How many times is the power of - 9999 to the power of - 9999

How many times is the power of - 9999 to the power of - 9999

〖0.25〗^(100 )x4^101=〖0.25〗^(100 )x4^100x4=(〖0.25x4)〗^(100 )x4=4
〖(-1)〗^9999x5x99=-1x5x99=-495
Given the function f (x) = 2Sin ^ 2x + 2sinxcosx, X ∈ [0,2 π], find the set of X that makes f (x) positive
Given the function f (x) = 2Sin ^ 2x + 2sinxcosx, X ∈ [0,2 π], find the set of X that makes f (x) positive
f(x)=2sin^2 x+2sinxcosx
=1-cos2x+sin2x
=sin2x-cos2x-1
=√2sin(2x-45)-1
F (x) > 0, that is sin (2x-45) > 2 / 2
That is 45 ＜ 2x-45 ＜ 135 to get π / 4 ＜ x ＜ π / 2
0 to a quarter of a π
sin100°cos 110° +cos100°sin110°
sin100°cos 110° +cos100°sin110°
=sin(100°+110°）
=sin210°
=sin（180°+30°）
=-sin30°
= - 1/2
Simplification: (2cos10 ° - sin20 °) / sin70 ° is
The original is = (2cos (30-20) ° - sin20 °) / sin70 ° = lonely
(2012 the second mock exam of Pudong New Area) the known function f (x) =2sinxcosx+2cos2x. (1) asks for the monotonous increasing interval of function f (x); (2) after the function y=f (x) image is shifted to 4 units of the right, we get the image of function y=g (x) and find the solution of equation g (x).
(1) The function f (x) of the function f (x) is the function f (x) = 2 SiNx cosx x + 2cos2x = 2x + cos2x + 1 = 2 sin (2x + π 4) + 1, and by 2K π - π (2k π - π 2 ≤ 2x + π 4 ≤ 2K π + π 4 ≤ 2K + π 2 (K ∈ z) we get: K π - 3 π 8 ≤ 3 π 8 ≤ x ≤ K π + π + π 8 (K ∈ z), then the monotone increasing interval of F (x) is [K π π - 3 π 3 π 8, π 8, K π 8, K π + π + 8] (K \\\\\\\\\\\\\\sin (2x - π 4) = 0,  2x- If π 4 = k π (K ∈ z), then x = k π 2 + π 8 (K ∈ z)
How to find (cos80 - cos20) / (sin80 + sin20)?
Sum difference product
Cos80 - cos20 = - 2sin50 ° sin30 ° = - sin50 °
Sin80 degree + sin20 degree = 2sin50 ° cos30 degree = √ 3sin50 degree
(COS 80-cos 20) / (sin 80 + sin 20)
=-sin50°/√3sin50°
=-√3/3
The value of 2cos10 °− sin20 ° sin70 ° is ()
A. 12B. 32C. 3D. 2
The original formula is: 2cos (30 °− 20 °) - sin20 ° sin70 ° = 2 (cos30 °· cos20 ° + sin30 °· sin20 °) - sin20 ° sin70 ° = 3cos20 ° cos20 ° = 3
The known function f (x) = 2sinxcosx-2cos & # 178; X + 1
1. If f (θ) = 3 / 5, find the value of Cos2 (π / 4-2 θ)
Function f (x) = 2sinxcosx-2cos & # 178; X + 1,1. If f (θ) = 3 / 5, find the value of Cos2 (π / 4-2 θ): F (x) = 2sinxcosx-2cos & # 178; X + 1 = sin2x-cos2x = √ 2Sin (2x - π / 4) ∵ f (θ) = √ 2Sin (2 θ - π / 4) = 3 / 5 = = = > sin (2 θ - π / 4) = 3 √ 2 / 10 ∵ sin (π / 4-2 θ) ∵
f(x)=2sinxcosx-2cos²x+1
=sin2x - cos2x
=Radical 2 * sin (2x - π / 4)
If f (θ) = 3 / 5, then there is:
π / 2 / 3
That is: sin (2 θ - π / 4) = 3 (radical 2) / 10
So:
cos²（π/4-2θ）
=(2) unfold
f(x)=2sinxcosx-2cos²x+1
=sin2x - cos2x
=Radical 2 * sin (2x - π / 4)
If f (θ) = 3 / 5, then there is:
Radical 2 * sin (2 θ - π / 4) = 3 / 5
That is: sin (2 θ - π / 4) = 3 (radical 2) / 10
So:
cos²（π/4-2θ）
=cos²（2θ - π/4）
=1- sin²（2θ - π/4）
=1 - [3 (radical 2) / 10] &;
=1 - 18/100
=41 / 50 "put away
F (x) = sin2x cos2x = radical 2Sin (2x - π / 4)
And f (θ) = 3 / 5, that is, the root sign 2Sin (2 θ - π / 4) = 3 / 5
Sin (2 θ - π / 4) = 3 radical 2 / 10
Sin (- 2 θ + π / 4) = - 3 radical 2 / 10
Cos2 (π / 4-2 θ) = 2cos square (π / 4-2cos2 θ) - 1 = 2 (1-sin square (π / 4-2 θ)) - 1 = 19 / 25
If it is right, please accept it.
First, we simplify the function and get FX = root sign 2Sin (2x - π / 4)
The formula is expanded by double angle formula, that is, Cos2 (π / 4-2 θ) = 1-2sin ^ 2 (π / 4-2 θ)
This is the way of thinking. Do you understand
How to change (sin20 ° - sin80 °) / (cos20 ° - cos80 °)
=I'm not very smart
molecule
=sin(50-30)-sin(50+30)
=sin50cos30-cos50sin30-sin50cos30-cos50sin30
=-2cos50sin30
Denominator = cos (50-30) - cos (50 + 30)
=cos50cos30+sin50sin30-cos50cos30+sin50sin30
=2sin50sin30
Approximate points
The original formula = - cos50 / sin50 = - cot50
Help me
The original formula = (2cos10 ° - sin20 °) / sin70 °
=(2cos10°-sin20°)/cos20°
=[cos10°+(cos10°-cos70°)]/cos20°
=[cos10°+2sin40°*sin30°]/cos20°
=[cos10°+2*1/2*sin40°]/cos20°
=[cos10°+cos50°]/cos20 °
=2cos30°*cos20°/cos20°
=2cos30°
=Root 3