It is proved that Tan (3a / 2) - Tan (A / 2) = [2Sin (A / 2)] / [cos (3a / 2)]

It is proved that Tan (3a / 2) - Tan (A / 2) = [2Sin (A / 2)] / [cos (3a / 2)]

tan(3a/2)-tan(a/2)
=sin(3a/2)/cos(3a/2)-sin(a/2)/cos(a/2)
=[sin(3a/2)cos(a/2)-sin(a/2)cos(3a/2)]/[cos(3a/2)*cosa]
=sina/[cos(3a/2)*cos(a/2)]
=2sin(a/2)cos(a/2)/[cos(3a/2)*cos(a/2)]
=【2sin（a/2）】/【cos（3a/2）】
Get the certificate

It is proved that 2Sin β / cos α + cos β = Tan (α + β) / 2-tan (α - β) / 2

Right = sin (α + β) / 2 / cos (α + β) / 2-sin (α - β) / 2 / cos (α - β) / 2
=[sin(α+β)/2*cos(α-β)/2-sin(α-β)/2*cos(α+β)/2]/[cos(α-β)/2*cos(α+β)/2]
=2Sin β / (COS α + cos β) (difference angle formula, product sum difference)

To prove the fourth power of sin α - cos the fourth power α = the second power - 1 of 2Sin, we need to solve the problem concretely

sin^4a-cos^4a=2sin²a-1
It is proved that the left formula = (sin? A-cos? A) (sin? 2A + cos? A)
=sin²a-cos²a
=sin²a-(1-sin²a)
=2sin²a-1
Right formula = 2Sin? A-1
So left = right
Conclusion

It is proved that the quartic power of sin a-cos is the square A-1 of 2Sin

Because sin? A + cos? A = 1
So cos? A = 1-sin? A
So the left = (sin? 2A + cos? A) (sin? 2a-cos? A)
=1*[sin²a-(1-sin²a)]
=2Sin? A-1 = right
Proof of proposition

It is proved that sin 4 power a + cos square a + sin square ACOS square a = 1

Sin ^ 4A + cos + sinacosa = (sin ^ 4A + sinacosa) + cosa = Sina (Sina + COSA) + cosa = Sina + cosa = 1!

If f (x) = cos Λ 2 (x / 4) - 2Sin Λ 2 (x / 4) + 3 √ 3sin (x / 4) * cos (x / 4) x ∈ (2 / 3 faction, 3 / 4 faction), find the maximum value of F (x)

F (x) = cos (x / 2) - Sin ^ 2 (x / 4) + 3 radical 3 / 2Sin (x / 2)
=Cos (x / 2) - (1-cos (x / 2)) / 2 + 3 radical 3 / 2sinx / 2
=3 / 2cosx / 2 + 3 radical 3 / 2sinx / 2-1 / 2
=3sin(x/2+Pai/6)-1/2
2/3Pai=

f(x)=2sin(x/4)*cos(x/4)-2√3sin²(x/4)+√3 (1) Finding the minimum positive period (2) finding the maximum value and the set of X when obtaining the maximum value (3) the symmetric central axis of symmetry

In this paper, we give the following results: (x) 2Sin (x / 4) * cos (x / 4) - 2Sin (x / 4) - 2 √ 3sin (x / 4) + 3 = SiNx / 2 / 2 - √ 3 (1-cosx / 2) + 3 = 3 = SiNx / 2-2 - √ 3 + 3cosx / 2 + 3 = SiNx / 2 + 3 + 3 + cosx / 2 / 2 = 2 (SiNx / 2cos π / 3 + cosx / 2 / 2Sin π / 3) = 2Sin (x / 2 + π / 3) the minimum positive period is 2 π * 2 = 4 π when x x is x / 2 = 4 π when x x x x = 4 π * 2 = 4 π, when x x x x x x x is x, x = 4 π, when x/ 2 + π / 3 = (2k + 1 / 2)

The period of the function f (x) = cos squared X-1 / 2

F (x) = cos squared X-1 / 2 = (2cos ^ 2x-1) / 2 = (cos2x) / 2
T=2π/2=π

F (x) = the square of COS + X / 4 + 1 / 2sinx / 2 + 1 / 2 minimum positive period, e.g,

F (x) = the square of COS + X / 4 + 1 / 2sinx / 2 + 1 / 2?
Can you explain the title clearly? F (x) = cos ^ 2x / 4 + 1 / 2sinx / 2 + 1 / 2?
If so, the answer is this
F (x) = cos ^ 2x / 4 + 1 / 2sinx / 2 + 1 / 2 = (1 + cosx / 2) / 2 + 1 / 2sinx / 2 + 1 / 2 = radical 2 / 2Sin (x / 2 + pi / 4) + 1
Then the minimum positive period = 2pi / w = 2pi / 1 / 2 = 4Pi
Finish the explanation

How to find the period of y = (COS x) square? Not just results! Or commentary!

The original image is up and down the x-axis. After squared, all parts less than 0 are symmetrical to the top with respect to the x-axis. The period is half of the original, π