It is proved that arctan radical (x ^ 2-1) + arcsin1 / x = π / 2 when x ≥ 1

It is proved that arctan radical (x ^ 2-1) + arcsin1 / x = π / 2 when x ≥ 1

Let arctan radical (x ^ 2-1) = a
arcsin1/x=b
Then Sina = radical (x ^ 2-1) / X cosa = 1 / X
SINB = 1 / X CoSb = radical (x ^ 2-1) / X
Sin (a + b) = sinacosb + cosasinb = radical (x ^ 2-1) / X * radical (x ^ 2-1) / x + 1 / X * 1 / x = 1 = sin π / 2
(a+b)=π/2

X / (1 + x) + arctan root sign x-radical sign 2-radical x derivation under y = xarcsin radical sign

In this paper, we give the following derivation: dy / DX = arcsin, and get the derivation: dy / DX = arcsin √ [x / (1 + x)] + X {[x / (1 + x)]] + X {[x / (1 + x)]]} ′ / / [1-x / (1 + x)] + [(x - / (2 + (2)]) / (1 + X - / (1 + x)] + [(x - / (1 + x)] / (1 + X - / (1 + x)] / (1 + X-1 + x)] / (x [x (1 + x)] / (x [x / (1 + x)] / 2 √ [x / (1 + x)] / (x / (1 + x)] / (1 + 1 + x)] / (1 + x)] / (1 + X / (1 + x√ [1 / (1 + x)] + {1 / [2 √ (x

TaNx = root 2 / 2, find the angle X, why did I find the inverse trigonometric function? And, which is bigger, 5 / 8 π or π - arccos (root 6 / 2)?

There is no special angle such that TaNx = root 2 / 2, sin (π / 4) or cos (π / 4) is equal to root 2 / 2
5 / 8 π < π - arccos (Radix 6 / 2)
Therefore, arccos (Radix 6 / 2) < arccos (Radix 3 / 2) = π / 6 π - arccos (Radix 6 / 2) > 5 / 6 π

What is the root of arccos-3

125.264 degrees

4 / 5-5 + 1 / 6 (5 / 3 + 3 / 5) × 20 49% of root number 2 + root number 108 - root number 12

4 / 5-5 + 5 + 6 = 6 = - 3 / 5, 5 + 1 / 6, 6 = - 3 / 5, 5 + 1 / 6, 6 = - 3 / 5, 5 + 1, 6 (5 / 3 + 3) × 20 = 100 / 3 + 12 = 10 / 3, 3 + 3 = 16 / 3, 49, 108, 12, 12, 3, 3 = 7 / 5, 2, 3 = 7 / 5, 5, 5, 6, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3

How to express 5 / 4 of the root on the number axis?

Draw a rectangle with length of 2 and width of 1, draw its two diagonals, mark the intersection point of diagonal lines as O, and mark the corner of the lower left corner of the rectangle as a, and draw the bottom edge on the number axis, where a corresponds to the origin,
Draw a circle with point a as the center, ao as the radius, and the intersection axis as point B. then the length of AB is 5 / 4 of the root

(4 times the root 3 divided by 5 times the 2 / 5 times the root 2 and then multiplied by 1 / 4 times the root 3) and then added (6 minus 2 times the root 2)

(4 root number 3 ÷ 5 root number 2 / 2 × root number 3 / 4) + (6-2 root number 2)
=(8 root number 3 / (5 root number 2) × root number 3 / 4 + 6-2 root number 2
=3 root number 2 / 5 + 6-2 root number 2
=6-7 roots 2 / 5
=(30-7 root number 2) / 5

Calculate Tan [arccos (- 2 / 2) - TT / 6] Hope to provide the process. Thank you

If the cosine value is positive, it is in the first and fourth quadrants. You are wrong in the quadrant where the angle belongs to ha... Let the angle a = arccos (- 2 / 2) ok, the angle range of the anti cosine function belongs to [0180 degree], then a is equal to 3tt / 4, then Tana = - 1. The formula of tangent sum used is: Tan

Is cos ^ α equal to cos α under radical

√ cos? α must be positive - otherwise, there will be embarrassing situations such as √ 2 being negative? Scientists are shameless
Cos α is not necessarily positive
So we can't say equal
It's like: √ (± 25) 2 = 5, but √ (± 25) 2 ≠ - 5

-What is the root of COS 2 / 2?

-1