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 inner 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.
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.
10. In acute triangle ABC, A, B and C are opposite sides called A, B and C respectively, C is 60 degrees, c= root number 7, and the area of triangle ABC is 3* root number 3/2. Find the value of a+B.
S triangle ABC = 1/2 ABS Inc.
3√3/2= 1/2absin60
ab=6
cosine theorem
cosC=(a？ +b？ -c？ )/(2ab)
cos60=(a？ +b？ -7)/(2×6)
Answer? +b？ =7+6
Answer? +b？ = 13
(a+b)？ -2ab= 13
(a+b)？ =25
a+b=5
Because a>0, b>0
1 1, triangle ABC, COSA = 12/ 13, COSB =-3/5, then sinC.
cosA= 12/ 13
So a is an acute angle.
Sin? A= 1-cos？ a = 1- 144/ 169 = 25/ 169
Sina =5/ 13
cosB=-3/5
So b is an obtuse angle.
Sin? B= 1-cos？ B= 1-9/25= 16/25
sinB=4/5
sinC = sin( 180-A-B)= sin(A+B)= Sina cosb+cosa sinb
=5/ 13×(-3/5)+ 12/ 13×4/5
=- 15/65+48/65
=33/65
involve