Given TaNx = root 7, find the value of (1 -- cos2x + sin2x) / (1 + cos2x + sin2x) (please write the procedure)

Given TaNx = root 7, find the value of (1 -- cos2x + sin2x) / (1 + cos2x + sin2x) (please write the procedure)

(1 -- cos2x + sin2x) / (1 + cos2x + sin2x) = (1 - (1-2sin ^ x) + 2sinxcosx) / (1 + (2cos ^ x-1) + 2sinxcosx) = (2Sin ^ x + 2sinxcosx) / (2cos ^ x + 2sinxcosx) = (SiNx / cosx) [(SiNx + cosx) / (cosx + SiNx)] = SiNx / cosx = TaNx = radical 7

If TaNx = radical 3, then sin2x / 1 + cos ^ x

Let t = sin2x / 1 + cos? X = 2sinxcosx / sin ^ x + 2cos ^ X
1/t=sin^x+2cos^x/2sinxcosx=(sinx/2cosx)+cosx/sinx
=tanx/2+ctgx
Put TaNx = root 3, ctgx = 1 / root 3
Is this simple?

If the period of the function y = 2Sin (2aX + π / 3) (a > 0) is π, then the inclination angle of the line ax + y + C = 0 is?

Using the formula 2 π / 2A = π, we get a = 1, the slope of the line is - a = Tan B, and the angle B is 135 degrees

The minimum positive period of the function y = 2Sin (x / 3 - π / 4) + 1 is

General trigonometric (sine) functions can be written in the form of asin (ω x + φ) + B
Here, ω = 1 / 3, φ = - π / 4, a = 2, B = 1
According to the formula t = 2 π / ω = 6 π
So the period is six PI
That's it!

Find the minimum positive period of the function y = 2Sin (π / 3-x / 2)

Minimum positive period of sine function
Formula
T=2π/|ω|
T=2π/1/2
T=4π

What condition does "a = 1" mean that the minimum positive period of the function y = cos ^ 2aX -- sin ^ 2aX is π "?

Y = cos ^ 2aX -- sin ^ 2aX = cos4ax the minimum positive period is 2 π / (4 | a |) = π / | a|
When a = 1, the minimum positive period is π, so it is a sufficient condition
When the minimum positive period is π, a = 1 or a = - 1, so it is unnecessary
So it's a sufficient and unnecessary condition

What is the minimum positive period of the function y = 2tanx / (1-tanx ^ 2)? The minimum positive period of y = 2tanx / (1-tanx ^ 2) is π / 2! The answer is wrong

Although it can be converted into tan2x, the denominator must not be zero. Secondly, the key to the problem is that TaNx should be meaningful, that is, X ≠ π / 2 + K π. In other words, in the function image, all x = π / 2 + K π are hollow, while tan2x is just zero when x = x = π / 2 + K π

The minimum positive period of y = tanx-1 / TaNx

y=-1/tanx+tanx
=-cosx/sinx+sinx/cosx
=-(cos^2 x-sin^2 x)/(sinxcosx)
=-2cos2x/sin2x
=-2cot2x
T=π/2

The minimum positive period of y = TaNx - (1 / TaNx) is

Y = TaNx - (1 / TaNx) = TaNx Cotx = SiNx / cosx cosx / SiNx = (SiNx * SiNx cosx * cosx) / SiNx * cosx = - 2cos2x / sin2x = - 2cot2x, so the minimum positive period is π / 2

If the period of trigonometric function becomes 2 times, how to change trigonometric function? Take y = sin (AX + b) as an example

The period becomes twice the angular velocity and a becomes the original 1 / 2, i.e. y = sin (A / 2x + b) because the period = 2 pails / angular velocity