sin^4A+cos^2A+sin^2Acos^2A=?

sin^4A+cos^2A+sin^2Acos^2A=?

sin^4a+sin^2acos^2a+cos^2a
=sin^2a(sin^2a+cos^2a)+cos^2a
=sin^2a+cos^2a
=1

If cosa + cos ^ 2A = 1, what is sin ^ 2A + sin ^ 6A + sin ^ 8A If cosa + cos ^ 2A = 1, then what is sin ^ 2A + sin ^ 6A + sin ^ 8A equal? The wrong letter is typed in the above question

Because of the reasons for cos a + cos ^ 2A = 1 and because of sin ^ 2A + cos ^ 2A = 1, then: sin ^ 2A = cosa: sin ^ 2A + sin ^ 6A + sin ^ 8A = sin ^ 2A + (sin ^ 2a) ^ 3 + (sin ^ 2a) ^ 4 = cosa + cos ^ 3A + cos ^ 4A = cosa + cos ^ 2A (COSA + cos ^ 2a) = cosa + cos ^ 2A (sin ^ 2A + cos ^ 2a) = cosa + cos ^ 2A (sin ^ 2A + cos ^ 2a) = cosa + cos ^ 2A = cosa + cos ^ 2A = cosa + cos ^ 2A = cos a + cos ^ 2A = cos a + cos ^ 2A = cos a + cos ^ 2A = sin ^ 2A = sin ^ 2A = sin ^ 2A = sin + cos ^ 2A = sin + cos ^ 2A = sin ^ 2A = sin ^ 2A = sin ^ 2A =

Evaluation of cosin-6cos-6s-1 As the title The numerator is the part before / and the denominator is the part after

It turns out to be 1.5, right?
1 = (sin ^ 2A + cos ^ 2a) ^ 2 = sin ^ 4 + 2Sin ^ 2A × cos ^ 2A + cos ^ 4A?
1-sin^6a-cos^6a sin^4a+2sin^2a×cos^2a+cos^4a-(sin^6a+cos^6a)
--------------- = --------------------------------------------
1-sin^4a-cos^4a sin^4a+2sin^2a×cos^2a+cos^4a-(sin^4a+cos^4a)
sin^4+2sin^2a×cos^2a+cos^4a-(sin^6a+cos^6a)
= --------------------------------------------①
2sin^2a×cos^2a
According to the cubic sum formula:
sin^6a+cos^6a=(sin^2a)^3+(cos^2a)^3
=(sin^2a+cos^2a)(sin^4a-sin^2a×cos^2a+cos^4a)
=sin^4a-sin^2a×cos^2a+cos^4a②
By replacing sin ^ 6A + cos ^ 6A in ① with that in ②, we can get the following results:
sin^4+2sin^2a×cos^2a+cos^4a-(sin^4a-sin^2a×cos^2a+cos^4a)
Original formula=-------------------------------------------------------------
2sin^2a×cos^2a
3sin^2a×cos^2a
= -----------------=1.5
2sin^2a×cos^2a
I don't know if the result is right. But the idea is correct
The typesetting is very good when inputting, but it is out of order as soon as it is released
Can you read it? If you can't understand it, you can send me a message
It's a little messy. If you look at it carefully, you can understand it
--------It's the score line... I'm sorry~

Simplify 1-cos ^ 4a-sin ^ 4A / 1-cos ^ 6-sin ^ 6A Cos ^ 6 should be cos ^ 6A

1-cos ^ 6-sin ^ 6A = 1 - (sin ^ 2 + cos ^ a) (COS ^ 4a-cos ^ 2asin ^ 2A + sin ^ 4a) = 1-cos ^ 4A + cos ^ 2asin ^ 2a-sin ^ 2A = 3cos ^ 2asin ^ 2a-sin ^ 4A = cos ^ 2A cos ^ 4A + sin ^ 2A cos ^ 4A = 2cos ^ 2asin ^ 4A = 2 / 3

Given cos2a = (1 / 4), find the value of COS ^ 4A + sin ^ 4 + sin ^ 2acos ^ 2A

cos(2a)=1/4
[sin(2a)]^2=1-[cos(2a)]^2=1-1/16=15/16
(cosa)^4+(sina)^4+(sina)^2(cosa)^2
=[(cosa)^2+(sina)^2]^2-(sina)^2(cosa)^2
=1-sin(2a)^2/4
=1-(15/16)/4
=1-15/64
=49/64

The increasing range of the function y = sin 2 π x-cos 2 π x + 1 is

The increasing range of y = - cos2paix + 1 is 2kpai

Simplification of (sin 2 α / 1-cos α) - cos α

[sin²α/(1-cosα)]-cosα
=[(1-cos²α)/(1-cosα)]-cosα
=(1+cosα)(1-cosα)/(1-cosα) -cosα
=1+cosα-cosα
=1

Simplify cos ⁴ α + sin? α (1 + cos? α)

cos?α+sin2α(1+cos2α)=cos²α（cos²α+sin²α）+sin²α=cos²α+sin²α=1

Simplification of sin 2 (a + π) cos (π + a) sin (- A-2 π) / sin (π - a) cos 3 (- A - π)

sin²（a+π)cos(π+a)sin(-a-2π)/sin(π-a)cos³(-a-π)
=sin^2a(-cosa)(-sina)/[sina(-cosa)]
=-sin^2a

Simplification (sin α + cos α) 2

sin²α+cos²α+2sinαcosα
1+sin2a