Sin ^ 2x + cos ^ 2x = 1 why and how?

Sin ^ 2x + cos ^ 2x = 1 why and how?

Place the acute angle a in the plane rectangular coordinate system, its vertex coincides with the origin o, its starting edge coincides with the positive half axis of X axis, and its terminal edge is always in the first quadrant. Take any point P (x, y) on the terminal edge of angle a, and the distance from P to the origin r = root sign (x ^ 2 + y ^ 2) > 0, and make a vertical line of X axis through P. if the perpendicular foot is m, the length of segment OM is x, and the length of segment MP is y
Then Sina = MP / op = Y / R,
cosa=OM/OP=x/r
So sin ^ 2 A + cos ^ 2 a = (x ^ 2 + y ^ 2) / R ^ 2 = 1
What is the minimum value of (sin ^ 2x + 1 / sin ^ 2x) (COS ^ 2x + 1 / cos ^ 2x)
Let t = (SiNx) ^ 2, then the original formula = (T + 1 / T) [1-T + 1 / (1-T)] = T-T ^ 2 + T / (1-T) + (1-T) / T + 1 / (T-T ^ 2) = T-T ^ 2-2 + 1 / (1-T) + 1 / T + 1 / (T-T ^ 2), denoted as f (T), f '(T) = 1-2T + 1 / (1-T) ^ 2-1 / T ^ 2 - (1-2T) / (T-T ^ 2) ^ 2, Let f' (T) = 0, get (1-2T) (T-T ^ 2) ^ 2 + T ^ 2 - (1-T) ^ 2 +
How to change sin (2x + 60 °) into cos (2x + 60 °)
It's coordinate system translation
The coordinate system shifts π / 4 units to the right
Let f (x) = cos3x2cosx2-sin3x2sinx2. (I) find the minimum positive period of function f (x); (II) find the zero point of function f (x) when x ∈ [π 2, π]
(I) f (x) = cos3x2cosx2-sin3x2sinx2 = cos (3x2 + x2) = cos2x, (4 points) ∵ ω = 2, ∵ t = 2 π 2 = π, then the minimum positive period of function f (x) is π; (5 points) (II) Let f (x) = 0, that is, cos2x = 0, and ∵ x ∈ [π 2, π], (7 points) ∵ 2x ∈ [π, 2 π], (9 points) ∵ 2x = 3 π 2, that is, x = 3 π 4, then x = 3 π 4 is the zero point of function f (x). (12 points)
If f (COS x) = cos 3x, then f (SiN x)=
If f (COS x) = cos 3x, then f (SiN x) = has a process
sin x=cos(π/2-x)
f(sin x)= f(cos(π/2-x))= cos3(π/2-x)=sin 3x
Try to find the monotone decreasing interval of the function y = 1 / 2 * cos (π x + π / 3) - sin (π x + 5 π / 6)
If y = 1 / 2 * cos (π x + π / 3) - sin [π - (π x + 5 π / 6)] = 1 / 2 * cos (π x + π / 3) - sin (- π x + π / 6) = 1 / 2 * cos (π x + π / 3) - cos [π / 2 - (- π x + π / 6)] = 1 / 2 * cos (π x + π / 3) - cos (π x + π / 3) = - 1 / 2 * cos (π x + π / 3) y decreases, then cos increases and cosx increases in the range of (2k - π, 2K
For definite integral, the integrand is x * sin (x) / (2 + cos (x)), and the integral interval is 0 to PI. How to solve it?
Because the integral interval is 0 to π, we can use trigonometric function to determine the property of integral: ∫ x f (SiNx) DX = pi / 2 ∫ f (SiNx) DX, the integral interval is 0 to π, sin (x) / (2 + cos (x)) can be regarded as f (SiNx)
∫ X*sinx/（2+cosx)dx
=π/2 ∫ sin(x)/（2+cosx)dx
=π/2 ∫ (2+cosx)^(-1) d(-cosx)
= - π/2 ∫ (2+cosx)^(-1) d(2+cosx)
=-The integral interval of π / 2 * ln (2 + cosx) x is [0, π]
=(π * ln3)/2
Welcome to ask!
Find the monotone increasing interval of function y = − cos (x2 − π 3)
The monotone decreasing interval of y = cos (X2 - π 3) is the monotone increasing interval of y = - cos (X2 - π 3). From 2K π ≤ X2 - π 3 ≤ 2K π + π (K ∈ z), the monotone increasing interval of y = - cos (X2 - π 3) is 2 π 3 + 4K π ≤ x ≤ 8 π 3 + 4K π (K ∈ z), and the monotone increasing interval of y = - cos (X2 - π 3) is [2 π 3 + 4K π, 8 π 3 + 4K] (K ∈ z)
What is the increasing range of the function y = sin (x / 2) + cos (x / 2) in (- 2 π, 2 π)?
y=√2*sin(x/2+π/4)
-π/2+2kπ≤x/2+π/4≤π/2+2kπ
So the increasing interval is ∪ [- 3 π / 2 + 4K π, π / 2 + 4K π],
Where k ∈ Z
What is the increasing interval of the function y = sin (x / 2) + cos (x / 2) on (- 2 faction, 2 faction)?
WHY?
y=sin(x/2)+cos(x/2)
=(radical 2) * sin (x / 2 + pi / 4). (introducing auxiliary angle, PI is PI)
So:
The increasing interval on (- 2pi, 2pi) is:
Satisfy - pi / 2 + 2 * k * pi