2 cosa(cosa sinb+Sina cosb)-sin(A+B)= 0
Sin(A+B)(2cosA- 1)=0
cosA= 1/2
A=60
2. Prove: (1+sinα+cosα+2sinα cosα)/(1+sinα+cosα) = sinα+cosα.
& lt = =>1+Sina +cosa+2 Sina cosa = Sina +cosa+ (Sina +cosa)?
& lt= = = & gt 1+Sina+cosa+2 Sina cosa = Sina+cosa+ 1+2 Sina cosa
& lt = =>0 = 0 hold.
The above steps are reversible and the original proposition holds.
Certificate of completion
3. In △ABC, sinB*sinC=cos? (A/2), what is the shape of △ABC?
Simbuxin (180-a-b) = (1+COSA)/2.
2sinBsin(A+B)= 1+cosA
2 sinb(Sina cosb+cosA sinb)= 1+cosA
sin2BsinA+2cosAsin？ B-cosA- 1=0
sin2BsinA+cosA(2sin？ B- 1)= 1
sin2BsinA-cosAcos2B= 1
cos2BcosA-sin2BsinA=- 1
cos(2B+A)=- 1
Because a and b are the interior angles of a triangle.
2B+A= 180
Because A+B+C= 180.
So B=C
Triangle ABC is an isosceles triangle
4. Find the maximum and minimum values of the function y=2-cos(x/3), and write the set of x that makes this function get the maximum and minimum values respectively.
- 1≤cos(x/3)≤ 1
- 1≤-cos(x/3)≤ 1
1≤2-cos(x/3)≤3
range
When cos(x/3)= 1, that is, x/3=2kπ, that is, x=6kπ, the minimum value of y is 1 {x | x = 6kπ, k ∈ z}.
When cos(x/3)=- 1, that is, x/3=2kπ+π, that is, x=6kπ+3π, the minimum value of y is 1 {x | x = 6kπ+3π, k ∈ z}.
5. Given △ABC, if (2c-b)tanB=btanA, find the angle a..
[(2c-b)/b]sinB/cosB=sinA/cosA
Sine theorem c/sinC=b/sinB=2R substitution
(2sinC-sinB)cosA=sinAcosB
2 sin(A+B)cosA = Sina cosb+cosA sinb
2sin(A+B)cosA-sin(A+B)=0
sin(A+B)(2cosA- 1)=0
sin(A+B)≠0
cosA= 1/2
A=60 degrees
6. Verification of known 2cosx=3cosy: 3cosx-2cosy/2siny-3sinx = tan (x+y)
Proof: 3cosx-2cosy/2siny-3sinx = tan (x+y)
& lt= = & gt(3 cosx-2 cosy)/(2 siny-3 sinx)= sin(x+y)/cos(x+y)
& lt= = & gt(3 cosx-2 cosy)/(2 siny-3 sinx)=(sinx cosy+cosx siny)/(cosx cosy-sinx siny)
& lt= = & gt3cos？ xcosy-3cosxsinxsiny-2cosxcos？ y+2 sinxcosxsiny = 2 sinxsinysy+2 sin？ ycosx-3sin？ xcosy-3sinxcosxsiny
& lt= = & gt3cos？ xcosy+3sin？ xcosy=2sin？ ycosx+2cos？ ycosx
& lt= = & gt3cosy (sin? x+cos？ x)=2cosx(sin？ y+cos？ y)
& lt= = & gt3cosy=2cosx known.
So the above steps are reversible.
The original claim was established.
7. It is known that in △ABC, sinB+sinC=√2sinA and side length a=4. If S△ABC=3sinA, find the value of cosA.
Sine theorem a/sinA=b/sinB=c/sinC=2R(R is the radius of the circumscribed circle of a triangle).
sinA=a/2R，sinB=b/2R，sinC=c/2R
Substitute b/2R+c/2R=4√2/2R.
b+c=4√2( 1)
1/2bcsinA=3sinA
BC =6(2)
(1) square
b？ +2bc+c？ =32
b？ +c？ =20
Cosine theorem cosA=(b? +c？ -a？ )/(2bc)=(20- 16)/ 12 = 1/3
8. In the triangle abc, the opposite side of the angle ABC is known by ABC as sin2 * 2c+sin2csinc+cos2c = 1, while a+b = 5 and c = follow the number 7 to find the size of the angle c (1). (2) The area of the triangle ABC.
Sin? 2C+sin2CsinC+cos2C= 1
sin2CsinC+cos2C=cos？ 2C
2sin？ CcosC+cos2C( 1-cos2C)=0
2sin？ CcosC+2sin？ Ccos2C=0
C is not 0.
therefore
cosC+cos2C=0
2cos？ C+cosC- 1=0
(2cosC- 1)(cosC+ 1)=0
CosC= 1/2 or cosC=- 1 (omitted)
C=π/3
cosine theorem
cosC=(a？ +b？ -c？ )/(2ab)
1/2=[(a+b)？ -2ab-c？ ]/(2ab)
3ab= 18
ab=6
S triangle ABC =1/2absinc =1/2× 6× sin60 = 3 √ 3/2.
9、π/4 & lt； a & lt3π/4
0 & lta-π/4 & lt； π/2
A-π/4 is the first quadrant angle.
So sin(a-π/4)=√[ 1-cos? (a-π/4)]=√48/7
cos(a-π/4)= 1/7
-π/4 & lt； b & ltπ/4
π/2 & lt； 3π/4+b & lt； Pi?
So 3π/4+b is the second quadrant angle.
So cos (3 π/4+b) =-75/ 14.
sin(a+b)=-cos(a+b+π/2)=-cos(a-π/4+b+3π/4)
= sin(a-π/4)sin(b+3π/4)-cos(a-π/4)cos(b+3π/4)
=√48/7× 1 1/ 14+ 1/7×√75/ 14
=√3/2
π/4 & lt； a & lt3π/4
-π/4 & lt； b & ltπ/4
0 & lta+b & lt； Pi?
So a+b=π/3 or 2π/3.