Sin (a + 2K) = Sina, what are cosa and Tana?

Sin (a + 2K) = Sina, what are cosa and Tana?

cosa=cos(a+2kpai) tana=tan(a+kpai)
tan（∏+a）=tana
cos（-a）=cosa
sin（∏/2±a）=cosa
Given sin (π + a) = 1 / 2, find the value of Tana cosa
Sin (π + a) = 1 / 2, = = > Sina = - 1 / 2, so angle a is the third or fourth quadrant angle,
1) When a is the third quadrant angle, cosa = - √ 3 / 2, Tana = √ 3 / 3
tana-cosa=√3/3+√3/2=5√3/6
2) When the fourth angle is 3 - / Tana = 3
tana-cosa=-√3/3-√3/2=5√3/6= - 5√3/6
Given cosa = 4 / 5, a belongs to (0, π / 2), find the value of Tana and sin (π - a)
Because a ∈ (0, π / 2), then Sina > 0, cosa > 0, Tana > 0 and: Sin & # 178; a + cos & # 178; a = 1 and cosa = 4 / 5, then: Sin & # 178; a = 1-cos & # 178; a = 9 / 25, since Sina > 0, then: Sina = 3 / 5, Tana = Sina / cosa = 3 / 4sin (π - a) = Sina = 3 / 5
tana=3/4 sin(pi-a)=-3/5
Cosa = 4 / 5, a belongs to (0, π / 2)
sina=3/5
tana=3/4
sin(π-a)
=sina
=3/5
sina>0
(sina)^2+(cosa)^2=1
(sina)^2=9/25
sina=3/5
tans=sina/cosa=(3/5)/(4/5)=3/4
sin(π-a)=sina=3/5
A square plus the square of B = (a + b) (a-b), then how much is a cube plus B cube
First of all, I'll correct a small mistake for you. Your expression should be the square of the square difference formula a minus the square of B, that is: A ^ 2-B ^ 2 = (a + b) (a-b). You asked the sum of cubes formula: A ^ 3 + B ^ 3 = (a + b) (a ^ 2-AB + B ^ 2). By the way, I'll tell you the difference formula: A ^ 3-B ^ 3 = (a-b) (a ^ 2 + AB + B ^ 2)
（a+b)(a^2-ab+b^2)
Find the minimum positive period of F (x) = (SiNx cosx) SiNx, X ∈ R
solution
y=sin²x-sinxcosx
=1/2-1/2cos2x-1/2sin2x
=1/2-√2/2[cos2xcos45°+sin2xsin45°]
=1/2-√2/2cos[2x-45°]
∴ T=2π÷2=π
Minus 2 times the absolute value of a minus 1 plus a plus B plus 1 under the root sign, plus the square of C-2 under the root sign equals zero?
It is - 2 √ (A-1) + A + │ B + 1 │ + √ (C-2 ^ 2) = 0,
-2√(a-1)+a+│b+1│+√(c-2^2)=0,
Obviously, a ≥ 1, C ≥ 4
│b+1│+√(c-2^2)=2√(a-1)-a≥0,
2 √ (A-1) ≥ a, square of both sides
（a-2）^2≤0,a=2.
The original formula can be reduced to: B + 1 + √ (C-2 ^ 2) = 0
From the definition of nonnegative number, B = - 1, C = 4
So a + B ^ 2 + C ^ 3 = 2 + 1 + 64 = 67
Simplify the period f (x) = (SiNx + sin3x) / (cosx + cos3x)
The result is Tan 2x, but the period is π. The answer is correct
I know how to simplify, but I don't know why the period is π, not π 2
F (x) = (SiNx + sin3x) / (cosx + cos3x) = {2Sin [(x + 3x) / 2] cos [(x-3x) / 2]} / {2cos [(x + 3x) / 2] cos [(x-3x) / 2]} = sin (2x) / cos (2x) = Tan (2x) period is π
According to the cube of a + cube of B = (a + b) (a square - AB + b square), we prove that 1000001 is a composite number
ten million and one
=100^3+1^3
=(100+1)(10000-100+1)
=101*9901
Because apart from 1 and itself, 1000001 has at least two factors 101 and 9901
So 1000001 is a sum
one million and one
=1000000+1
=100³+1
=(100+1)×(100²-100+1)
=101×9901
So 1000001 is a sum
Because 1000001 = 1000000 + 1 = 100 cube + 1 cube = (100 + 1) * (10000-100 + 1) = 101 * 9901, 1000001 is a composite number
From the natural number sequence cubic sum formula 8 ^ 8 + 8 ^ 8 + 8 ^ 8 + +N ^ 8 = [n ^ 8 * (n + 8) ^ 8] / 8: 8 ^ 8 + 8 ^ 8 + 8 ^ 8 + +8118 ^ 8 = 8 / 8 * 8118 ^ 8 * first look at 8111118 / 818 = 9918, there are eight in the middle, 1 81111111118 / 818 = 99119918, and there are nine in the middle, 1 811111111111118 / 818 = 99119911
：1=√3·cosx+sinx
I'm sorry I forgot to say it's for X
3 is the root 3
This problem is done with the formula of unity
1=√3·cosx+sinx
x=2kπ + π/6 - π/3 = 2kπ - π/6
Given that a > 0, b > 0, it is proved that: (a + b) × (cube of a + cube of B) ≥ (square of a + square of B) ≥ (square of a + square of B)