How to prove cos3x. SiNx = 1 / 2 (sin4x-sin2x)?

How to prove cos3x. SiNx = 1 / 2 (sin4x-sin2x)?

According to the integral sum difference formula cosa * SINB = (sin (a + b) - sin (a-b)) / 2, the left side = 1 / 2 (sin4x sin (2x) = the right side, it is proved that the integral sum difference formula sin (a + b) - sin (a-b) = sianacosb + Cosa SINB - sinacosb + Cosa SINB = 2cosa SINB

Find the limit sin 5x / sin 3x. When x is infinitely close to 0, find the limit value
Lim sin5x / sin3x, X -- & gt; 0, find the limit value

Using Lim SiNx / x = 1 (when x is infinitely close to 0) formula
lim sin5x/sin3x=lim (5x/3x)（sin5x/5x)/(sin3x/3x)=5/3

The minimum positive period of function f (x) = sin3x + sin5x

The minimum positive period of F (x) = sin3x is 2 Π / 3;
The minimum positive period of F (x) = sin5x is 2 Π / 5;
2 Π / 3 and 2 Π / 5 are divided into 10 Π / 15 and 6 Π / 15, and the least common multiple of 10 and 6 is 30,
Then the minimum positive period of F (x) = sin 3x + sin 5x is 30 Π / 15, that is 2 Π

Let x, y ∈ R, then the minimum value of (3-4y-cosx) 2 + (4 + 3Y + SiNx) 2 is ()
A. 4B. 5C. 16D. 25

∵ (3-4y-cosx) 2 + (4 + 3Y + SiNx) 2 = ([(3-4y) - cosx] 2 + [(4 + 3Y) - (- SiNx)] 2) 2, by analogy with the distance formula between two points | ab | = (x1-x2) 2 + (y1-y2) 2, and 3 (3-4y) + 4 (4 + 3Y) - 25 = 0, the formula obtained is from a point on the straight line 3x + 4y-25 = 0 to a point on the circle x2 + y2 = 1

Who can solve a math problem? The maximum value of y = SiNx cosx + sinxcosx (x belongs to 0 to 2 π)

Let t = SiNx cosx t on - 1 to 1
sinxcosx=(1-t^2)/4
y=t+(1-t^2)/4
=-(t^2)/4+t+1/4
=-[(t-2)^2]/4+5/4
Because t is from - 1 to 1
It is known that the symmetry axis of parabola is y = 2
So substituting x = - 1 has a minimum of - 1, substituting x = 1 has a maximum of 1
If not, there is no corresponding maximum value

Mathematical problem 2cos (SiNx cosx) + 1
The detailed process of simplification thank you

2cos(√2sin(x-π/4))+1
2 [cos(sinx)cos(cosx)+(sinsinx)(sincosx)]+1
Make sure the title is correct?

Tell me why (2 ^ x) ^ 2 equals 4 ^ X

（2^x)^2=2^（2X）=4^x

The function f (x) = sin1 / 4x * sin1 / 4 (x + 2 * π] * 1 / 2 (x + 3 * π) is arranged into a sequence {an} from small to large
The function f (x) = sin1 / 4x * sin1 / 4 (x + 2 * π] * 1 / 2 (x + 3 * π) is arranged into a sequence {an} from small to large
(1) Finding the general term formula of sequence {an}

First, f (x) = 1 / 4 (x + 3 π) sin1 / 2x
Derivation f '(x)=
Let f '(x) = 0
Find out the value of X, arrange and find out the law
The general formula is obtained

In this paper, all extreme points of function f (x) = sin14x · sin14 (x + 2 π) · sin12 (x + 3 π) in the interval (0, + ∞) are arranged into the sequence {an} (n ∈ n *) (1) the general term formula of the sequence {an}; (2) let BN = 2nan and the sum of the first n terms of the sequence {BN} be TN, the expression of TN is obtained

(1) F (x) = sin14x · sin14 (x + 2 π) · sin12 (x + 3 π) = sin14x · cos14x · (− cos12x) = 12 · sin12x · (− cos12x) = − 14sinx. According to the properties of sine function, its extreme point is x = k π + π 2 (K ∈ z), and all its extreme points in (0, + ∞) form an arithmetic sequence with π 2 as the first term and π as the tolerance. The general formula of the sequence {an} is & nbsp; An = π 2 + (n − 1) · π = 2n − 12 π (n ∈ n *) (6 points) (2) from (1), we can get BN = 2nan = π 2 (2n − 1) · 2n (8 points) ‖ TN = π 2 [1.2 + 3.22 + +(2n − 3) · 2n − 1 + (2n − 1) · 2n], two sides multiplied by 2, 2tn = π 2 [1.22 + 3.23 +...] +By subtracting (2n − 3) · 2n + (2n − 1) · 2n + 1], we get − TN = π 2 [1.2 + 2.22 + 2.23 +...] +2 · 2n − (2n − 1) · 2n + 1] = π 2 [2 + 8 (1 − 2n − 1) 1 − 2 − (2n − 1) · & nbsp; 2n + 1] = π 2 [− 6 + (3 − 2n) 2n + 1] = - π [(2n-3) · 2n + 3] ‖ TN = π [(2n-3) · 2n + 3] (12 points)

Is cos (1 + 2) equal to cos1 + Cos2?