First simplify and then evaluate: the square of A-1 / A + 2A + 1 minus A-1 / 1, where a = root 3 + 1 fast

First simplify and then evaluate: the square of A-1 / A + 2A + 1 minus A-1 / 1, where a = root 3 + 1 fast

solution
(a²+2a+1)/(a²-1)-a/(a-1)
=(a+1)²/(a-1)(a+1)-a/(a-1)
=(a+1)/(a-1)-a/(a-1)
=(a+1-a)/(a-1)
=1/(a-1)
=1/(√3+1-1)
=1/√3
=√3/3

Given cos (π / 6 + a) = radical 3 / 3, find the value of COS (7 π / 6 + a)

cos(7π/6+a)
=cos[π+（π/6+a)]
=-cos(π/6+a)
=- radical 3 / 3

Given that cos (π / 6 - α) = root 3 / 3, the value of COS (5 π / 6 + α) - Sin 2 (- α + 7 π / 6) is obtained,

Cos (π / 6 - α) = radical 3 / 3
cos(5π/6+α)-sin²(-α+7π/6)
=cos[π-(π/6-a)]-sin²[π+(π/6-a)]
=-cos(π/6-a)-sin²(π/6-a)
=-√3/3-[1-cos²(π/6-a)]
=-√3/3-(1-1/3)
=-√3/3-2/3

30 ° is known

Root sign (COS β - cos α) 2 - | cos β - radical 3 / 2 | + | 1-cos α|=____ 1-radical 3 / 2
30°

Given that a is equal to the root sign π, what is sin? A + cos? A?

Regardless of the value of a: sin? A + cos? A = 1

Cos a? = root 25 / 58 find a sin B? = root 9 / 58 find B (cos a)² (cos b)²

Cosa = root 5 / root 28, a = arccos root (5 / root 28), the one below is similar to ha

In mathematics of senior one, sin (α + π / 2) = - radical 5 / 5, α ∈ (0, π) is known. The value of COS 2 (π / 4 + α / 2) - cos 2 (π / 4 - α / 2) / sin (π - α) + cos (3 π + α) is obtained

sin(α+π/2)
=cosα
=-√5/5
∴α∈(π/2,π)
sinα=2√5/5
[cos²(π/4+α/2)-cos²（π/4-α/2）]/[sin（π-α）+cos（3π+α）]
=[cos(π/4+α/2)+cos(π/4-α/2)][cos(π/4+α/2)-cos(π/4-α/2)]/(sinα-cosα)
=√2cos(α/2)*(-√2sin(α/2))/(sinα-cosα)
=-2sin(α/2)cos(α/2)/(sinα-cosα)
=-sinα/(sinα-cosα)
=-2/3

Find the definition field. Root sign 2sinx / root sign (1 + cos square x-sin square x)

1+cos²x-sin²x>0
2cos²x>0
cos²x>0
Namely
cos²x≠0
x≠kπ+π/2 ,k∈Z
The domain is defined as {x ∈ R | x ≠ K π + π / 2, K ∈ Z}

How to resolve sin square x cos square x + double root three SiNx? Just dissolve it into a form

2√3sinx-cos2x

The known function f (x) = Radix 3sin X / 4 cos X / 4 + cos squared X / 4 + 1 / 2 The known function f (x) = root sign 3sin X / 4 cos X / 4 + cos squared X / 4 + 1 / 2. (1) find the monotone increasing interval of F (x). (2) in the acute triangle ABC, the opposite sides of a, B, C are a, B, C respectively, and satisfy (2a-c) CoSb = bcosc, find the value range of F (2a)

1 / 2 = 3 / 2sinx / 2 + 1 / 2cosx / 2 + 1 / 2cosx / 2 + 1 = sin (π / 6 + X / 2) + 1 increase interval (4K π - 4 π / 3,4k π + 2 π / 3), K belongs to z2sinacosos, B = sin (b + C) = Sina, B = π / 3f (2a) = sin (π / 6 + 6 + a) + 1 minimum value = 1 / 2 + 1 = 3 / 2, maximum value = 3 / 2, maximum value = 1 / 2 + 1 = 3 / 2, maximum value = 1 / 2 + 1 = 3 / 2, maximum value = 1 / 2 + 1 = 3 / 2, maximum value = 1 / 2 + 1 = 3 / 2, maximum value = 1 / 2 + 1 = 3 / 2 = 3 / 2, maximum 1 + 1 = 2