The third power of the integral sign cos times x times DX

The third power of the integral sign cos times x times DX

∫cos^3 x dx
=∫cosxcos^2xdx
=∫cosx(1-sin^2)dx
=∫cosxdx-∫cosxsin^2xdx
=-sinx-1/2∫sin2xsinxdx
=-sinx-1/2∫(-1/2(cos(3x)-cosx)dx
=-sinx+1/4∫cos3xdx-1/4∫cosxdx
=-sinx+1/4*1/3∫cos3xd3x+1/4sinx
=-3/4sinx-1/12sin3x+c

To prove: sin α + sin β = 2Sin (α + β) / 2 * cos (α - β) / 2

sina-sinb=sin(a/2+a/2+b/2-b/2)-sin(a/2-a/2+b/2+b/2)
=sin[(a+b)/2+(a-b)/2]-sin[(a+b)/2-(a-b)/2]
=sin(a+b)/2cos(a-b)/2+cos(a+b)/2sin(a-b)/2-[sin(a+b)/2cos(a-b)/2-cos(a+b)/2sin(a-b)/2]
=2cos(a+b)/2sin(a-b)/2

Can 2 * 2 matrix (COS a, - sin a; sin a, cos a) ^ n = (Cosn a, - Sinn a; Sinn a, Cosn a) be proved by Euler formula
How to prove (COS a, - sin a; sin a, cos a) ^ n = (COS Na, - Sin Na; sin Na, cos Na) by using e ^ ina = cos Na + I * sin Na

In my opinion, the most convenient way to prove this conclusion is to use mathematical induction
If you must use Euler formula, you can use the following observation:
The second order orthogonal matrix with determinant 1 can always be expressed as [cos (θ), - sin (θ); sin (θ), cos (θ)], denoted as s (θ)
It is very easy to prove, only when the columns of orthogonal matrix constitute a group of standard orthogonal bases, and the determinant is 1
The characteristic polynomial of matrix s (θ) is X & # 178; - 2cos (θ) x + 1 = 0, and the eigenvalues are e ^ (I θ) and e ^ (- I θ), corresponding to eigenvectors (1, - I) 'and (1, I)' respectively
So for the invertible matrix T = [1,1; - I, I], T ^ (- 1) · s (θ) t = [e ^ (I θ), 0; 0, e ^ (- I θ)] (for any θ)
Then T ^ (- 1) · s (θ) ^ n · t = [e ^ (I θ), 0; 0, e ^ (- I θ)] ^ n = [e ^ (in θ), 0; 0, e ^ (- in θ)] = T ^ (- 1) · s (n θ) t
From t invertibility, we obtain s (θ) ^ n = s (n θ), which is the proof
Another view is to define the exponential function exp (x) = ∑ {0 ≤ K} x ^ k / K
It can be proved that the series converges to any square matrix and has the following properties
If x and y are square matrices of order n and satisfy xy = YX, then exp (x) exp (y) = exp (x + y)
As a corollary, exp (x) ^ n = exp (NX)
Consider the matrix J = [0, - 1,1,0], it is easy to verify J & # 178; = - E, so J ^ (2k) = - 1) ^ k · e, J ^ (2k + 1) = - 1) ^ k · J
So exp (θ J) = ∑ {0 ≤ K} (- 1) ^ k · θ ^ (2k) / (2k)! · e + ∑ {0 ≤ K} (- 1) ^ k · θ ^ (2k + 1) / (2k + 1)! · J
= (∑{0 ≤ k} (-1)^k·θ^(2k)/(2k)!)·E+(∑{0 ≤ k} (-1)^k·θ^(2k+1)/(2k+1)!)·J
= cos(θ)E+sin(θ)J
= S(θ).
So s (θ) ^ n = exp (θ J) ^ n = exp (n θ J) = s (n θ), which is the proof
Finally, a little more, in fact, the complex number a + bi can correspond to the second order real matrix [a, - B; B, a], which can be verified that this correspondence preserves the operation (algebraic homomorphism)
The matrix corresponding to the complex e ^ (I θ) is s (θ), and e ^ (in θ) = (e ^ (I θ)) ^ n corresponds to s (n θ)
From this correspondence preserving operation, s (n θ) = s (θ) ^ n

It is known that sinm + Sinn = 1 / 4, COSM + Cosn = 1 / 2?
In Chinese. I don't understand that
It's cos, not cos

I'm angry. I don't want to speak any more. But I'm very angry when I see so many wrong solutions above. Let's talk about it: first of all, the correct solution is 3 / 5. For those figures with other solutions, I can't help asking if sinm ^ 2 + Sinn ^ 2 can be equal to 1!? why? Even if one is wrong, there will be endless mistakes at the bottom They say

Given sin power θ + cos power θ = 1, find the value of sin power θ + cos power θ (n belongs to n)

SIN3 power + cos3 power = 1 = SIN3 power + cos power
Sin cubic + cos cubic sin cos square = 0
SIN-1 + COS-1 = 0
And because sin is greater than or equal to 0, SIN-1 and COS-1 are less than or equal to 0
So when sin = 1, cos = 0 or sin = 0, cos = 1
So the answer is 1

Which quadrant are CoS / Tan / sin 1,2,3,4 in? How to judge?

Sin is in quadrant one and two +, cos is in quadrant one and four +, Tan is in quadrant one and three+

If f (x) = {- cos π x, X ＞ 0, f (x + 1) + 1, X ＜ = 0, then the value of F (4 / 3) + F (- 4 / 3) is equal to?

f(4/3)+f(-4/3)
f(-4/3)=f(-4/3+1)+1
=f(-1/3)+1
= f(-1/3+1)+1
=f(2/3)+1
=-cos[π*(2/3)] +1
=1/2+1=3/2
f(4/3) =-cos[π*(4/3)]=1/2,
Then f (4 / 3) + F (- 4 / 3) = 3 / 2 + 1 / 2 = 2

Let X be less than 0, f (x) = [a (1-cos x)] / x ^ 2, and X be greater than or equal to 0
f(x)=x^2+bx+1
Finding the value of a and B
It is continuous on positive and negative infinity and differentiable everywhere

f(0)=1
Because cosx = [cos (x / 2)] ^ 2 - [sin (x / 2)] ^ 2
So x

Half angle formula of sine

The sine, cosine and tangent formula of double angle sin2 α = 2Sin α cos α Cos2 α = Cos2 α - sin2 α = 2cos2 α - 1 = 1-2sin2 α
2tanα
tan2α＝—————
1－tan2α
Can it solve your problem?

Who invented calculus, Newton or Leibniz?

Newton and Leibniz invented them respectively
Leibniz invented calculus from 1673 to 1676, and published his paper in 1684; Newton invented calculus from 1665 to 1666, and published it in the great work "mathematical principles of natural philosophy" in 1687. Who invented calculus is a public case in the history of world science