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sinx*sin2x*sin3x=?
If it is multiplication, then it is:
sinx*sin2x*sin3x
=sinX*2sinX*cosX*sin(2X+X)
=2sinX*sinX*cosX*(sin2X*cosX+cos2X*sinX)
=2sinX*sinX*cosX*[2sinX*cosX*cosX+(cosX*cosX-sinX*sinX)*sinX]
=2sinX*sinX*cosX*[sinX*cosX*cosX+sinX*cosX*cosX+cosX*cosX*sinX-sinX*sinX*sinX
=2sinX*sinX*cosX*(3sinX*cosX*cosX-sinX*sinX*sinX)
=?
I'll go back and read first!
Find the mathematical integral ∫ (SiNx * sin2x * sin3x) DX
Sinxsin3xsin2x = - 1 / 2 (cos4x-cos2x) sin2x = 1 / 2 (sin2xcos4x-sin2xcos2x) = 1 / 2 * [1 / 2 (sin6x-sin2x) - 1 / 2sin4x] = 1 / 4 * sin6x-1 / 4 * sin2x-1 / 4 * sin4x original formula = 1 / 4 ∫ sin6xdx-1 / 4 ∫ sin2xdx-1 / 4 ∫ sin4xdx = - cos6x / 24 + cos2x / 8 + cos4x / 16 + c
sin3x-sin2x+sinx=0
sin3x
=sin(2x+x)
=sin2xcosx+cos2xsinx
=2sinxcosxcosx+cos2xsinx
=2sinxcos ² x+cos2xsinx
sin3x-sin2x+sinx
=2sinxcos ² x+cos2xsinx-sin2x+sinx=0
If SiNx = 0, then x = k π
If SiNx ≠ 0, get:
2cos^2x+cosx+1=0
It is equivalent to the solution of quadratic equation 2x ^ 2 + X + 1 = 0
It is easy to know that the equation has no solution (discriminant △ = 1-8
Who can help calculate the sum of SiNx + sin2x + sin3x + + sinnx, which is a key step
Analysis: the structural feature of this sum formula is that the change of the angle of each sinusoidal function forms an equal difference sequence. We wonder whether it can be compared with + +... + = (1 -) + (-) +... + (-) = and try to divide each term in the sum formula into the difference of two terms, so that all intermediate terms can be offset exactly, so as to obtain the result. In fact
Given the function y = 2sinx (SiNx + cosx) - 1, find the minimum positive period and maximum value of function f (x)
Y = 2sinx (SiNx + cosx) - 1, which can be simplified as y = sin2x cos2x by double angle formula
Further simplify y = 2 ^ 0.5sin (2x - Π / 4) and know that the minimum positive period is Π and the maximum value is 2 ^ 0.5. To solve this kind of problem, we should be familiar with the basic formula of sine and cosine
Known function f (x) = 2sinx (SiNx + cosx) 1. Find the minimum positive period and maximum value of function f (x)
f(x)=2sinx(sinX+cosX)
=2sinxsinx+2sinxcosx
=1-cos2x+sin2x
=√2sin(2x-π/4)+1
So the minimum positive period of F (x) = 2 π / 2 = π
Max = 1 + √ 2
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