Find the logarithm of the monotone increasing interval (1) y = log with 1 / 2 as the base cos2x of the following function

Find the logarithm of the monotone increasing interval (1) y = log with 1 / 2 as the base cos2x of the following function

Log1 / 2 (x) decreasing
So increasing y means decreasing the true number
The decreasing interval of cosx is (2k π, 2K π + π)
Then 2K π

Given sin (3 π + a) = 2Sin (3 π / 2 + a), find the value of sina-4cosa / 5sina + 2cos2a

Sin (3 π + a) = 2Sin (3 π / 2 + a), left = sin (3 π / 2 + 3 π / 2 + a) = sin (3 π / 2) cos (3 π / 2 + a) + sin (3 π / 2 + a) cos (3 π / 2) = - cos (3 π / 2 + a) = cos (π / 2 + a) = - Sina, right = 2Sin (3 π / 2 + a) = - 2Sin (π / 2 + a) = - 2cosa, so - Sina = - 2cosatana = 2 denominator is not

Given sin (3 π + a) = 2Sin (3 π / 2 + a), find the value of sina-4cosa / 5sina + 2cosa

sin(3π+a)=2sin(3π/2+a)
That is - Sina = - 2cosa
∴tana=2
(sina-4cosa)/(5sina+2cosa )
The numerator and denominator are divided by cosa
=(tana-4)/(5tana+2)
=-1/6
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Given that sin (PAI / 4 2a) · sin (PAI / 4-2a) = 1 / 4, a belongs to (PAI / 4, Pai / 2), find the value of 2sina ^ 2 tana-1 / tana-1

Sin (π / 4 + 2a) sin (π / 4-2a) = sin [π / 2 - (π / 4-2a)] sin (π / 4-2a) = cos (π / 4-2a) sin (π / 4-2a) = [cos (4a)] / 2 = 1 / 4 cos (4a) = 1 / 2, because a belongs to (π / 4, π / 2), so 4A belongs to (π, 2 π), so 4A = 5 π / 3, so a = 5 π / 12, so 2 (Sina) ^ 2 + Tana

Given Tana = 2, find (4sina-2cosa) / (5cosa + 3sina) find the square of 2sina + 3sinacosa + 5cosa

The first question is: (4sina-2cosa) / (5cosa + 3sina) = (4tana-2) / (5-3tana), that is, the upper and lower parts of the fraction are cosa, equal to 6 / 11
The second question: 2Sin & # 178; a + 3sinacosa + 5cos & # 178; a = (2Sin & # 178; a + 2cos & # 178; a) + 3sinacosa + 3cos & # 178; a = 1 + 3 / 2sin2a + 3 (cos2a + 1) / 2 = 7 / 2 + 3 / 2sin2a + 3 / 2cos2a
tana=2,tan2a=sin2a／cos2a=2tana／(1-tan²a)=-4／3
sin²2a+cos²2a=1
The solution is sin2a = 4 / 5, cos2a = - 3 / 5
Original formula = 7 / 2 + 3 / 2 × 4 / 5 + 3 / 2 × (- 3 / 5) = 19 / 5

Given (tana-3) * (SiNx + cosx + 3) = 0, find (4sina COSA) / (3sina + 5cosa)

SiNx + cosx + 3 ≥ 3 - radical 2 > 0
So Tana = 3
(4sina-cosa)/(3sina+5cosa)
=(4tana-1) / (3tana + 5) = self calculation

Given Tana = 3, find the value of (2) (4sina COSA) / (3sina + 5cosa) of (1) Tan (a + Pai / 4)

Analysis: question (1) directly uses the tangent formula of the sum of two corners. Question (2) divides the numerator and denominator by cosa at the same time, The original formula is changed into the formula (1) Tan (a + π / 4) = (Tana + Tan π / 4) / (1-tana * Tan π / 4) = (3 + 1) / (1-3 * 1) = 4 / (- 2) = - 2 (2) (4sina COSA) / (3sina + 5cosa) molecular denominator, and divided by cosa = (4sina / cosa-1) / (3sina / cosa + 5) = (4tana-1) / (3tana + 5) = (4 * 3-1) / (3 * 3 + 5) = 11 / 14. I hope it can help you,

If sin (a + K Π) = - 2cos (a + K Π) (K ∈ z), find 1 / 4sina + 2 / 5 (COSA) ^ 2

If sin (a + K Π) = - 2cos (a + K Π) (K ∈ z), find 1 / 4sina + 2 / 5 (COSA) ^ 2
sin（kπ＋α）＝sinkπ cosα＋coskπ sinα
-2cos(a+k∏)=-(cosαcosk∏－sinαsin k∏)
sinkπ cosα＋coskπ sinα+cosαcosk∏－sinαsin k∏=0
====>sinkπ(cosα－sinα)+cosk∏(cosα+sinα)=0
cosk∏(sinα+cosα)-sinkπ(sinα-cosα)=
Note that sink π (COS α - sin α) = sink π (- sin α + cos α) = - sink π (sin α - cos α)

Given sin (a + KPAI) = 2cos (a + KPAI) (k belongs to Z), find the square of 0.25sina + the square of 0.4cosa

According to the induced formula to simplify the known conditions, the sum of squares of sine and cosine values of the same angle is equal to 1
Original formula = 7 / 25

Given sin (K π + a) = - 3cos (K π + a) (K ∈ z), then (4sina + COSA) / (2sina COSA)=

The formula of Tana is equal to - 3 is obtained by dividing the numerator and denominator by cosa