For any real number x, y ∈ R, SiNx + cosy = 2Sin (X-Y / 2 π / 4) cos (x y / 2 - π / 4), then sin13 π / 24. Cos5 π / 24 = =? Which formula is used in this problem? I don't want it. I need the formula

For any real number x, y ∈ R, SiNx + cosy = 2Sin (X-Y / 2 π / 4) cos (x y / 2 - π / 4), then sin13 π / 24. Cos5 π / 24 = =? Which formula is used in this problem? I don't want it. I need the formula

This problem uses the sum difference product formula of trigonometric function: the original formula is: sin α + sin β = 2Sin [(α + β) / 2] * cos [(α - β) / 2]
SiNx + cosy = SiNx + sin (π / 2 - y) regards x as α in the formula and π / 2 - y as β in the formula
There is SiNx + cosy = SiNx + sin (π / 2 - y) = 2Sin [(X-Y) / 2 + π / 4] cos [(x + y) / 2 - π / 4]
Do you understand?

The length of an arc is 8.37dm, and the center angle of the circle is 60 degrees

Six arc lengths are a whole circle
6 arc length = circumference = 50.22dm
Perimeter = diameter * 3.14
Diameter = 50.22 / 3.14 = 15.99 = 16
The radius is 8 DM

5.21-mathematics 3 / 10. Is there a pair of real numbers (x, y) such that SiNx = cosy and arcsinx = arccosy hold simultaneously? Why?

No, it is explained by image method
The first quadrant is a straight line connecting (0, π / 2), (π / 2,0),
The image of the latter is the unit circle, and the two do not coincide

It is known that the radius of the circle where an arc is located is 8 cm, and the central angle of the circle is 60 °, then the arc length is_______ .

8×2×3.14×60°/360°≈8.37cm

If real numbers x and y satisfy | X-1 | + (x + y) & sup2; = 0, then the value of XY is equal to

x=1 x+y=0 y=-1
xy=-1

It is known that the arc with a length of 50cm is 200 ° and the radius of the circle where the arc is located (accurate to 1cm) can be calculated. Thank you for the process

The arc length of n ° center angle is L = n ° π R △ 180 °
The radius of the circle where this arc is located is
180×l÷（n°π）
=180×50÷（200×3.14）
=9000÷628
≈14cm

Known: x 2 + XY + y = 14, y 2 + XY + x = 28, find the value of X + y

∵ x2 + XY + y = 14 (1), Y2 + XY + x = 28 (2), ∵ ① + 2, we get: x2 + 2XY + Y2 + X + y = 42, ∵ (x + y) 2 + (x + y) - 42 = 0, ∵ (x + y + 7) (x + y-6) = 0, ∵ x + y + 7 = 0 or x + y-6 = 0, the solution is: x + y = - 7 or x + y = 6

How much m / S is 1kn / h?

One meter in one second
3.6km in one hour
3.6km/h
That's what our teacher explained

Let sinxy + e ^ xy = ax, find dy / DX

Sinxy + e ^ xy = ax is derived from X twice
(y+xdy/dx)cosxy+(y+xdy/dx)e^xy=a
(y+xdy/dx)(cosxy+e^xy)=a
From this we get
dy/dx=a/[(cosxy+e^xy)x]-y/x

1m/s=______ km/h．

1m / S = 1 × 11000km13600h = 3.6km/h