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Anonymous users2024-02-11Summary. f（x）＝-cos2x+sin2x
f(x)=sin x-cos x+2sinx·cosx simplification.
f（x）＝-cos2x+sin2x
Can you take a look at the process?
OK. Can this end up in this form?
OK. f(x) -cos2x+sin2x root number 2sin(2x-4).
Thank you. That process didn't happen.
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Anonymous users2024-02-10The way I think it is, I guess someone else has a better idea.
f(x)=1+cos2x-[1-cos(2x-π/3)]=cos2x-cos(2x-π/3)
cos2x-[cos2xcos(-π3)-sinxsin(-π3)]
1/2cos2x-√3/2sinx
sin(π/6)cos2x-cos(π/6)sin2x=sin(π/6-2x)
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Anonymous users2024-02-09The formula is rented by the number of doublings: f(x)=(1 2)cos2x+(1 2)sin2x+1 2
By the auxiliary angle formula: f(x)=(2 closed slip2)sin(2x+ 4)+1 potato state mega2
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Anonymous users2024-02-08<>Please don't have a cavity to dry up and annihilate the round.
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Anonymous users2024-02-07<> ginseng Wang Wu and the ants are in ambush.
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Anonymous users2024-02-06Why did -2 be extracted.
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Anonymous users2024-02-05Use the formula cos a=1+cos2a 2, sin a=1-cos2a 2
According to induction equation five, the result is sin2x
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Anonymous users2024-02-04^f（x）=2sin(x/4)cos(x/4)-2√3sin²(x/4)+√3
f(x) = sin(x 2) - root number 3 (1-cos(x 2)) + root number 3f(x) = sin(x 2) + root number 3cos(x 2) f(x) = 2sin (x 2 + arctan [root number 3]) f(x) = 2sin (x 2 + pai 3).
The formula is used: asinx+bcosx=root(A2+B2)sin(X+arctan(b a)).
x should have a range, right? y=f(x)=(1+cos2x+8sin 2x) sin2x =(1+cos2x+4-4cos2x) sin2x =(5-3cos2x) sin2x =(5-3cos2x) [1-(cos2x) 2] and then solve according to the range Follow-up: When 0 is basically the same as the following.
Since cos(2a)=cos2(a)-sin 2(a)=2cos2(a)-1, then 2cos2(a)=cos(2a)+1 >>>More
f'(x)=2-1 x 2=(2x 2-1) x 2, let f'(x)=0: x= 2 2 x (0, 2 2 ) f'(x)<0,x ( 2 2, + f'(x) >0, so f(x) decreases on (0, 2 2) and increases on (2 2, +).
Solution: f(x)=cos x-(3)cosxsinx+1
f(x)=2[cos²x-(√3)cosxsinx+1]/2 >>>More
f(2a)=f(b+3)
That is, 4a-3 = 2b+3 >>>More