Given that cosx = 1 / 7, cos (x + b) = - 11 / 14, and X, B belong to (O, Wu / 2), find the value of CoSb (need detailed process)

Given that cosx = 1 / 7, cos (x + b) = - 11 / 14, and X, B belong to (O, Wu / 2), find the value of CoSb (need detailed process)

x. B belongs to (O, Wu / 2)
x+B0
sin(x+B)=√(1-cos^2(x+B))=5√3/14
cosx=cos(x+B-B)
=cos(x+B)cosB+sin(x+B)sinB
=-11/14 cosB+5√3/14 sinB=1/7
-11cosB+5√3sinB=2
5 √ 3sinb = 2 + 11cosb squared
75sin^2B=75-75cos^2B=4+44cosB+121cos^2B
196cos^2B+44cosB-71=0
∵cosB>0
∴cosB=(-44+√(44^2+4*196*71))/2*196
=(-44+240)/2*196
=1/2

It is known that cosx = 1 / 7, cos (X-Y) = 13 / 14, and 0

(1) SiNx can be obtained from cosx. Sin2x and cos2x can be obtained from SiNx and cosx, and tan2x can be obtained by dividing them
(2) If cos (X-Y) = 13 / 14 is divided into cosxcosy + sinxsiny = 13 / 14, and the values of SiNx and cosx are substituted in, the equation of Y is left. By combining sin and COS, y can be obtained
Detailed explanation (2)
Because 0 < y < x < π / 2, (SiN x) ^ 2 + (COS x) ^ 2 = 1
So SiN x = (4 * 3 ^ (1 / 2)) / 7
cos(x-y)=cos(y-x)=13/14
And - π / 2

cosX=1/7 cos(X+Y)=-11/14 0
cosX=1/7 cos(X+Y)=-11/14 0

0

Given cos (a-b) cos (a + b) = 1 / 2, ` find the square of SINB - the square of cosa

There are cos (a-b) cos (a + b) = 1 / 2
(cosa*cosb+sina*sinb)(cosa*cosb-sina*sinb)=1/2
Open (COSA) ^ 2 * (CoSb) ^ 2 - (Sina) ^ 2 * (SINB) ^ 2 = 1 / 2
Substitute (Sina) ^ 2 = 1 - (COSA) ^ 2 (SINB) ^ 2 = 1 - (CoSb) ^ 2 into the above formula
There is 1 - (CoSb) ^ 2 - (COSA) ^ 2 = - 1 / 2
Because (SINB) ^ 2 - (COSA) ^ 2 = 1 - (CoSb) ^ 2 - (COSA) ^ 2
So (SINB) ^ 2 - (COSA) ^ 2 = - 1 / 2

If cosa = radical 5 / 5, then cos (a - π / 4)

Cos (a - π / 4) = cosacos π / 4 + sinasin π / 4 = radical 2 / 2 (Sina + COSA)
Cosa = radical 5 / 5, Sina square + cosa square = 1, so we can get Sina = 2 radical 5 / 5 or - 2 radical 5 / 5,
So the original formula = 3 radical 10 / 10 or - radical 10 / 10

Given that a belongs to (0, Pai / 2), B belongs to (PAI / 2, PAI), cosa = 3 / 5, SINB = 5 / 13, find cos (a + b)

0

It is known that cos (a + π / 4) = 1 / 3,0

cosa=[(√2)²+(√10)²-2²]/(2√2*√10)=2√5/5

If cosa = - 12 / 13, a belongs to (Wu, 3 / 2 Wu), then cos (a + Wu / 4)=

Cosa = - 12 / 13, a belongs to (Wu, 3 / 2 Wu). Sina = - 5 / 13
Cos (a + Wu / 4) = cosacos Wu / 4-sinasin Wu / 4 = - 12 / 13 * √ 2 / 2 - (- 5 / 13) * √ 2 / 2 = - 7 √ 2 / 26

The maximum value of the function y = sin (Wu / 2 + x) cos (Wu / 6-x)

First use the induction formula
y=cosx cos（π/6-x）
open
y=√3/2 cos²x+1/2 sinxcosx
Power reduction formula + multiple angle formula
y=√3/4 (1+cos2x)+1/4 sin2x
=1/2(√3/2 cos2x+1/2 sin2x ) + √3/4
=1/2cos(2x-π/6)+√3/4
Maximum: 1 / 2 + √ 3 / 4

Cos (a + π / 3) = - 3 / 5 and π / 6 are known

cos（a+π/3)=-3/5cosacosπ/3-sinasinπ/3 = -3/51/2 cosa - √3/2 sina = -3/55cosa - 5√3sina = -65cosa + 6 = 5√3sina(5cosa + 6)^2 = (5√3sina)^225cos^2a+60cosa+36 = 75sin^2a25cos^2a+60cosa+36 = 75(1-...