The value range of y = / root sign 3sinx cosx / - 4

The value range of y = / root sign 3sinx cosx / - 4

Y = | radical 3sinx-cosx | 4
=2|sinxcosπ/6-cosxsinπ/6|-4
=2|sin(x-π/6)|-4
So:
When | sin (x - π / 6) | = 1, y has a maximum value = - 2
When | sin (x - π / 6) | = 0, y has a minimum value = - 4
Therefore, the range of function is y ∈ [- 4, - 2]

What is the value range of y = (√ 3sinx) / cosx + 2? It is 3 under the root sign

There is a universal formula and let Tan (x / 2) = tmolecule = V3 * 2T / (1 + Tan ^ 2t) = 2v3t / (1 + T ^ 2) denominator = 2 - (1-tan ^ 2t) / (1 + Tan ^ 2t) = 2 - (1-T ^ 2) / (1 + T ^ 2) y = 2v3t / (1 + 3T ^ 2) (t belongs to R) (*) when t = 0, y = 0 (a) when t is not equal to 0, * the numerator and denominator are divided by T, y = 2v3 / (1 / T + 3T)

The value range of the function y = radical 3sinx + cosx, X ∈ [- 6 parts π, 6 parts π] is y=2(√3/2sinx+1/2cosx) =2(sinxcosπ/6+cosxsinπ/6) How did you come back from this step

Because cos π / 6 = √ 3 / 2, sin π / 6 = 1 / 2
Write √ 3 / 2 as cos π / 6 and 1 / 2 as sin π / 6
It can be concluded that y = 2 (√ 3 / 2sinx + 1 / 2cosx)
=2(sinxcosπ/6+cosxsinπ/6)

The function y = radical 3sinx + cosx can be reduced to - with a maximum value of - and a minimum value of——

y=√3sinx+cosx
=2(√3/2sinx+1/2cosx)
=2(sinxcosπ/6+cosxsinπ/6)
=2sin(x+π/6)
The maximum value is: 2; the minimum value is: - 2

If the radical 3sinx + cosx-y = 0 and X belongs to [0, π], find the monotone interval of function y I convert it to y = 2Sin (x + π / 6), and then according to the monotone increasing interval of sin [2K π - / 2,2k π + / 2], I get that x belongs to [- 2 / 3 π, 1 / 3 π], and because x belongs to [0, π], but the answer is wrong. What's wrong?

What's your answer? You get that x belongs to [- 2 / 3 π, 1 / 3 π], which is monotonically increasing, and the definition domain of this function is
X belongs to [0, π], which is the intersection of these two sets. Isn't it increasing at (0, π / 3)? Similarly, we can find that the monotone decreasing interval is
(π / 3, π), don't you figure it out? Isn't that the answer?

If we know that the root of [cosin + S] is an interval, then it is a function of π

f(x)=2sin(x+π/6)
Its increasing range is 2K π - π / 2=

The known function f (x) = 2 3sinx•cosx+cos2x-1(x∈R)． (1) Find the monotone increasing interval of function y = f (x); (2) If x ∈ [- 5 π 12，π 3] Find the value range of F (x)

The solution (1)  f (x) = 23sinx · cosx + cos2x-1,  f (x) = 2Sin (2x + π 6) - 1. (2 points) the solution 2K π - π 2 ≤ 2x + π 6 ≤ 2K π + π 2, K ∈ Z. the monotonic increasing interval of the function y = f (x) is [K π - π 3, K π + π 6], K ∈ Z

Reduction of F (x) = 2cosx (radical 3sinx cosx) + 1

f(x)=2√3sinxcosx-2cos²x+1
=√3sin2x-(2cos²x-1)
=√3sin2x-cos2x
=2sin(2x-π/6)

Simplify / sin ^ 2 70 ° given - SiNx = 3 / 2cosx. Find the value of 2cos ^ 2x - sin2x

The formula is too long to play, so I'll divide it into two parts
sin50°=cos40°
cos40°(1+√3tan10°)=cos40°*2*（1/2+√3/2tan10°）
=2cos40°*（sin30°cos10°+cos30°sin10°）/cos10°
=2cos40°sin40°/cos10°
=sin80°/cos10°=1
Sin ^ 2 70 ° = (cos20 °) ^ 2 = (cos40 ° + 1) / 2 (angle doubling formula)
Then the original formula = 2
∵ -sinx=3/2cosx
∴ sin² x=9/4cos² x ∵sin² x+cos² x =1
The results show that cos? X = 4 / 13 sin? X = 9 / 13
Original formula = 2cos? X - sin2x
=2cos² x -2sinxcosx
=2cos² x+2*3/2cosxcosx
=5cos² x=20/13
Oh, finally, I've finished writing. If you don't understand, you can bring it up

Prove SiNx (COS ^ 2 2x sin ^ 2 2x) + 2cosx cos2x sin2x = sin 5x

sinx(cos^2 2x-sin^2 2x) + 2cosx cos2x sin2x
=sinxcos4x+cosxsin4x
=sin(x+4x)
=sin5x