One right angle side is 3 meters, the other right angle side corresponding angle is tan22, 5 degrees, find the length of this right angle side, please advise

One right angle side is 3 meters, the other right angle side corresponding angle is tan22, 5 degrees, find the length of this right angle side, please advise

Tan22.5 ° = another right angle side △ 3M right angle side
So the other right angle side = 3M right angle side × Tan 22.5 degree
Approximately equal to 1.242640687

Given Tana = 2, find Sina, cosa, tan22,5 degrees and cot22,5 degrees

Sina = 2 / 5 root 5
Cosa = root 5 of 5

Tan67 ° 30 ′ = 1 / tan22 ° 30 ′ why,

tanx=cot(90°-x)=1-tan(90°-x)
Let x = 67 ° 30 'be substituted

Simplification: Sina ^ 2 / (Sina COSA) - (Sina + COSA) / (Tan ^ 2-1)

It's not very convenient for me to write about it, so I'll write it in small steps~
（1）tan^2-1=（sina^2-cosa^2）/（cosa^2）=(sina-cosa)(sina+cosa)/（cosa^2）
(2) (Sina + COSA) / (Tan ^ 2-1) = (COSA ^ 2) / (Sina COSA)
(3) Finally, the original formula = Sina ^ 2 / (Sina COSA) - (COSA ^ 2) / (Sina COSA) = (Sina ^ 2-cosa ^ 2)
/(sina-cosa)=sina+cosa
That is to say, the original formula is simplified to Sina + cosa
I'm sorry, but I hope I can help you solve the problem

In this paper, sin a × cos A is reduced to a formula with Tan a

He doesn't know, I know!
Because 1 = Sina ^ 2 + cosa ^ 2; so
Original formula = (sin a × cos a) / 1
=(sin a × cos a)/(sina^2+cosa^2);
The numerator and denominator are divided by cos a ^ 2 at the same time
The original formula is Tan A / (Tan a ^ 2 + 1)
That's all

Given 3sin2b = sin (2a + b), it is proved that Tan (a + b) = 2tana

3sinB=sin(2A+B)
3sin(A+B-A)=sin(A+B+A)
3sin(A+B)cosA-3cos(A+B)sinA=sin(A+B)cosA+sinAcos(A+B)
2sin(A+B)cosA=4cos(A+B)sinA
tan(A+B)=2tanA

Given the function f (x) = − 2sin2x + 23sinxcosx + 1. (1) find the minimum positive period and symmetry center of F (x); (2) if x ∈ [− π 6, π 3], find the maximum and minimum of F (x)

(1) Let sin (2x + π 6) = 0, then x = k π 2 X2X + cos2x = 3ssin2x + cos2x = 2ssin (2x + cos2x) 2 (2x + π 6) = 0, then x = k π 2 − π 12 (K ∈ z) and the center of symmetry of the (3) x (x) is (K π 2 − π 12, 0), (K \\\\\6 (2x + π 6 (2x + π 6) 6 (t = 2) (2) \\\\ (2 \\\\\\\\\\\\\\\when − π 6, the minimum value of F (x) is - 1; When x = π 6, the maximum value of F (x) is 2

f(x)=1/2cos²x-sinxcosx-1/2sin²x
(1) Finding the minimum positive period of F (x)
(2) Finding the equation of symmetry axis of F (x) image
(3) Finding monotone decreasing interval of F (x)
Please use cos

F (x) = 1 / 2cos2x-1 / 2sin2x = radical 2 / 2cos (2x + Π / 4)
(1) Π (2) axis of symmetry is a straight line x = k Π / 2 + Π / 8 (K ∈ R) (3) [- Π / 8 + K Π, 3 Π / 8 + K Π] (K ∈ R)

-How to transform 2Sin (x / 2) [1-2cos ^ 2 (x / 4)] into 2Sin (x / 2) [2cos ^ 2 (x / 4) - 1]?

Put the minus sign in parentheses
That is - [1-2cos ^ 2 (x / 4)] = [2cos ^ 2 (x / 4) - 1]

F (x) = 1 / 2cos squared x-sinxcosx-1 / 2Sin squared x for FX period, axis of symmetry equation, monotone interval

f(x)=(1/2)[(cosx)^2-(sinx)^2]-(1/2)sinxcosx
=(1/2)cos2x-(1/2)sin2x
=1/2[cos2x-sin2x]
=-√2/2sin(2x-π/4)
The period is π;
The equation of symmetry axis is 2x - π / 4 = k π + π / 2,
x=kπ/2+3π/8 k∈Z；
It is a decreasing function on (K π - π / 8, K π + 3 π / 8)
It is an increasing function on (K π + 3 π / 8, K π + 7 π / 8)