Given Sina + cosa = a, find the values of (1) Sina * cosa (2) SIN3 power a + cos3 power a

Given Sina + cosa = a, find the values of (1) Sina * cosa (2) SIN3 power a + cos3 power a

2sina*cosa=(sina+cosa)²-(sin²a+cos²a)=a²-1
So Sina * cosa = a 2 / 2-1 / 2
sin³a+cos³a
=(sina+cosa)(sin²a-sinacosa+cos²a)
=a*（1-a²/2+1/2）
=a*(3/2-a²/2)
=3a/2-a³/2

For any angle θ, Cos4 θ - SiN4 θ = Cos2 θ

It is proved that for any angle θ,
∵cos4θ-sin4θ=（cos2θ+sin2θ）•（cos2θ-sin2θ）
=cos2θ-sin2θ=cos2θ，
The result shows that cos 4 θ - Sin 4 θ = cos 2 θ holds

The fourth power of cosin 2a is known

It is known that (sin2a) ^ 2 = 1 - (cos2a) ^ 2 = 16 / 25,
So (Sina) ^ 4 + (COSA) ^ 4 = [(Sina) ^ 2 + (COSA) ^ 2] ^ 2-2 (sinacosa) ^ 2 = 1 - (sin2a) ^ 2 / 2 = 1-8 / 25 = 17 / 25

Why is the fourth power of cosa Sina equal to cos2a

cos^4(a)-sin^4(a)
=[cos^2(a)+sin^2(a)][cos^2(a)-sin^2(a)]
=1*[cos^2(a)-sin^2(a)]
=cos2a

It is proved that: (1-cosa to the 4th power of sina) / (1-cosa to the 6th power of sina) = 2 / 3 Eleven

(1－cos^4a－sin^4a)/(1－cos^6a－sin^6a)＝[1－(cos^4a＋sin^4a＋2cos^2asin^2a)＋2cos^2asin^2a]/[1－((cos^2a)^3＋(sin^2a)^3)]＝[1－(cos^2a＋sin^2a)^2＋2cos^2asin^2a]/[1－(cos^2a＋sin^2a)(cos^4a＋sin^4a－...

Let a > 0, the function f (x) = x + a 2 / x, and f (- 1) = - 5.1. Find the value of a; 2. Prove that f (- x) + F (x) = 0

Solution:
(1)f(-1)=-1-a²=-5,∴a=±2
And ∵ a > 0, ᙽ a = 2
∴f(x)=x+4/x
(2) Proof:
f(-x)+f(x)=-x-4/x+x+4/x=0
f(-x)+f(x)=0

The function f (x) = x / x 2 + 1 is an odd function defined on (- 1,1). It is proved by definition that f (x) is an increasing function on (- 1,1)

The derivation of F (x) = x / x? + 1 shows that f (x) = - 2x? / (x? + 1) 2 + 1 / x? + 1 = (1-x?) / (x? + 1) 2 on (- 1,1), 1-x? > 0; (x? + 1) 2 > 0, so f (x) '> 0, so f (x) is an increasing function on (- 1,1)

Given the function f (x) = x? - 3x + 2 (1), it is proved that the function y = f (x) is an increasing function on (1, + ∞)? (2) It is proved that the equation f (x) = 0 has no root greater than 1? The function should be f (x) = x 3 x + 2,

There is a serious problem with this topic. Both questions are wrong,
First of all, (1) this function is an increasing function on (3 / 2, + ∞) and a decreasing function in (1,3 / 2) interval
(2) The equation has two roots X1 = 1; x2 = 2; both > = 1; the original problem should be changed to have no root less than 1

The known function f (x) = x2 + xsinx + cosx (1) Find the minimum value of F (x); (2) If the curve y = f (x) is tangent to the straight line y = B at the point (a, f (a)), find the values of a and B (3) If the curve y = f (x) and the straight line y = B have two different intersections, find the value range of B

(1) By F (x) = x2 + xsinx + cosx,
F '(x) = 2x + SiNx + xcosx - SiNx = x (2 + cosx) (1 point)
Let f '(x) = 0, then x = 0 (2 points)
The list is as follows:
  … (4 points)
The function f (x) decreases monotonically on the interval (- ∞, 0),
Monotonically increasing on the interval (0, + ∞),
/ / F (0) = 1 is the minimum value of F (x) (5 points)
(2) ∵ the curve y = f (x) is tangent to the line y = B at the point (a, f (a)),
∴f′（a）=a（2+cosa）=0，b=f（a），… (7 points)
A = 0, B = f (0) = 1 (9 points)
(3) When B ≤ 1, there is only one intersection point between the curve y = f (x) and the straight line y = B at most;
When b > 1, f (- 2b) = f (2b) ≥ 4b-2b-1 > 4b-2b-1 > b, f (0) = 1 < B,
There exists x1 ∈ (- 2b, 0), X2 ∈ (0, 2b), such that f (x1) = f (x2) = B (12 points)
Because the function f (x) is monotone on the interval (- ∞, 0) and (0, + ∞),
When b > 1, the curve y = f (x) and the straight line y = B have and only two different intersections (13 points)
To sum up, if the curve y = f (x) and the straight line y = B have two different intersections, then the value range of B is (1, + ∞) (14 points)

Is cos? X-sin? X + cosx odd or even?

=cos2x+cosx
Even function
Or not to simplify, direct judgment
F (x) = f (- x), so it is even function]