It is known that the square root 6 * 1 / 2-m of inequality f (x) = 3 * root 2 * SiNx / 4 * cosx / 4 + root 6 * (cosx / 4) is less than or equal to 0, For any - 5 π divided by 6 less than or equal to x less than or equal to π / 6, what is the value range of real number m

It is known that the square root 6 * 1 / 2-m of inequality f (x) = 3 * root 2 * SiNx / 4 * cosx / 4 + root 6 * (cosx / 4) is less than or equal to 0, For any - 5 π divided by 6 less than or equal to x less than or equal to π / 6, what is the value range of real number m

F (x) = 3 * root 2 * SiNx / 4 * cosx / 4 + square of root 6 * (cosx / 4) - root 6 * 1 / 2-m is less than or equal to 0
∴(3√2/2)sin(x/2)+(√6/2)[1+cos(x/2)]-√6/2≤m
∴(3√2/2)sin(x/2)+(√6/2)cos(x/2)≤m
∴√6【(√3/2)sin(x/2)+(1/2)cos(x/2)】≤m
∴√6sin[(x/2)+π/6]≤m
∵-5π/6≤x≤π/6 ∴-π/4≤[(x/2)+π/6]≤π/4
∴√6sin[(x/2)+π/6]≤√6sin(π/4)=√3
∴m≥√3

It is known that f (x) = 2 (SiNx + cosx) + 3 under the root sign of sin2x-2, X belongs to [- 3 π / 4, π / 4] (1) If f (x) = 8 / 9, find the value of sin2x

F (x) = (SiNx + cosx) ^ 2-1-2, 2 (SiNx + cosx) + 3 under the root, let (SiNx + cosx) = t, then f (x) = T ^ 2-2, root 2T + 2 = (T-root 2) ^ 2 = 8 / 9, so t = 7, root 2 / 4 = under the root (SiNx + cosx), so (SiNx + cosx) ^ 2 = 2401 / 64 = 1 + sin2x, so sin2x = 2337 / 64

Given TaNx = root 2, find the value of 2cos2 (x / 2) - sinx-1 / SiNx + cosx

Known TaNx = root 2
Then [2cos2 (x / 2) - sinx-1] / (SiNx + cosx)
=(cosx-sinx)/(sinx+cosx)
=(1-tanx)/(tanx+1)
=(1-√2)/(1+√2)
=(1-√2) ²/ [(1+√2)(1-√2)]
=(3-2√2)/(1-2)
=2√2-3

Given that x belongs to (- π / 2,0), SiNx + cosx = - radical 2 / 31, find the value of COS (x + π / 4). 2. Find (cos2x) / [TaNx + 1 / TaNx]

SiNx + cosx = √ 2Sin (x + π / 4) = - radical 2 / 31, and then the cosine value cosx is obtained through the sine angle formula. Therefore, the fraction can be obtained. Below Tan can be expressed by sin / cos, and then the denominator is multiplied by the common fraction. The sum of sin square + cos square and a cosxsinx sin square + cos square is equal to 1. Therefore, the whole formula becomes cos2x

It is known that 3sin X / 2-cosx / 2 is equal to 0 (1) find the value of TaNx (2) find the value of cos2x / radical 2cos (Pie / 4 + x) SiNx

3sinx/2-cosx/2=0
TaNx / 2 = 1 / 3
tanx=2tanx/2/(1-tan^2x/2)=(2/3)/(1-1/9)=6/(9-1)=3/4
Cos2x / radical 2cos (PAI / 4 + x) SiNx
=Cos2x / root 2 (cosx * root 2 / 2-sinx * root 2 / 2) SiNx
=cos2x/(cosxsinx-sin^2 x)
=(cos^2x-sin^2x)/(cosxsinx-sin^2x)
=(1-tan^2x)/(tanx-tan^2x)
=(1-9/16)/(3/4-9/16)
=(16-9)/(12-9)
=7/3

Find the period of y = cos2x + sin2x and y = radical 3cosx + SiNx

y=cos2x+sin2x
=√2(√2/2cos2x+√2/2sin2x)
=√2(sinπ/4cos2x+cosπ/4sin2x)
=√2sin(2x+π/4)
Period: 2 π / 2 = π
y=√3cosx+sinx
=2(√3/2cosx+1/2sinx)
=2*(sin(π/3)cosx+cos(π/3)sinx
=2*sin(x+π/3)
Period: 2 π