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Find the minimum positive period of the function y = sin2x + sin3x It requires a process
The minimum positive period of Y is the least common multiple of the minimum positive period of two factors
In addition, u = sin2x, Tu = π
In addition, v = sin3x, TV = (2 / 3) π
The least common multiple of Tu and TV is 2 π
So the minimum positive period of Y is t = 2 π
The known function f (x) = 2sinxcosx + cos2x (x belongs to R) 1. Find the minimum positive period and maximum value of F (x) 2. If θ is an acute angle and f (θ + π / 8) = the root of 3, find the value of tan2 θ
f(x)=sin2x+cos2x
=√2sin(2x+π/4)
So t = 2 π / 2 = π
Maximum = √ 2
f(θ+π/8)=√2sin(2θ+π/4+π/4)
=√2cos2θ
=√2/3
cos2θ=1/3
When the angle of θ is acute, sin2 θ > 0
sin²2θ+cos²2θ=1
sin2θ=2√2/3
tan2θ=2√2
The known function f (x) = 2sinxcosx + cos2x (x belongs to R) (1) Find the maximum value of the positive period of the sum of (f);
f(x)=sin2x+cos2x
=√2(√2/2*sin2x+√2/2cos2x)
=√2(sin2xcosπ/4+cos2xsinπ/4)
=√2sin(2x+π/4)
So t = 2 π / 2 = π
Maximum = √ 2
The known function f (x) = 2sinxcosx + cos2x (x ≡ R) Find the minimum positive period and maximum value of F (x)
The original formula = sin2x + cos2x = √ 2Sin (2x + π / 4)
So the minimum positive period T = 2 π / 2 = π
The maximum value is √ 2
The function f (x) = 2sinxcosx + cos2x (x) is known (1) Find the minimum positive period and maximum value of F (x); (2) if (the sign with a bar in 0) is an acute angle and f (that sign + Pai / 8) = root 2 / 3, find the value of tan2
f(x)=sin2x+cos2x=(√2)sin(2x+π/4)
Tmin=π,ymax=√2,
F (θ + π / 8) = (√ 2) sin (2 θ + π / 2) = (√ 2) Cos2 θ = 2 / 3, so Cos2 θ = (√ 2) / 3
Thus, Tan 2 θ = √ (7 / 2) = (√ 14) / 2
Given the function f (x) = 2sinxcosx + cos2x, find: a ∈ (0, π), f (A / 2) = √ 2 / 2, find the value of Tana / 2
If (x) = 2 SiNx cosx + cos2x = sin2x + cos2x = sin2x + cos2xf (A / 2) = Sina + cos2xf (A / 2) = Sina + cosa = (√ 2) × sin (a + π / 4) = 1 / 2, so sin (a + π / 4) = 1 / 2, because a ∈ (0, π), so a + π / 4 ∈ (π / 4,5 π / 4), so a + π / 4 = 5 π / 6, that is, a = 7 π / 12, so Tana / 2 = Tan (7 π / 24) = Tan (7 π / 24) = Tan (π / 8 / 8 / 8 / 8 8 8, 8, 8 π / 12, so Tana / 2 = Tan (2 = Tan (7 π / + π / 6) = [Tan (...)
If x is greater than 0 and less than π / 4, then the function f (x) = (1 + cos2x) / (sin2x-2sin ^ 2x) is the minimum value I've simplified it, but I can't find the minimum
So 1 + cos2x = 2cos? X-1, so 1 + cos2x = 2cos? X denominator sin2x-2sinx = 2sinx (cosx SiNx) molecular grave and divide 2cos? X to get f (x) = 1 / [TaNx (1-tanx)] because x is between 0 and π / 4, 2cos? X can be safely divided into two parts
The function y = sin2x-2 (SiNx + cosx) + A ^ 2 is known. Let t = cosx + SiNx, what is the minimum value of function y? If the minimum value of function y is 1, try to find the value of a?
T = radical 2 * sin (PI / 4 + x)
y=sin2x-2(sinx+cosx)+a^2
=1+sin2x-2(sinx+cosx)+a^2-1
=sin^2(x)+cos^2(x)+sin2x-2(sinx+cosx)+a^2-1
=(sin(x)+cos(x))^2+a^2-1
=(Radix 2 * (Radix 2 / 2 * sin (x) + Radix 2 / 2 * cos (x))) ^ 2 + A ^ 2-1
=2*(sin(pi/4+x))^2+a^2-1
=t^2+a^2-1
Because 0
The minimum positive period of the function f (x) = sin2x (2cos ^ 2x-1) is?
f(x)=sin2xcos2x
=(1/2)sin4x
The minimum positive period is:
T=2π/4=π/2
Given the function f (x) = 2cos ^ 2 (x + π / 12) + sin2x2, find the minimum positive period of F (x) and the value range 1 on the interval [0, π / 2], if f (a) = 3 / 2, a ∈ (0, π / 2), find a
f(x)=2cos²(x+π/12）+sin2x
=2[1+cos(2x+π/6)]/2+sin2x
=cos2xcosπ/6-sin2xsinπ/6+sin2x+1
=cos2xcosπ/6-1/2*sin2x+sin2x+1
=cos2xcosπ/6+1/2*sin2x+1
=cos2xcosπ/6+sin2xsinπ/6+1
=cos(2x-π/6)+1
(1) The minimum positive period is t = 2 π / 2 = π
When 0 ≤ x ≤ π / 2
-π/6≤2x-π/6≤π-π/6
-π/6≤2x-π/6≤5π/6
When 2x - π / 6 = 5 π / 6, the minimum value of 1 - √ 3 / 2 is obtained
When 2x - π / 6 = 0, the maximum value of 1 is obtained
So the range is [1 - √ 3 / 2,1]
（2）
a∈(0,π/2)
-π/6≤2a-π/6≤5π/6
f(a)=3/2
cos(2a-π/6)=1/2
2a-π/6=π/3
2a=3π/6
a=π/4
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