1) is an odd function, find the value of real number M

f(x)=(2x-m)/x²+1
f(-x)=(-2x-m)/x²+1
∵ f (x) is an odd function
∴f(x)+f(-x)=0
∴2x-m-2x-m=0
∴m=0

If a given even function FX is monotone in the interval (0, positive infinity), then f (x-2x-1) = f (x + 1) is satisfied
The sum of all x is

X & # 178; - 2x-1 = x + 1 or X & # 178; - 2x-1 = - X-1
∴x1+x2=3,x3+x4=1
∴x1+x2+x3+x4=4

If the even function FX is monotonically increasing in the interval [0, + infinity], then f (2x-1) is satisfied

It is equivalent to that the absolute value 2x is less than 1 because the original function is even and monotonically increasing in [0, + infinity). According to these, we can also know that the function is monotonically decreasing in [0, - infinity). We can draw a function image satisfying the appeal condition for better understanding

The function f (x) = MX ^ 2 + (2m-1) x + M-1 has at least one zero point on the left side of the origin. Find the value range of real number M
I've seen all of them on the Internet. It's not right,

1. When m = 0, Let f (x) = 0, separate x, make it less than 0, solve the inequality, and then see whether the result includes M = 0; 2. When m is not equal to zero, we know that the quadratic function must have two zeros by analysis. In this case, we only need f (x) > 0 or F (x) < 0, that is, f (x) is not equal to zero, and solve it. Finally, we make union of 1 and 2