If α ∈ (O, π / 2), 3sin α = 2cos α / 2, then Cos2 α =?

If α ∈ (O, π / 2), 3sin α = 2cos α / 2, then Cos2 α =?

3sinα=2cos(α/2)
That is: 6sin (α / 2) cos (α / 2) = 2cos (α / 2)
The results show that sin (α / 2) = 1 / 3
Then: cos α = 1-2sin & # 178; (α / 2) = 7 / 9
cos2α=2cos²α-1=17/81

Given SiNx + siny = 13, find the maximum value of siny cos 2x

From the known condition that siny = 13 − SiNx and siny = 13 − SiNx ∈ [− 1,1] (combined with SiNx ∈ [- 1,1]), we get − 23 ≤ SiNx ≤ 1, and siny cos2x = 13 − sinx-cos2x  sin2x − SiNx − 23, let t = SiNx (− 23 ≤ t ≤ 1), then the original formula = T2 − t − 23 (− 23 ≤ t ≤ 1). According to the properties of quadratic function, we get the maximum value of the original formula is 49

Given SiNx + siny = 1 / 2, cosx + cosy = 1 / 3, find cos & # 178; ((X-Y) / 2)

(sinx+siny)^2=1/4 ①
(cosx+cosy)^2=1/9 ②
①+ ②：2cosxcosy+(cos x) ²+(cos y) ²+2sinxsiny+(sinx) ²+(siny) ²=13/36
2cosxcosy+2sinxsiny+2=13/36
cosxcosy+sinxsiny=-59/72
That is cos (X-Y) = - 59 / 72
And COS & # 178; ((X-Y) / 2) = [1 + cos (X-Y)] / 2 = 13 / 144

Given SiNx + siny = 2 / 3, find the value range of 2 / 3 + siny cos & # 178; a

From SiNx + siny = 2 / 3, we get siny = 2 / 3-sinx,
2/3+siny-cos²x=1/3-sinx-cos²x=1/3-sinx+sin²x=(sinx-1/2)²+1/12,
Because 0 ≤| SiNx | ≤ 1,0 ≤ (sinx-1 / 2) &# 178; ≤ 9 / 4
So 1 / 12 ≤ 2 / 3 + siny cos & # 178; X ≤ 7 / 3

Given SiNx + siny = 1 / 3, find the minimum and maximum value of siny - (cosx) ^ 2, and seek the guidance of high people
The answer given by the teacher is: because SiNx + siny = 1 / 3, that is, siny = 1 / 3 - SiNx
And - 1 ≤ siny ≤ 1
Then - 1 ≤ 1 / 3 - SiNx ≤ 1
That is - 1 ≤ SiNx - 1 / 3 ≤ 1
-2/3≤sinx≤4/3
And - 1 ≤ SiNx ≤ 1
So - 2 / 3 ≤ SiNx ≤ 1
Because Sin & # 178; X + cos & # 178; X = 1, siny = 1 / 3 - SiNx
So siny - (cosx) &;
=1/3-sinx-(1-sin²x)
=sin²x-sinx-2/3
=(sinx-1/2)²- 11/12
Because - 2 / 3 ≤ SiNx ≤ 1
When SiNx = 1 / 2, siny - (cosx) & # 178; has a minimum value of - 11 / 12
When SiNx = - 2 / 3, siny - (cosx) &# 178; has a maximum value of 4 / 9
I don't understand why 1 ≤ 1 / 3 - SiNx ≤ 1
That is - 1 ≤ SiNx - 1 / 3 ≤ 1
-2/3≤sinx≤4/3
And - 1 ≤ SiNx ≤ 1
So if we change SiNx to siny, isn't the range of siny also - 2 / 3 ≤ siny ≤ 1? Why don't we solve the equation directly with - 1 ≤ SiNx ≤ 1, we have to find the range of SiNx. On the other hand, if we find the range like SiNx, isn't it possible for siny to find the range like that? That's - 2 / 3 ≤ siny ≤ 1. At this time, the range of SiNx doesn't change again? I don't understand

I don't understand why - 1 ≤ 1 / 3 - SiNx ≤ 1
That is - 1 ≤ SiNx - 1 / 3 ≤ 1
-2/3≤sinx≤4/3……
And - 1 ≤ SiNx ≤ 1
So - 2 / 3 ≤ SiNx ≤ 1
In this case, if SiNx is replaced by siny, isn't the range of siny - 2 / 3 ≤ siny ≤ 1? That is - 2 / 3 ≤ siny ≤ 1. At this time, the range of SiNx will change again? I don't understand
Why don't we solve the equation directly with - 1 ≤ SiNx ≤ 1, we have to find the range of SiNx, but on the other hand, if we find the range as SiNx, isn't it possible for siny to find that range as well 0) is "when x equals - B / (2a), y has the maximum value (4ac-b ^ 2) / 4A ^ 2, that is, the vertex of quadratic function is the extreme point". If the range of X is limited, X may not get - B / (2a), and the maximum value of Y may not be (4ac-b ^ 2) / 4A ^ 2, >

If SiNx + siny = 1, then the maximum value of (cosx + cosy) ^ 2 is

(cosx+cosy)^2=cos²x+cos²y+2sinxcosx+1-1=cos²x+cos²y+2sinxcosx+sin²x+cos²x+2sinxcosx-1=2+2(sinxcosx+sinxcosx)-1=1+2cos(x-y)∵cos(x-y)∈[-1,1]∴1+2cos(x-y)∈[-1,3]∴（cosx+...

Given SiNx + siny = 1 / 3, the range of Z = SiNx cos ^ 2Y is obtained
Please write down the specific process!

It is known that SiNx + siny = 1 / 3
Then z = SiNx cos ^ 2Y = 1 / 3-siny-cos2y
=1/3-siny-1+2sin^2y
=2[sin^2y-0.5siny-(1/3)]
=2(siny-1/4)^2-1/8-2/3
=2(siny-1/4)^2-19/24
Because - 1 ≤ siny ≤ 1, then - 1-1 / 4 = - 5 / 4 ≤ siny-1 / 4 ≤ 3 / 4
9/8≤2(siny-1/4)^2≤25/8
1/4≤2(siny-1/4)^2-19/24≤7/3
That is, 1 / 4 ≤ Z ≤ 7 / 3

It is known that sina + cosa = 1 / 5 and (0

Sina + cosa = 1 / 5, the simultaneous square of both sides is: Sin & # 178; a + 2sinacosa + cos & # 178; a = 1 / 251 + 2sinacosa = 1 / 252, sinacosa = - 24 / 25sinacosa = - 12 / 25 (Sina COSA) &# 178; a-2sinacosa + cos & # 178; a = 1-2sinacosa = 1 - (- 24 / 25) = 1 + 24 / 25 = 49 / 25

It is known that sinacosa = 3 / 8 and 0

sinacosa=3/8
2sinacosa=3/4
sin²a+cos²a=1
therefore
（cosa-sina）²=1-3/4
（cosa-sina）²=1/4
Because 0

How to transform Sina and cosa in triangle?
What is the relationship between the sine and cosine of the same angle?

sinA=√1-cos^2(A)
sinA=cosA·tanA
tanA=sinA/cosA
sin^2(A)+cos^2(A)=1