Is sin - x equal to - SiNx? What about the rest?

Is sin - x equal to - SiNx? What about the rest?

sin(-x)=-sinx
cos(-x)=cosx
tan(-x)=-tanx
cot(-x)=-cotx

SiNx + SiNx = 1, what is SiNx equal to?

0

0

f'(x)=(sinα-sinx)'=0-cosx=-cosx
f'(α)=-cosα

Is sin x equal to SiNx times SiNx? I'm confused

Yes. It's equal to (SiNx) ^ 2

If SiNx = 2cosx, then 1 1+tanx 2−1 1−tanx 2=______ .

From SiNx = 2cosx, TaNx = 2,
Another one
1+tanx
2−1
1−tanx
2=−2tanx
Two
1−tan2x
2=-tanx=-2，
So the answer is - 2

Convert sinx-2cosx into? Sin (? + or -?) Cos (? + or -?),

0

0

One
y=sin(x/4)cos(x/4)-1=(1/2)sin(x/2) -1
The minimum positive period T = 2 π / (1 / 2) = 4 π
When sin (x / 2) = 1, y has a maximum value ymax = 1 / 2 - 1 = - 1 / 2; when sin (x / 2) = - 1, y has a minimum value, Ymin = - 1 / 2-1 = - 3 / 2
Two
Y = SiNx + 2cosx = √ 5sin (x + a), where Tana = 2
Minimum positive period T = 2 π / 1 = 2 π
When sin (x + a) = 1, y has a maximum value ymax = √ 5; when sin (x + a) = - 1, y has a minimum value, Ymin = - √ 5
Three
y=2cosxsin(x+π/3)-√3sin²x+sinxcosx
=2cosx[sinxcos(π/3)+cosxsin(π/3)]-√3sin²x+sinxcosx
=2cosx[(1/2)sinx+(√3/2)cosx]-√3sin²x+sinxcosx
=sinxcosx+√3cos²x-√3sin²x+sinxcosx
=√3(cos²x-sin²x)+2sinxcosx
=√3cos(2x)-sin(2x)
=2[(√3/2)cos2x-(1/2)sin(2x)]
=2cos(2x+π/6)
Minimum positive period T = 2 π / 2 = π
When cos (2x + π / 6) = 1, y has the maximum value ymax = 2; when cos (2x + π / 6) = - 1, y has the minimum value Ymin = - 2

Given the vector M = (2cos ^ 2x, SiNx), n = (1,2cosx) (1) if M ⊥ N and 0, try to find the equation of symmetry axis and the center of symmetry of F (x)

1. (1,2cosx) = 2cosx + 2sinxcosx = cos2x + 2sinxcosx = cos2x + 1 + sin2x = cos2x + 1 + sin2x = cos2x + 1 + sin2x = √ 2 [(2 / 2 / 2) sin2x + ((2 / 2) cos2x + (2 / 2) cos2x] + 1 = √ 2Sin (2x + π / 4) + 1 = 0  sin (2x + π / 4) = - √ 2 / 2 ȼ 2x + π / 4 = 5 π / 4 = 5 π / 4 + 2K π or 7 7 7 7 (7) 7 (7) 7 (7) 7 (7) or 7 (4 + 2K) π (4 + 2K) or 7 π / 4 +

Given the vector n = (2cosx, √ 3 SiNx), vector M = (cosx, 2cos), Let f (x) = vector n times vector m + a If x ∈ [0, PI] and a = - 1, the equation f (x) = B has two unequal real roots X1 and x2. The value range of B and the value of X1 + x2 are obtained

When a = - 1, f (x) = 2Sin (2x + 2x + π / 6) is because x ∈ (0, π) = = = > 2x + π / 6 ∈ (π / 6) because x ∈ (0, π) = = = > 2x + π / 6 ∈ (π / 6,13 π / 6) = = = > 2Sin (2x + π / 6) ∈ (π / 6,13 π / 6) = = = = > 2Sin (2x + π / 6) ∈ [- 2,2] because f (x) = b = = = = > 2Sin (2x + π / 6) ∈ [- 2,2,2] because f (x) = b = b = = > 2, 2, 2] because f (x) = b = = > 2, 2,(2x + π / 6) = B has two unequal real roots-2

The vector M = (2cos 2 (x - π / 6, SiNx) n = (1,2sinx) function f (x) = vector m × vector n When x ∈ [0,5 π / 12], the value range of function f (x) is? F (x) = sin (2x + π / 6) has been calculated

x∈【0,5π/12】
2x∈【0,5π/6】
2x+π/6∈【π/6,π】
=>0≤sin(2x+π/6)≤1/2