F (SiNx) = cos2x + 1 is it OK to find f (cosx)? One is 2Sin ^ 2x and the other is 1-cos2x

F (SiNx) = cos2x + 1 is it OK to find f (cosx)? One is 2Sin ^ 2x and the other is 1-cos2x

2 (SiNx) ^ 2 = 1-cos2x, which is identical

Find the maximum and minimum values of the function y = cos2x SiNx, X ∈ [-- 3 π / 4, π / 6]

y=cos2x-sinx
=1-2sin²x-sinx
=-2sin²x-sinx+1
=-2(sinx+1/4)²+9/8
When SiNx = - 1 / 4, the maximum value is 9 / 8
When SiNx = 1, there is a minimum value of - 2

The maximum value of the function y = SiNx + cos2x

y=sinx+cos2x
=sinx+1-2sin²x
=-2sin²+sinx+1
=-2【sin²-2sinx×1/4+（1/4）²-（1/4）²】+1
=-2【(sinx-1/4)²-1/16】+1
=-2(sinx-1/4)²+1/8+1
=-2(sinx-1/4)²+9/8
≤9/8
So the maximum value is 9 / 8, where SiNx = 1 / 4

Find the value of y = cos2x + sinx-2 (maximum and minimum)

Cos2x = 1-2sinx2 is the square of SiNx. Write it as a square plus a number. Pay attention to the value range of SiNx. Do it yourself. It should be easy

Y = 2cos ^ 2x + sin2x how to get 2cos ^ 2x = 1 + cos2x

Because cos (2x) = cos ^ 2 (x) - Sin ^ 2 (x) and COS ^ 2 (x) + sin ^ 2 (x) = 1, then sin ^ 2 (x) = 1-cos ^ 2 (x), then cos (2x) = 2cos ^ 2 (x) - 1, then 2cos ^ 2x = 1 + cos2x y = 2cos ^ 2x + sin2x = cos2x + sin2x + 1 = [2 ^ (1 / 2)] cos (2x-45) + 1, where [2 ^ (1 / 2)] is equal to the root 2

sinx+cosx=tanx(0

Multiple choice questions with several special values
h(x)=sinx+cosx-tanx
h(pi/6)>0,h(pi/4)>0,h(pi/3)

Take the derivative f (x) = (pi * TaNx * secx) ^ 6, and f (x) = arcsin (SiNx + 1 / 2),

1,f(x)=(πtan sec x)^6
f'(x)=[6(πtan sec x)^5]×[πsec^2(sec x)]×[secx tanx]
=6π^6(tan secx)^5×(sec secx)^2×secx×tanx
Let g (x) = π Tan sec X
f(x)=[g(x)]^6
Derivation rules of composite functions
f'(x)=6[g(x)]^5×g'(x)
g'(x)=πtan'(sec x)×sec'x
={π[sec (sec x)]^2}×(secx tanx).
2,f(x)=arcsin(sinx+1/2)
f'(x)={1/√[1-(sinx+1/2)^2]}×cos x.
Let g (x) = SiN x + 1 / 2
Here we should pay attention to the domain of F (x) - 1 ≤ SiN x + 1 / 2 ≤ 1
That is - 1 ≤ SiN x ≤ 1 / 2
The definition domain is: [- π / 2 + 2K π, π / 6 + 2K π] ∪ [5 π / 6 + 2K π, 2 π + 2K π]
f'(x)=1/√{1-[g(x)]^2}×g'(x)
g'(x)=cos x.

It is proved that SiNx (1 + TaNx * tan2 / x) = TaNx As the title

Left = SiNx (1 + TaNx * tan2 / x)
=sinx[1+(sinxsinx/2)/(cosxcosx/2)]
=sinx[sinxsinx/2+cosxcosx/2]/(cosxcosx/2)]
=sinx[cosx/2]/(cosxcosx/2)]
=sinx/cosx
=tanx
=Right
So SiNx (1 + TaNx * tan2 / x) = TaNx

The X of [(2 + TaNx) ^ 10 - (2-sinx) ^ 10] / SiNx approaches the limit of 0

This is a limit of type 0 / 0, which can be used with the law of rotita:
=lim[10*(2+tanx)^9 *(secx)^2 - 10*(2-sinx)^9 *(-1*cosx)]/1
=lim[10*(2+0)^9 * 1^2 + 10 * (2-0)^9 * 1]/1
=20 * 2^9
=10 * 2^10
=10240

Given SiNx + cosx = 5 / 13 √ 2, and X ∈ (U / 4,3 μ / 4), find cosx and (1-tanx / 1 + TaNx)

Then we get (SiNx) ^ 2 + 2sinxcosx + (cosx) ^ 2 = 50 / 169 and (SiNx cosx) ^ 2 = (SiNx) ^ 2 + (cosx) ^ 2-2sinxcosx = 2 - 50 / 169 = 288 / 169, so SiNx - cosx = 12 / 13 √ 2 and SiNx + cosx = 5 / 13 √ 2