Calculate the 2012 power of 2-2, the 2011 power of 2 and the 2010 power of 2 -... - 2's 2-1 speed point

Calculate the 2012 power of 2-2, the 2011 power of 2 and the 2010 power of 2 -... - 2's 2-1 speed point

2 ^ 2012-2 ^ 2011 -... = (2-1) 2 ^ 2011 - 2 ^ 2010 -... = (2-1) 2 ^ 2010 - 2 ^ 2009 -... =... = 2 ^ 2-2-1 = (2-1) 2-1 = 2-1 = 1 or in binary, the first is 100.. 000 (2012 zeros) - (100.00 (2011 zeros) +...) = 100.. 000 (2012 zeros) - 11111... 111 (

How to calculate the minus 2 power of 2?

The negative 2nd power of 2 = 0.25, 2 born is 0 power, plus 2 power is negative, so it can be understood that adding 0,1.2 power is the third power, and the result is that the third power of 2 divided by 2 is equal to 0.25

Calculate (1 / 3) ° + √ 12cos30 ° - (1 / 2) negative first power

A:
(1 / 3) ° + √ 12cos30 ° - (1 / 2) negative first power
=1+√12*(√3/2)-(1/2)^(-1)
=1+2*√3*(√3/2)-1/(1/2)
=1+3-2
=2

It is faster to simplify the square (- a) - Tan (360 degree + a) / sin (- a) of COS It is {cos square (- a)} - {Tan (360 degrees + a) / sin (- a)} This is the second small question in question 7 on page 28 of the senior one mathematics compulsory four a edition

[cos²(-a)]-[tan(360º+a)/sin(-a)]=cos²a-1/cosa=(cos³a-1)/cosa

Simplification (1), sin (α + 180 degrees) cos (- α) sin (- α - 180 degrees) (2), SIN3 (- α) cos (2 π + α) Tan (- α - π)

Multiplication or simplification of separation
I'll teach you how to solve it all
For example, sin (α + 180 degrees), a = 30 A + 180 in the third quadrant, then sin is negative in the third quadrant,

Simplification: Tan α * (COS α - sin α) + [sin α (sin α + Tan α) / 1 + cos α] /It's a division sign (it should be a fractional line, but it can't be typed) * it's a multiplication There must be a solution process, thank you!

Is it Tan α (COS α - sin α) + [sin α (sin α + Tan α) / (1 + cos α)]?
Let's look at sin α (sin α + Tan α) / (1 + cos α), and the molecule is sin α (sin α cos α / cos α + sin α / cos α)
(sin α) ^ 2 (1 + cos α) / cos α. After reduction with denominator, (sin α) ^ 2 / cos α = Tan α sin α
The original formula is tan α (COS α - sin α) + Tan α sin α = Tan α cos α = sin α

Simplification (sin θ - cos θ) / (Tan θ - 1) A.tanθ B.sinθ C.-sinθ D.cosθ

Choose D
(sinθ-cosθ)/(tanθ-1)
=(sinθ-cosθ)/(sinθ/cosθ-1)
=(sinθ-cosθ)/[(sinθ-cosθ)/cosθ]
=1/(1/cosθ)
=cosθ

Simplification of sin (540 ° + a) * cos (- a) / Tan (a-180 °)

sin(540°+a）*cos(-a）/tan(a-180°）
= -sina*cosa/tana
=-sina*cosa*cosa/sina
=-cos^2a

If sin α * cos α > 0, and sin α * Tan α > 0, simplify: Cos (α / 2) * {under radical: [1-sin (α / 2)] / [1 + sin (α / 2)]} + cos (α / 2) * {under radical: [1 + sin (α / 2)] / [1-sin (α / 2)]}under radical sign

From sin α * cos α > 0
It can be concluded that a is located in the first and third quadrants
From sin α * Tan α > 0
We can know that sina * Sina / cosa > 0
It can be concluded that cosa > 0
So a can only be the angle of the first quadrant
That is 2K π

Simplification: cos quadratic (- @) - Tan (360 degree + @) / sin (- @)

The original formula = cos ^ 2 (α) + Tan α / sin α = cos ^ 2 (α) + 1 / cos α