2Sin ^ (A / 2) = 1-cos a ^ is the square of the formula Forget to ask online Can't you understand the first step?

2Sin ^ (A / 2) = 1-cos a ^ is the square of the formula Forget to ask online Can't you understand the first step?

It can be deduced from sum angle formula
cosa=cos(a/2+a/2)=cosa/2cosa/2-sina/2sina/2
=cos^a/2-sin^a/2=1-2sin^a/2
So 2Sin ^ (A / 2) = 1-cosa

If f (x) = SiNx cos ^ 2 x, then the range of F (x)

f(x)=sinx-[1-sin^2(x)]
=sin^2(x)+sinx-1
Let t = SiNx, then t ∈ [- 1,1]
The original function is as follows:
Y = T ^ 2 + T-1, parabola opening upward, symmetry axis: T = - 1 / 2,
The function y (T) first decreases and then increases on [- 1,1], and the decreasing interval is short and the increasing interval is long,
Therefore, the maximum value is:
y(max)=y(1)=1
y(min)=y(-1/2)= -5/4
So the range of the original function is: [- 5 / 4,1]

If cos (x + π 4) = 45, X ∈ (− π 4, 0), then SiNx=______ ．

∵ x ∈ (− π 4, 0) ∵ 0 ＜ x + π 4 ＜ π 4 ∵ sin (x + π 4) = 1 − 1625 = 35 ∵ SiNx = sin (x + π 4 - π 4) = sin (x + π 4) cos π 4-cos (x + π 4) sin π 4 = - 210, so the answer is: - 210

Given cos (π / 4 + x) = 4 / 5, X ∈ (- π / 2, - π / 4), find the value of SiNx

2/5=cosπ/4cosx-sinπ/4sinx=(cosx-sinx)/√2
That is, cosx SiNx = 2 √ 2 / 5
Because x ∈ (- π / 2, - π / 4), so cosx > 0, SiNx

Given SiNx = 4 / 5, and X ∈ (π / 2, π), find the value of COS (2x-60 °)

∵ given SiNx = 4 / 5, and X ∈ (π / 2, π)
∴cosx=-3/5
cos2x=2cos²x-1
=2·9/25-1
=-7/25
sin2x=2sinxcosx
=2·4/5(-3/5)
=-24/25
cos(2x-60º)
=cos2xcos60º+sin2xsin60º
=-7/25·1/2-24/25·√3/2
=-7/50-12√3/25

SEC & # 178; α + 1 = what are those formulas

secα =1/cosα
sec² α-1=sin² α/cos² α=tan² α

McLaughlin formula of e ^ x
Such as the title

1/N!*x^n

Let LG X be a function of (x-1). Maclaurin formula

Using the indirect expansion method, lgx = LNX / ln10
For the expansion of LNX, we can obtain the derivative first and then the integral
(lnx)′=1/x=1/(1+x-1))=1-(x-1)+(x-1)²-（x-1）³+.
lnx=x-(1/2)(x-1)²+(1/3)(x-1)³+.
lgx=lnx/ln10=1/ln10 {x-(1/2)(x-1)²+(1/3)(x-1)³+.}

The expansion term of Maclaurin Formula 1 / (1 + x)

1/(1+x) = 1 - x + x² - x³ + .+ (-1)^n * x^n + o(x^n)

Why is the McLaughlin expansion of e ^ x right?
The one with e ^ x = 1 + X + x ^ 2 / 2! +

Because McLaughlin expansion is derived from Taylor formula at x = 0,
All derivatives of e ^ X are e ^ X,
So f '(0), f' '(0) and so on are all 1,
So the McLaughlin expansion of e ^ x is e ^ x = 1 + X + x ^ 2 / 2! +