Given cosx = - 3 / 4 and X is the angle of the third quadrant, find the values of sin2x, cos2x and tan2x

Given cosx = - 3 / 4 and X is the angle of the third quadrant, find the values of sin2x, cos2x and tan2x

solution
cosx=-3/4
X is the angle of the third quadrant
∴sinx=-√7/4
∴sin2x=2sinxcosx=2 × (-√7/4) × (-3/4)=3√7/8
cos2x=2cos ² x-1=2 × (-3/4) ²- 1=-1/8
tan2x=sin2x/cos2x=-3√7

(1+sin2x-cos2x)/(1+sin2x+cosx)

(1+sin2x-cos2x)/(1+sin2x+cosx)
=(2sin ² x+2sinxcosx)/(2cos ² x+2sinxcosx)
=sinx(sinx+cosx)/[cosx(cosx+sinx)]
=tanx

(2005 • Anhui) when 0 ＜ x ＜ π 2, function f (x) = 1 + cos2x + 8sin2x The minimum value of sin2x is () A. 2 B. 2 three C. 4 D. 4 three

f(x)＝2cos2x+8sin2x
2sinxcosx＝4sin2x+cos2x
sinxcosx=4tan2x+1
tanx＝4tanx+1
tanx．
∵0＜x＜π
2，
∴tanx＞0．
∴4tanx+1
tanx≥2
4tanx•1
tanx＝4．
When TaNx = 1
2, f (x) min = 4
Therefore, C

When 0 < x < π / 2, the minimum value of function FX = (1 + cos2x + 8sin Λ 2x) / sin2x is

Can you provide methods

What is the derivative of F (x) = x cubic - 3 x square - 9x + 5? It's due today

F '(x) = 3x ^ 2-6x-9, which is derived directly. The formula used is that the derivative of x ^ n is NX ^ (n-1)

How to find the second derivative of y = x times e to the negative x power

y'=e^(-x)-xe^(-x)
So y '' = - e ^ (- x) - [e ^ (- x) - Xe ^ (- x)]
=(x-2)e^(-x)