Why is the maximum value of the function y = cos Λ 4x sin Λ 4x 1 because = cos2x, cos is preceded by 1,

Why is the maximum value of the function y = cos Λ 4x sin Λ 4x 1 because = cos2x, cos is preceded by 1,

y=cos∧4x-sin∧4x
=(cos^2x-sin^2x)(cos^2x+sin^2x)
=cos^2x-sin^2x
=cos2x
-1

FX = cos ^ 4x + 2sinxcosx sin ^ 4x find the minimum positive period of FX

FX = cos ^ 4x + 2sinxcosx sin ^ 4x = (COS & # 178; X + Sin & # 178; x) (cosx & # 178; - Sin & # 178; x) + sin2x = 1 × cos2x + sn2x; = √ 2 (cos2x × (√ 2 / 2) + sin2x × (√ 2 / 2)) = √ 2Sin (2x + π / 4); minimum positive period T = 2 π / 2 = π;

(3) Given that 0 ≤ a ≤ 2 Π, point P (sin a – cos a, Tan a) is in the first quadrant, then the value range of a is ()?
（∏/4 ,∏/2）∪(∏,5∏/4)

Tan a 〉 0 = = = > Sina / cosa > 0 = = = > Sina, cosa is the same sign
The value range of sin a – cos a 〉 0 = = 〉 A is: (Π / 4, Π / 2) ∪ (Π, 5 Π / 4)

If the point P (sin α - cos α, Tan α) is in the first quadrant, then the value range of α in [0,2 π] is ()
A. (π2，3π4)∪(π，5π4)*B. (π4，π2)∪(π，5π4)C. (π2，3π4)∪(5π4，3π2)D. (π2，3π4)∪(3π4，π)

∵ sin α − cos α ＞ 0tan α ＞ 0 {π 4 ＜ α ＜ 5 π 40 ＜ α ＜ π 2 π ＜ α ＜ 5 π 4 {α ∈ (π 4, π 2) ∪ (π, 5 π 4), so select B

Given that the root of the equation KX + 1 = 2x-1 is a positive number, then the value range of K is______ ．

∵ the root of equation KX + 1 = 2x-1 is a positive number, ∵ x = − 2K − 2 ＞ 0, that is, K-2 ＜ 0, the solution is k ＜ 2

The solution of the equation kx-k = 2x-5 of X is a positive number, and the value range of K is obtained
fast

kx-k=2x-5
kx-2x=k-5
x=(k-5)/(k-2)>0
k> 5 or K

Given that the solution of the equation KX + 1 = 2x-1 is a positive number, what is the range of values?
Find the value of K

If the transfer term is (2-k) x = 2, then x = 2 / (2-k). Because the solution of the equation is positive, so if x > 0, then x = 2 / (2-k)
(2-k) > 0, then K

The area of a right triangle is 90 cm. The length of one right side is 7.2 cm. What is the length of the other right side

25 cm 90x2 △ 7.2 = 25 cm answer another right angle side length is 25 cm

The area of a right triangle is 90 bisection centimeters, the length of one right side is 7.2cm, what is the length of the other right side

x*7.2/2=90
x=25cm
The other right angle side is 25 cm long

The area difference between the two squares is 5 square centimeters, and the area of the right triangle formed by the length of the two squares is 3 cm
The area difference between the two squares is 5 square centimeters. The area of the right triangle formed by taking the length of the two squares as the right side is 3 square centimeters. Find the length of the middle line on the third side of the right triangle