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What is the original function of cos2x / (cosx SiNx)
cos2x=cosx^2-sinx^2=(cosx-sinx)(cosx+sinx)
So the above formula is reduced to = cosx + SiNx, so the original function is SiNx cosx
The period of the function y = SiNx · cosx · cos2x
π/2
Cos2x + sin2x becomes √ 2 × sin (2x + π / 4) It is missed that cos2x + sin2x becomes √ 2 × sin (2x + π / 4)
cos2x+sin2x
=√2（√2/2*cos2x+√2/2*sin2x）
=√2（sinπ/4*cos2x +cosnπ/4*sin2x）
=√2sin（2x+π/4）
1+sin2x+cos2x=1+sin(2x+π/4), Why?
1+sin2x+cos2x
=1+√2(sin2x*√2/2+cos2x*√2/2)
=1+√2(sin2xcosπ/4+cos2xsinπ/4)
=1+√2sin(2x+π/4)
Simplify f (x) = sin (2x + 6 / π) - sin (2x-6 / π) - cos2x + 1 The function f (x) = sin (2x + π / 6) + sin (2x - π / 6) - cos2x + 1 is used to find the minimum positive period, symmetry axis, symmetry center and monotone increasing interval of F (x).
6 / π? It's usually π / 6. Let me solve it as π / 6
f(x)=sin(2x+π/6)-sin(2x-π/6)-cos2x+1
=(sin2xcosπ/6+cos2xsinπ/6)-(sin2xcosπ/6-cos2xsinπ/6)-cos2x+1
=2cos2xsinπ/6-cos2x+1
=cos2x-cos2x+1
=1
How to simplify [1 + cos2x + sin ^ 2x] / SiNx?
cos2x = (cosx)^2 -(sinx)^2 =1-2(sinx)^2
2 - (SiNx) ^ 2 / SiNx = 2 / SiNx - SiNx
Simplify cos2x / (cosxsinx sin ^ 2x) The first is cos 2x, and the last is sin squared X. don't get confused
cos2x/（cosxsinx-sin^2x）
=(cosxcosx-sinxsinx)/[（cosx-sinx）sinx]
=(cosx+sinx)/sinx
=1+1/tanx
（1+cos2x)/2cosx=sin2x/(1-cos2x)
It is known that sin2x = 2sinxcosx cos2x = (cosx) ^ 2 - (SiNx) ^ 2, so 1-cos2x = 2 (SiNx) ＾ 21 + cos2x = 2 (cosx) ^ 2, so (1 + cos2x) / 2cosx = sin2x / (1-cos2x) resolves to 2 (cosx) ^ / 2cosx = 2sinx cosx / 2 (SiNx) ＾ 2cosx = cosx / sinxsinx = 1x = arcsin1
If TaNx is equal to √ 2, then (COS? X-cos2x) / 1-sin? X =____ ?
（cos²x-cos2x）/1-sin²x =[cos²x-(2cos²x-1)]/cos²x=(1-cos²x)/cos²x =tan²x =2
Find the period of the function y = sin (60-2x) + cos2x
y=sin(60-2x)+cos2x
=sin60cos2x-cos60sin2x+cos2x
=(1 + radical 3) / 2 * cos2x-1 / 2 * sin2x
=Under root sign [((1 + radical 3) / 2) ^ 2 + (- 1 / 2) ^ 2] * sin (2x + W)
So the period is 2 π / 2 = π
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