Given Tana = - 1 / 3, find 1 / 2 sinacosa + cos & sup2; a

Given Tana = - 1 / 3, find 1 / 2 sinacosa + cos & sup2; a

tana=sina/cosa=-1/3
-3sina=cosa
9sin^2a=cos^2a=1-sin^2a
10sin^2a=1
sin^2a=1/10
cos^2a=9/10
Because Sina and cosa are different symbols
Sina * cosa

If Tana = - 1 / 2 and a ∈ (π / 2, π), then cos (a + π / 4)=

Tana = - 1 / 2, and a ∈ (π / 2, π), then Sina > 0, cosa

Finding 3sin a COS with known Tana = - 4

3sina cosa=3sina cosa/（sin²a+cos²a）
The numerator and denominator are simultaneously divided by cos & # 178; a
The original formula = 3tana / (Tan & # 178; a + 1)
= -12/17

Let the ratio of sina to Sina / 2 be 8:5, and find the value of cosa and Tana / 4

Since the ratio of sina to Sina / 2 is 8:5, 2sina / 2 * cosa / 2: Sina / 2 = 8:5
So cosa / 2 = 4 / 5
So cosa = 2cos squared a / 2-1 = 2 * (4 / 5) squared-1 = 7 / 25
Because cosa / 2 = 4 / 5, cosa = 7 / 25
So Sina / 2 is greater than 0
So Sina / 2 = 3 / 5
So Tana / 2 = (2tana / 4) / 1 - (Tana / 4) square = (Sina / 2) / (COSA / 2) = 3 / 4
Tana / 4 = 1 / 3 or Tana / 4 = - 3 (rounding off)

If a ∈ (- π / 2 + 2K π, 2K π), then Tana, Sina, cosa size

First draw the coordinate system, then give K value, take a few more values, you will understand

If a belongs to (π / 4, π / 2), then the order of sina, cosa and Tana is

Draw trigonometric function line or give special value:
tana＞sina＞cosa
Special value a = π / 3

If π / 4 "a" π / 2, then compare Sina, cosa, Tana

cosa

Given Tan (2a-b) = √ 3 (1 + m), Tan (a-b) - √ 3 [m-tan (2a-b) Tan (a-b)], find the value of Tana

tanA
=tan[(2A-B)--(A-B)]
=[ tan(2A-B)--tan(A-B) ] / [1+tan(2A-B)tan(A-B)]
=[√3(1+M)--tan(A-B)] / tan(A-B)-√3[M-tan(2A-B)tan(A-B)],
=1

If Tana = 2, Tan (β - a) = 3, then the value of Tan (β - 2A) is

Consider Tan (β - a) as a whole
tan(β-2a)=tan （(β-a)-a）=(tan(b-a)-tana)/(1+tan(b-a)tana)
=3-2/1+2*3=1/7

Verification: 1-cos2a / 1 + cos2a = Tan ^ 2A
It's a process

Because: cos2a = 2cos ^ 2a-1 = 1-2sin ^ 2a,
(1-cos2a)/(1+cos2a)=[1-(1-2sin^2a)]/[1+2cos^2a-1]=sin^2a/cos^2a=tan^2a.