Given the function f (x) = the square X of the root 3 cos + sin xconx - the root sign 3 of 2, find the value of F (8 th)

Given the function f (x) = the square X of the root 3 cos + sin xconx - the root sign 3 of 2, find the value of F (8 th)

F (x) = root 3cos ^ x + sinxcosx - radical 3 / 2
=Radical 3 * (1 + cos2x) / 2 + sin2x / 2 - radical 3 / 2
So f (Pie / 8)
=Radical 3 * (1 + cos Pai / 4) / 2 + sin (PAI / 4) / 2-radical 3 / 2
=Root 3 * (1 / 2 + root 2 / 4) + root 2 / 4 - root 3 / 2
=(root 6 + root 2) / 4

Cos (2 α - β) = - 2 / 2, sin (α - 2 β) = 2 / 2, the quarter is less than α less than half, 0 is less than β is less than quarter, find cos (α + β)

∵π/4

How many degrees is cos radical 2 equal to? Is arccos radical 2 written directly? Or how many degrees is it equal to? Cosine

No, it's not a real number. The cos range is - 1 to 1. The root 2 is not possible

COS is equal to root 2

The root two is about 1.414. When cosx x x = 45 degrees, it is root two divided by 2, so 2cos (45) = what you want

It is known that α is the second quadrant angle 1+sinα 1-sinα- 1-sinα The result of 1 + sin α is___ .

∵ α is the second quadrant angle
∴cosα＜0
Qi
1+sinα
1-sinα-
1-sinα
1+sinα
=
(1+sinα)(1+sinα)
(1-sinα)(1+sinα)-
(1-sinα)(1-sinα)
(1+sinα)(1-sinα)
=-1+sinα
cosα+1-sinα
cosα
=-2tanα

It is known that α is the third quadrant angle, which can be simplified as (1 + sin α / 2) / (1-sin α / 2) + under radical (1-sin α / 2) / (1 + sin α / 2)

first
（√a+√b）^2=a+b+2√(ab)
understand
√a+√b=√[a+b+2√(ab)]
The original formula = √ [(1 + sin (α / 2)) / (1-sin (α / 2))] + √ [(1-sin (α / 2)) / (1 + sin (α / 2))]]
=√[（1+sin（α/2））/（1-sin（α/2））+（1-sin（α/2））/（1+sin（α/2））+2√[（1+sin（α/2））/（1-sin（α/2））*（1-sin（α/2））/（1+sin（α/2））]]
=√[（1+sin（α/2））/（1-sin（α/2））+（1-sin（α/2））/（1+sin（α/2））+2]
=√{[(1+2 sin（α/2）+ (sin（α/2）)^2)+ (1-2 sin（α/2）+ (sin（α/2）)^2)+2*(1- (sin（α/2）)^2)]
/[1 - (sin (α / 2)) ^ 2]} radical
=√{4/[(cos（α/2）)^2]}
=2/| cos（α/2）|
Since α belongs to the third quadrant angle, let
If K belongs to the integer Z, then
α belongs to [π + 2K п, 3 п / 2 + 2K п]
Therefore, α / 2 belongs to [п / 2 + K п, 3 п / 4 + K п]
When k = 2m, m belongs to integer Z, α / 2 belongs to the second quadrant angle
cos（α/2）=0
Original formula = 2 / cos (α / 2)

The root sign (1 + sin α) + the root sign (1-sin α), α is the third quadrant angle

Root sign (1 + sin α) + root sign (1-sin α) = root sign (Sina / 2 + cosa / 2) ^ 2 + absolute value of root sign (Sina / 2 + cosa / 2) ^ 2 = (Sina / 2 + cosa / 2) ^ ∵ A is the third quadrant angle ∵ A / 2 is the second quadrant angle, Sina / 2 > cosa / 2 = Sina / 2 + cosa / 2 + Sina / 2-co

It is known that α is the third quadrant angle, sin α + cos α = negative radical 2 Find the quotient of (COS (3 π / 2 + α) cos (3 π - α)) / Tan (K π + α) divided by the quotient of COS alpha K belongs to Z

(sin α cos α) squared = 2, so 12sin α cos α = 2, sin2 α = 1 / 2,2 α = π / 62k π or 7 π / 62k π, so α = π / 12 K π or 7 π / 12 K π. Because α is in the third quadrant, α = π / 12 K π (k belongs to Z)

A symmetric center of the function sin χ cos χ + radical 3cos χ - a symmetric center of radical 3 A (2 π / 3, - radical 3 / 2) B (5 π / 6, - radical 3 / 2) C (- 2 π / 3, radical 3 / 2) D (π / 3, - radical 3)

1 / 2 * sin2x + √ 3 (1 + cos2x) 3 (1 + cos2x) / 2 - √ 3 = sin2xcos π / 3 + cos2xsin π / 3 - √ 3 / 2 = sin (2x + π / 3) - √ / 3 / 2Y = sin (2x + π / 3) symmetry center is the intersection point of X axis, then 2x + π / 3 = k π / 3 = k π x = k π / 2-π / 6K = 6K = 1 then x = 5 π / 6sin (2x + π / 3) - √ / 3 / 3 / 2 is the sin (2x + π / 3) - √ / 3 / 3 / 2 is the sin (2x + π + π + π + π π (3) - √ / 3 3) down √ / 3 / 2

In the triangle ABC, if sin (2 minus a) = 2 sin (minus b) and 3cosa = 2cos (b), then the size of angle a is urgent

sin(2∏-A)=-√2sin(∏-B)
-sinA=-√2sinB ①
√3cosA=-√2cos(∏-B)
√3cosA=√2cosB ②
① 2 ＋ 2
2cos²A=1
cosA=√2/2
∴A=∏/4.