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Known function f (x) = (SiNx + cosx) ²+ 2cos ² x 1. Find the maximum value of F (x) and the set of independent variables X at the maximum value 2. Find the monotonically increasing interval of function f (x)
Function f (x) = (SiNx + cosx) ²+ 2cos ² x=1+2sinxcosx+(2cos ² X-1) + 1 = sin2x + cos2x + 2 = root 2Sin (2x + π / 4) + 2 when 2x + Pai / 4 = 2kpai + Pai / 2, that is, there is x = KPAI + Pai / 8, the maximum value is: root 2 + 2 (2) 2K π - π / 2
Given the square · X of function f (x) = (SiNx + cosx) ^ 2-2sin, find the minimum positive period of function f (x)
F (x) = (SiNx + cosx) ^ 2-2sin square · x
=1+2sinxcosx-2sin^2x
=cos2x+sin2x
Therefore, it can be seen from the above: Tmin = 2pai / 2
The minimum positive period of F (x) is Tmin = Pai
Given that x is an acute angle and SiNx + cosx = 7 / 5, the value of root SiNx + root cosx
sin ² x+cos ² x = 1(sinx+cosx) ² = sin ² x+cos ² X + 2sinxcosx = 47 / 25, so 2sinxcosx = 22 / 25sinxcosx = 12 / 25 (you can guess that it is actually 3 / 5 and 4 / 5) (√ SiNx + √ cosx) ² = sinx+cosx +2√(sin...
If SiNx + cosx = root 2, find the value of sinxcosx
(SiNx + cosx) square = 2
So SiNx square + cosx square + 2sinxcosx = 2
Because SiNx square + cosx square = 1
So sinxcosx = 0.5
Given that x is an acute angle, SiNx * cosx = 2 times the root sign 3 / 7, then the value of TaNx
Answer:
X is an acute angle
sinxcosx=2√3/7=(2√3/7)(sin ² x+cos ² x)
Divided by sinxcosx on both sides:
1=(2√3/7)(tanx+ctanx)
tanx+1/tanx=7√3/6
Sorting:
6tan ² x-7√3tanx+6=0
According to the root finding formula of univariate quadratic equation:
tanx=[7√3±√(147-144)]/(2*6)
=(7√3±√3)/12
So:
TaNx = √ 3 / 2 or TaNx = 2 √ 3 / 3
Find the maximum value of F (x) = (1 / 2 + cosx) (root 3 + SiNx), where x is the acute angle Yes: F (x) = (1 / 2 + cosx) ((root 3) / 2 + SiNx)
The original function is transformed into
F (x) = root 3 / 4 + 1 / 2 * SiNx + root 3 / 2 * cosx + sinxcosx
=Radical 3 / 4 + [cos (π / 3) * SiNx + sin (π / 3) cosx] + (2sinxcosx) / 2
=Root 3 / 4 + sin (π / 3 + x) + 1 / 2 * sin2x
Therefore, only when sin (π / 3 + x) = 1, the original function has the maximum value, that is, π / 3 + x = 2K π + π / 2
Because: X is an acute angle, the original function has the maximum value only when x = π / 6
F (x) max = root 3 / 4 + 1 + 1 / 2 * sin (2 * π / 6)
=Root 3 / 4 + 1 + root 3 / 4
=1 + radical 3 / 2
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