How much is half plus one sixth plus one twelfth plus one twentieth plus one third plus one forty second,

How much is half plus one sixth plus one twelfth plus one twentieth plus one third plus one forty second,

=1/1×2+1/2×3+1/3×4+1/4×5+1/5×6+1/6×7
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
=1-1/7
=6/7
six-sevenths
Split term method: I want method, not result
2cos(α+β)cos(α-β)=cos2α+cos2β
cos2α=cos[(α+β)+(α-β)]=cos(α-β)cos(α+β)-sin(α-β)sin(α+β)
cos2β=cos[(α+β)-(α-β)]=cos(α-β)cos(α+β)+sin(α-β)sin(α+β)
The left is equal to the right
This is a formula in itself. If you want to prove it, you can prove it by yourself
Expand the left side with the sum difference angle formula. Just now it was a square difference formula
After expansion, we get a formula about sin α's Square, cos α's Square, sin β's Square and cos β's Square. Then we reduce all the squares here to power, and raise the angle to twice the angle.
The secret of trigonometric function lies in the movement of angle and function name!! ... unfold
This is a formula in itself. If you want to prove it, you can prove it by yourself
Expand the left side with the sum difference angle formula. Just now it was a square difference formula
After expansion, we get a formula about sin α's Square, cos α's Square, sin β's Square and cos β's Square. Then we reduce all the squares here to power, and raise the angle to twice the angle.
The secret of trigonometric function lies in the movement of angle and function name!! Put it away
Sum difference product
Because cos α + cos β = 2cos [(α + β) / 2] cos [(α - β) / 2]
So Cos2 α + Cos2 β = 2cos [(2 α + 2 β) / 2] cos [(2 α - 2 β) / 2]
=2cos(α+β)cos(α-β）
Using the integral sum difference formula cos α cos β = [cos (α + β) + cos (α - β)] / 2
2cos(α+β)cos(α-β)=2[cos(α+β+α-β)+cos(α+β-α+β)]/2=cos2α+cos2β
Two out of seven is the sum of three different fractions?
Mathematics 2cos θ Cos2 θ - cos θ =?
It is said that the answer seems to be 2cos θ Cos2 θ - cos θ = cos3 θ
Push backwards
cos3θ
=cos(θ+2θ)
=cosθcos2θ-sinθsin2θ
=cosθcos2θ-sinθ（2sinθcosθ）
=cosθcos2θ-（2sinθsinθ）cosθ
=cosθcos2θ-（1-cos2θ）cosθ
=cosθcos2θ-cosθ+cosθcos2θ
=2cosθcos2θ-cosθ
2cos0 * 2cos0-cos0 = 4cos0 * cos0-1 * cos0 = 3cos0 * cos0 = 3 times the square of cos0
(I'm ashamed that the Greek alphabet can't be typed.)
2cosAcos2A=cos(A+2A)+cos(A-2A)=cos3A+cosA,
So the answer is cos3a
This solution is within the requirements of high school syllabus. Because high school does not require triple angle formula.
2cosθcos2θ-cosθ=cosθ（2cos2θ-1）
=cosθ（4(cosθ)^2-3)
=4(cosθ)^3-3cosθ
=cos3θ
Two out of seven equals a fraction plus a fraction plus a fraction
Solution method, I confused a lot of numbers, or not, to the method!
The denominator is not the same!
I got it!
2/7=1/7-1/8+1/8+1/7
=（1/7-1/8）+1/8+1/7
=（1/7*1/8）+1/8+1/7
=1/56+1/8+1/7
I see~
Consider 7 / 2 as 16 / 56, 1 / 8 = 7 / 56, 16-7 = 9, 9 / 56, 1 / 7 = 8 / 56, 1 / 56, so 2 / 7 = 1 / 7 + 1 / 8 + 1 / 56
How to simplify 1 + 2cos ^ α - Cos2 α?
1+2cos^α-cos2α
=1+2cos^α-（2cos^α-1）
=2
1+2cos^α-cos2α=1+2cos^α-2cos^α+1=2
1+2cos^α-cos2α=1+2cos^α-cos^α+sin^α
=1+cos^α+sin^α
=1+1=2
The sum of two thirds and two thirds of a number is equal to 7
(7-2 / 3) △ 2 / 3
=(7-2 / 3) × 3 / 2
=3 / 7 × 2-2 / 3 × 3 / 2
=21-1 / 2
=19 / 2
21 / 4: Congratulations!
Wrong answer!
It is proved that 1 + 2cos ^ 2 Θ - Cos2 Θ = 2
Left = 1 + 2cos & sup2; Θ - (2cos & sup2; Θ - 1)
=1+2cos²Θ-2cos²Θ+2
=2
=Right
Proof of proposition
What is the sum of seven threes
Twenty-one
Evaluation: cos π / 7 * Cos2 π / 7 * cos3 π / 7
cosπ/7*cos2π/7*cos3π/7
=-cosπ/7*cos2π/7*cos4π/7
=-sin(π/7)*cos(π/7)*cos(2π/7)*cos(4π/7)/sin(π/7)
=-(1/2)sin(2π/7)*cos(2π/7)*cos(4π/7)/sin(π/7)
=-(1/4)sin(4π/7)*cos(4π/7)/sin(π/7)
=-(1/8)cos(8π/7)/sin(π/7)
Because sin (8 π / 7) = sin (π + π / 7) = - sin (π / 7)
=1/8