Simplification [2sin50 ° + sin80 ° (1 + √ 3tan10 °)] / cos5 °

Simplification [2sin50 ° + sin80 ° (1 + √ 3tan10 °)] / cos5 °

[2sin50°+sin80°(1+√3tan10°)]/cos5° =[2sin50°+cos10°(cos10°+√3sin10°)/cos10°]/cos5° =[2sin50°+2(1/2*cos10°+√3/2*sin10°)]/cos5°=[2sin50°+2(cos60°*cos10°+sin60°*sin10°)]/cos5°=[2si...

Under root sign (x-1) (xsquare-1) (- 1)

0

0

Let a = x / 3
y=sin(x/3)*cos(2x/3)+cos(x/3)*sin(2x/3)
=sina*cos(2a)+cosa*sin(2a)
=sina(2a+a)
=sina(3a)
=sinx
Then the function is symmetric about x = 2K π + π / 2 (K ∈ z)

The maximum value of the function f (x) = (sin2x-cos2x) ^ 2 Why should 4sina be equal to - 1?

f(x)=(sin2x)^2-2sin2xcos2x+(cos2x)^2
=1-sin4x
Because - 1 < = sin4x < = 1
So when sin4x = - 1, the maximum value of F (x) is 1 + 1 = 2

The sum of the maximum and minimum values of the function y = cosx - (1 / 2) cos2x (x ∈ R) is

y=cosx-(1/2)cos2x
=cosx-(2cos^2x-1)/2
=cosx-cos^2x+1/2
=-(cosx-1/2)^2+1/2+1/4
=-(cosx-1/2)^2+3/4
Cosx = 1 / 2 max = 3 / 4
Cosx = - 1 min = - 3 / 2
And = 3 / 4-3 / 2 = - 3 / 4

Given that 1 + cosx siny + sinxsiny = 0,1-cosx-cosy + sinxcosy + 0, find SiNx

From the known two formulas, it can be obtained that:
1+cosx=siny(1-sinx)——（1）
1-cosx=cosy(1-sinx)——（2）
From the sum of squares of the above two formulas:
(1) The square of (2) is obtained
2+2(cosx)^2=(1-sinx)^2
Let z = SiNx
Then (cosx) ^ 2 = 1 - (SiNx) ^ 2 = 1-z ^ 2
therefore
2+2（1-z^2）=(1-z)^2
z=(1±√10）/3
Then (1) + (2)
(siny+cosy)*(1-sinx)=2
Because of 1-sinx ≥ 0
So siny + cosy > 0
Because sin y + cosy = √ 2 (sin (y + π / 4)) ≤ 2
So 1-sinx ≥ √ 2
SiNx

Given vector M = (SiNx, 1), vector n = (√ 3cosx, 1 / 2), function f (x) = (M + n) · M. (1) find the minimum positive period T and monotone increasing interval of F (x) (2) Given that a, B, C are the opposite sides of the inner angle a, B and C of △ ABC, a is an acute angle, a = 2 √ 3, C = 4, and f (a) is the maximum value of the function f (x) on [0, π / 2], we find the area s of △ ABC

f(x)
= (m+n).n
= (sinx+√3cosx)sinx + (3/2)
= (sinx)^2 + √3/2sin2x + 3/2
= (1-cos2x)/2+ √3/2sin2x + 3/2
=-cos2x/2 + √3/2sin2x + 2
= sin(2x-30°) + 2
Minimum positive period = 90 degrees
Monotone increasing interval: 1
360°k -90°

If x equals root 2, what is the root 2 of x minus 3x plus 3 times

Is 2 (Radix 2) ^ 2-3 (Radix 2) + 3 (Radix 2) = 2

The solution set of inequality radical 2 (X-2) < radical 3x

√2x-2√2

If sin (α + β) = 1 / 4 and sin (α - β) = 1 / 3, then Tan α ∶ Tan β is equal to

Expand them separately, add and subtract to get - 7
sinacosb+cosasinb=1/4 --(1)
sinacosb-cosasinb=1/3 --(2)
(1)+(2):2 sinacosb=7/12 -(3)
(1)-(2):2 sinacosb=-1/12 -(4)
(3)/(4):tana/tanb=-7