If SiNx + cosx = 0.5, then sin ^ 3x + cos ^ 3x= Here sin ^ 3x + cos ^ x =? Reason, Sorry, the wrong number. Is sin ^ 3x + cos ^ 3x =?

If SiNx + cosx = 0.5, then sin ^ 3x + cos ^ 3x= Here sin ^ 3x + cos ^ x =? Reason, Sorry, the wrong number. Is sin ^ 3x + cos ^ 3x =?

The solution is obtained by SiNx + cosx = 0.5
The square result is sin? X + cos? X + 2sinxcosx = 0.5
That is, 1 + 2sinxcosx = 0.5
The solution is sinxcosx = - 1 / 4
From sin ^ 3x + cos ^ 3x
=(sinx+cosx)(sin²x-sinxcosx+cos²x)
=0.5×（sin²x+cos²x-sinxcosx）
=0.5×（1-（-1/4)）
=5/8

If SiNx + cosx = m, then the value of sin cube x + cos cube x is RT

(sinx)^3+(cosx)^3
=(sinx+cosx)[(sinx)^2-sinxcosx+(cosx)^2]
=m(1-sinxcosx)
Because (SiNx + cosx) = m ^ 2 = 1 + 2sinxcosx
So sinxcosx = (m ^ 2-1) / 2
The original formula = m {1 - [(m ^ 2-1) / 2]}
=m(1-m^2/2+1/2)
=m[(3/2)-(m^2/2)}
=3m/2-m^3/2
=(3m-m^3)/2

It is proved that cos square x + cos square (x + a) - 2cosa * cosx * cos (x + a) = sin square a

Cos square x + cos square (x + a) - 2cosa * cosx * cos (x + a)
=[cos2x+cos2(x+a)]/2+1-2cosa*cosx*cos(x+a)
=cos(2x+a)cosa+1-2cosa*cosx*cos(x+a)
=cosa*cosx*cos(x+a)-cosa*sinx*sin(x+a)+1-2cosa*cosx*cos(x+a)
=1-[cosa*cosx*cos(x+a)+cosa*sinx*sin(x+a)]
=1-cosa*cos(x+a-x)
=1-cos square a = answer

When 0

f(X)=cos²x/(cosxsinx-sin²x)
The numerator and denominator are divided by cos? X to obtain:
f(X)=1/(tanx-tan²x)
Zero

Sin ^ 3x cos ^ 3x > cosx SiNx, find the value range of X X belongs to (0,2 π)

Sin ^ 3x cos ^ 3x (SiNx cosx) (sin ^ 2x + sinxcosx + cos ^ 2x)
（sinx-cosx）(sin^2x+sinxcosx+cos^2x)>cosx-sinx
（sinx-cosx）(sin^2x+sinxcosx+cos^2x)-(sinx-conx)>0
(sinx-cosx)*sin2x>0
Radical 2 * sin (x - π / 4) * sin2x > 0
Do it yourself

Given that SiNx + cosx = 0.2 and X belongs to (0, π), find sin ^ 3x cos ^ 3x

From (SiNx + cosx) ^ 2 = 1 / 25, 2sinxcosx = - 24 / 25,
(SiNx cosx) ^ 2 = 48 / 25, SiNx cosx = - 4 √ 3 / 5,
So sin ^ 3x cos ^ 3x = (SiNx cosx) (1 + sinxcosx) = - 52 √ 3 / 125

SiNx + cosx = m (absolute value of M

sinx+cosx=m
Square the two sides
1+2sinxcosx=m^2
sinxcosx=(m^2-1)/2
sin^3x+cos^3x
=(sinx+cosx)[(sinx)^2-sinxcosx+(cosx)^2]
=m*[(sinx+cosx)^2-3sinxcosx]
=m*(m^2-3(m^2-1)/2)
=m*(-m^2/2+3/2)
=-m^3/2+3m/2
sin^4x+cos^4x
=[(sinx)^2+(cosx)^2]-2(sinxcosx)^2
=1-2[(m^2-1)/2]^2
=1-(1/2)(m^4-2m^2+1)
=-(1/2)(m^4-2m^2+1)

How to translate the image of function y = cos (x-3) + 2 to get the image of y = cosx, How to translate the image of function y = log2 (X-2) + 3 to get y = = log2x, the answer is (- 2,

The first problem is to shift two units left along the X axis and then two units down the Y axis. The second one is to shift two units left along the X axis and three units down the Y axis

The image of the function y = cos x (x ∈ R) shifts π to the left After 2 units, the graph of function y = g (x) is obtained, then the analytic expression of G (x) should be () A. -sin x B. sin x C. -cos x D. cos x

The image of the function y = cos x (x ∈ R) shifts π to the left
After 2 units, the function y = cos (x + π) is obtained
2) The image of = - SiNx,
Therefore, a

The image of function y = f (x) × cosx is translated by vector a = (π / 4,1) to get the image of function y = 2sinx ^ 2, then the function f (x) can be a cosx Change the function y= The image of F (x) × cosx is translated by vector a = (π / 4,1) to get the image of function y = 2sinx ^ 2. Then the function f (x) can be a cosx B 2sinx C SiNx D 2cosx How is it left or right? Math homework help users 2016-11-30 report Use this app to check the operation efficiently and accurately!

That is to say, first shift pi / 4 units to the right, and then 1 unit up → the function after translation is: y = f (x-pi / 4) * cos (x-pi / 4) + 1 = 2Sin ^ 2x → f (x-pi / 4) * cos (x-pi / 4) = 2Sin ^ 2x-1 = - cos2x. Therefore, substituting a, B, C, D in turn, we can get that B holds 2Sin (x-pi / 4) cos (x-pi / 4) = sin (2x)