Simplification: F (x) = (sin2x cos2x + 1) / (1 + Cotx)

Simplification: F (x) = (sin2x cos2x + 1) / (1 + Cotx)

f(x)=(sin2x-cos2x+1)/(1+cotx)
=(2sinxcosx+2sinx^2)/(1+cosx/sinx)
=2sinx^2(cosx+sinx)/(sinx+cosx)
=2sinx^2

Simplify 2cos ^ 4x-2cos ^ 2x + 1 / 2 Such as the title. Urgent

The original formula = 2cos? X (COS? X-1) + 1 / 2 = 2cos? X (- sin? 2x) + 1 / 2 = - 2 (sinxcosx) 2) + 1 / 2 = - 1 / 2 * (2sinxcosx) x2 + 1 / 2 = - 1 / 2 * sin? 2X + 1 / 2 = - (1-sin? 2x) / 2 = - (COS? 2x) / 2

If f (SiN x) = 3-cos 2x, then f (COS x) = () A. 3-cos 2x B. 3-sin 2x C. 3+cos 2x D. 3+sin 2x

f（sinx）=3-cos2x=3-（1-2sin2x）=3-1+2sin2x=2sin2x+2，
Then f (cosx) = 2cos2x + 2 = 2cos2x-1 + 3 = 3 + cos2x,
Therefore, C

If f (SiNx) = 3-cos2x, then f (cosx) is equal to

f（sinX）=3—cos2x=3-(1-2sin^2 x)=2sin^2 x+2
∴f(cosX)=2cos^2x+2=cos2x+3

Cos [2 (π - x)] = - cos2x, right? π - x is in the third quadrant, cos is negative, so - cos2x is right?

cos(2pai-2x)=cos2x
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Cos ^ 4 (x) - Sin ^ 4 (x) is equal to Cos ^ 2 (x) or cos2x

cos^4(x)-sin^4(x)
=[cos ^ 2 (x) - Sin ^ 2 (x)] [cos ^ 2 (x) + sin ^ 2 (x)] (where [cos ^ 2 (x) + sin ^ 2 (x)] = 1)
=[cos ^ 2 (x) - Sin ^ 2 (x)] (is the expansion of cos2x)
=cos2x

If cos (π / 2 + x) = 4 / 5, then cos2x=

cos(π/2+x)=4/5
therefore
-sinx=4/5
sinx=-4/5
Then cos2x = 1-2sin? X = 1-2 × (- 4 / 5) 2 = 1-2 × 16 / 25 = - 7 / 25
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Let f (x) = a (sin2x + cos2x), and f (π / 4) = - 1 (1) Find the value of a (2) find the maximum value of function f (x) and the set of independent variables X when taking the maximum value (3) find the monotone increasing interval of function f (x)

1) If (x) = (x) = 2 (sin2xcos π / 4 + cos2xsin π / 4) + 2A = √ 2Sin (2x + π / 4) + 2a, the minimum positive period T = 2 π / 2 = π 2 = π 2), ∵ 0 ≤ x ≤ x ≤ π / 2  π / 4 ≤ 2x + π / 4 ≤ 5 π / 4 ﹤ 2 / 2 ≤ sin (2x + π / 4) ≤ 1, then f (x) min = √ 2 * (- √ 2 / 2) + 2A = 2A = 2A = 2A = 2a-1 = - 1 = - 1 = - 1 = - 1 = - 2 * (- 2 / 2 / 2) + 2A = 2A = 2A = 2A = 2A = 2a-1 = - 1 = - 1 = 2 ν a = - 1 / 2

It is known that f (x) = sin2x + cos2x (1) Find the period, amplitude and phase of F (x) (2) Make the image of F (x) with five point drawing method [just write the coordinates of five points] PS: the "2" of sin2x and cos2x in the title is not square, it is twice X

The formula of auxiliary angle of F (x) = (radical 2) sin (2x + π / 4) period is π, amplitude is root 2 phase π / 4 five points are drawn: 2x + π / 4 is regarded as a whole, five points are taken as follows: 0, π / 2, π, 3 π / 2, 2 π, and then the corresponding values of X are calculated as follows: - π / 8, π / 8, 3 π / 8, 5 π / 8, 7 π / 8

(sin2x) ^ 2 + sin2x + cos2x = 1 find angle X

sin²2x＋sin2x＋cos2x=1sn2x＋cos2x=1－sin²2x=cos²2xsin2x=cos²2x－cos2x ---------------------------------------------(1)(sin2x)²=(cos²2x－cos2x)²sin²2x=(cos²...