How to convert 1 + cos square x-sin square x into 2cos square x

How to convert 1 + cos square x-sin square x into 2cos square x

1 + cos²x - sin²x
= sin²x + cos²x + cos²x - sin²x
= 2cos²x
Because sin? X + cos? X = 1

Simplify f (x) = 2cos (x / 2) · (sin (x / 2) + cos (x / 2)) - 1 (1) Simplify f (x); (2) Find the value of F (π / 12)

（1）f(x)=2cos(x/2)·(sin(x/2)+cos(x/2))-1
=2cos(x/2)·sin(x/2)+2cos^2(x/2)-1
=SiNx + cosx (double angle formula)
=√2sin(x+π/4)
（2）f(π/12)=√2sinπ/3=√6/2

Simplification: 1 + sin (X-2 π) cos (X-7 π / 2) - Tan (π - x) Tan (3 π / 2-x) - 2cos ＾ 2 (- x)

The original formula = 1 + (- SiNx) * cos (x + π / 2) - (- TaNx) * Tan (π / 2-x) - 2cos? X = 1 + sin? X + TaNx * cotx-2cos? X
= sin²x+2(1-cos²x)=3sin²x；

Simplification: sin (540 degrees - x) / Tan (900 degrees - x) * 1 / Tan (450 degrees - x) Tan (810 degrees - x) * cos (360 degrees - x) / sin (- x)

I can't understand the operation of the title, but the following simplification should help you, just take the right seat ~ sin (540 degrees - x) = sin (180 degrees - x) = sinxtan (900 degrees - x) = Tan (180 degrees - x) = - TaNx = - SiNx / cosxtan (450 degrees - x) = Tan (90 degrees - x) = cosx / sinxtan (810 degrees - x) = Tan (9

It is proved that Cos2 β = 2cos2 α = 2cos 2 (π + θ)

Cos 2 β = 1-2 sin 2 β = 1-2 sin θ cos θ = 1-sin 2 θ because 4 sin 2 α = 1 + 2 sin θ cos θ = 1 + sin 2 θ, sin 2 θ = 4 sin 2 α - 1, so Cos2 β = 1 - (4 sin 2 α - 1) = 2 (1-2 sin 2 α) = 2 Cos2 α

Simplify (sin α) ^ 2 * (sin β) ^ 2 + (COS α) ^ 2 (COS β) ^ 2-1 / 2cos2 α Cos2 β

Cos2 α Cos2 β = (COS? α - sin? α) (COS? β - sin? β) = cos? α cos? β - cos? α sin? α cos? β + sin? α sin? α, so the original formula = (2cos? α cos? β + 2Sin? α)

How to get the formula Cos2 α = cos ^ 2 (α) - Sin ^ 2 (α) = 2cos ^ 2 (α) - 1 = 1-2sin ^ 2 (α)

The formula cos (x + y) = cosxcosy sinxsinycos2a = cos (a + a) = Cosa × cosa Sina × Sina = cos? A-SiN? A = cos? A - (1-cos? A) = 2cos? 2a-1 = (2-2sin? A) - 1 = 1-2sin? A

Given the vector a = (sin2x, root 3). B = (1, - cos2x), X belongs to R, 1, if a is perpendicular to B and 0

（1）
A is perpendicular to B
1 * sin2x radical 3 * cos2x = 0
1 / 2 * sin2x - radical 3 / 2 * cos2x = 0
cos(π/3)sin2x-sin(π/3)cos2x=0
sin(2x-π/3)=0
Zero

Given the vector a = (1, sin2x) B = (cos2x, 1), X ∈ R, f (x) = a · B, if f (A / 2) = 3 radical 2 / 5, and a ∈ (π / 2, π), try to find the value of sina

F (x) = (radical 2) * sin (2x + π / 4)
F (A / 2) = (radical 2) * sin (a + π / 4) = 3 radical 2 / 5
sin（a+π/4）=3/5
Sina + cosa = 3 roots 2 / 5
(Sina) square + (COSA) square = 1

Cos2x / (1 + sin2x) = 1 / 5 to find TaNx

Cos2x / (1 + sin2x) = 1 / 5 (COS ^ 2x-sin ^ 2x) / (SiNx + cosx) ^ 2 = 1 / 5 (COS ^ 2x-sin ^ 2x) / (SiNx + cosx) ^ 2 = 1 / 5 (1-tan ^ 2x) / (TaNx + 1) ^ 2 = 1 / 55 (1-tan ^ 2x) = Tan ^ 2x + 2tanx + 13tan ^ 2x + 13tan ^ 2x + tanx-2 = 0 (3tanx-2) (TaNx + 1) = 0tanx = 2 / 3 or TaNx = -1.when the TaNx = 180k-45, 2x 2x = 180k-45, 2x = 180k-45, 2x = 180k-45, 2x = 180k-45, 2x = 1.tanx = - 1.when