The negative quadratic power of 8

The negative quadratic power of 8

The negative quadratic power of 8 is 1 / 64
The negative power is first written as the square of (1 / 8), that is, if the power has a negative sign, it is first written as the reciprocal and then squared

Figure guessing idioms 3.5 (); 5 10 (); 3 / 4 (); 7 / 8 ()
Why is three fourths of the time?

3.5 -- no three no four
5 10 -- all in all
Three quarters -- morning and evening
Seven eighths

1256789 guessing idioms

12345609 scattered
1256789 missing
1 + 2 + 3 in succession
333 555 in groups
3.5 no three no four
9 inches + 1 inch = 1 foot

If Tana = 2, then the square of 2sina - 3sinacosa =?
.

The square of 2sina-3sinacosa = the square of cosa * (the square of 2tana-3tana) = the square of 2cosa, then the square of cosa + the square of sina = 1, the square of cosa = 1 / 5, that is, the original formula = 2 / 5

Given that f (x) = the square X of half cos2x + root sign 3sinxcosx-2cos, find the maximum value of F (x)

F (x) = the square of half cos2x + radical 3sinxcosx-2cos, x = 1 / 2 * cos2x + √ 3 / 2 * (2sinxcosx) - (2cos & # 178; x-1) - 1 = 1 / 2 * cos2x + √ 3 / 2 * sin2x-cos2x-1 = √ 3 / 2 * sin2x-1 / 2 * cos2x-1 = sin (2x - π / 6) - 1 ∫ sin (2x - π / 6) - 1

What is the original function of COS (x) to the fourth power·····

In this paper, we will be able to get the original formula of (cosx) 8747; (cosx) ∫ (cosx) 8747; (cosx) (cosx) ∫ (cosx) (cosx) ^ 4DX = 8747; (1 + cos2x) / 2] ^ 2DX = 1 / 4 ∫ (1 + cos2x) / 2 (2) and [1 + 2cos2x + (cos2x (cos2x) and (cos2x2x) (cos2x) and (cos2x) [1 + cos2x2x2x [1 + cos2x2x2x (cos2x) in [1 + cos2x2x2x2x [1 + cos2x2x2x (cos2x (2) in [1 / 4 [1 + cos2x) 874747474747474747 [1 + cos2x; (cos4x) / 2DX = 3x / 8 +

Why is 1 + cos α equal to 2cos power 2 α
I have the wrong number. Why is 1 + Cos4 α equal to 2cos square 2 α

Yeah, why? You must have the wrong number,
The two are certainly not equal
It should be 2 (COSA / 2) ^ 2

Known 0

[e ^ (x) - e ^ (- x)] &# 178; = [2Sin θ / cos θ] &# 178; e ^ (2x) - 2 + e ^ (- 2x) = [4sin & # 178; θ / cos & # 178; θ] e ^ (2x) + 2 + e ^ (- 2x) = [4sin & # 178; θ / cos & # 178; θ] + 4 [e ^ (x) + e ^ (- x)] &# 178; = 4 / cos & # 178; θ because: 0

Why is ρ = 2cos θ equal to x + y = 2x? Isn't x equal to ρ cos θ?

Yes
X is equal to ρ cos θ
The reduction of this formula ρ = 2cos θ
The first step is to multiply both sides by ρ
We get ρ = 2 ρ cos θ
Because ρ = x + y
2ρcosθ=2X
So ρ = 2cos θ can be reduced to x + y = 2x
The watchtower owner adopts

3/10+(4/11)-(6/5)-(-12/7)

1733 out of 770