If Tana = 3, then sinacosa + (COSA) ^ 2= Like the title,

If Tana = 3, then sinacosa + (COSA) ^ 2= Like the title,

It is known that Tana = 3
So sinacosa + cos & # a
=(sinacosa + cos & # 178; a) / (Sin & # 178; a + cos & # 178; a) [numerator denominator divided by cos & # 178; a]
=(tana+1)/(tan²a+1)
=(3+1)/(3²+1)
=4/10
=2/5
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If Tana = & # 189;, then what is cos (2a + & # 165 / 2)?

tana=½,
cos(2a+¥/2)
=cos2a
=(1-tan^2a)/(1+tan^2a)
=3/5

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

cos(2a+π/4)=cos(π/4)cos(2a）-sin(π/4)sin(2a)=(√2/2)(cos（2a)-sin（2a））
=(√2/2)（（1-tan²a）/（1+tan²a）-2tana/（1+tan²a））
tana=1/2
cos(2a+π/4)=(√2/2)（（1-1/4)/(1+1/4)-(2*1/2)/(1+1/4))=-√2/10

If Tana = 1 / 2, then cos (2a + π / 2) =?
Urgent, now!

-4/5
cos(...)= -sin2a= -2tana/1 tan^2a= -4/5

Tana = 2, then cos ^ 2A = 1
Tana = 2, then cos ^ 2A=

cos^2 a
=1/sec^2 a
=1/(1+tan^2 a)
=1/(1+4)
=1/5

A belongs to (0. Pai / 2) B belongs to (PAI / 2, PAI) sin (a + b) = 33 / 65, CoSb = - 5 / 13, Sina =?

A + B belongs to (PAI / 2, Pai / 2 * 3), cos (a + b) = - 56 / 65, SINB = 12 / 13
sin(a)=sin(a+b-b)=sin(a+b)cons(b)-cos(a+b)sin(b)=33/65*(-5/13)-(-56/65)*(12/13)=3/5

It is known that a belongs to (0, Pai, 2), B belongs to (0, PAI) Sina + CoSb = - 8 / 65, sin (a + b) = - 33 / 65, and find SINB + cosa

sinA+cosB=-8\65
(sinA+cosB)^2=（-8\65）^2
(sinA)^2+(cosB)^2+2sinAcosB=(8/65)^2 (1)
sin（A+B）=-33\65
sinA*cosB+sinB*cosA=-33/65
sinB+cosA =K
(sinB+cosA)^2 =K^2
(sinB)^2+(cosA)^2+2sinBcosA=K^2 (2)
(1)+(2):
[(sinA)^2+(cosB)^2+2sinAcosB]+[(sinB)^2+(cosA)^2+2sinBcosA]=(8/65)^2+K^2
[(sinA)^2+(cosB)^2+(sinB)^2+(cosA)^2]+2(sinA*cosB+sinB*cosA)=(8/65)^2+K^2
2+2*(-33/65)=(8/65)^2+K^2
K^2=64/65-64/65^2
K^2=64/65(1-1/65)
K^2=64/65*64/65
A belongs to (0, Pai, 2), B belongs to (0, PAI)
sinB>0 cosA>0
sinB+cosA =K>0
K^2=(64/65)^2
K=64/65
sinB+cosA =K=64/65

Given Sina + CoSb = 1, find y = given Sina + CoSb = 1, find the value range of y = Sin & sup2; a + CoSb

sina+cosb=1
cosb=1-sina
y=sin²a+1-sina
y=(sina-½)²+¾
sina∈[-1,1]
y∈[¾,3]

Given 2sina + cosa = 0, then the value of (1 + 2sinacosa) / (sin2a cos2a) is

Two sides of 2sina + cosa = 0 are divided by cosa to get 2tana + 1 = 0, Tana = - 1 / 2, so sin2a = 2tana / (1 + Tan & # 178; a) = - 4 / 5cos2a = (1-tan & # 178; a) / (1-tan & # 178; a) = 3 / 5, so (1 + 2sinacosa) / (sin2a cos2a) = (1 + sin2a) / (sin2a cos2a) = (1-4 / 5)

Given that a is an acute angle, Tana = 1 / 2, find (2sina + 3cosa) / (2sina-3cosa)

It is known that a is an acute angle, Tana = 1 / 2,
（2sinA+3cosA）/（2sinA-3cosA）
The denominator of Fen is also divided by cosa
=（2tanA+3）/（2tanA-3）
=(1+3)/(1-3)
=-2;
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