急~高二數學三角函數一題

答案僅供參考 未必完全正確









已知sinA+sinB=1, cosA+cosB=0，求
(1)cos平方A+ cos平方B=?
(2) cos三次方A+ cos三次方B=?



sinA=1-sinB

cosA=-cosB

(sinA)^2+(cosA)^2=1

(1-sinB)^2+(-cosB)^2=1

1-2sinB+(sinB)^2+(cosB)^2=1

1-2sinB+1=1

-2sinB=-1

sinB=1/2

sinA=1/2

so 角A=30度時角B=150度

角A=150度時角B=30度

(1)(cosA)^2+(cosB)^2

=(cosA+cosB)^2-2(cosA)(cosB)

=0-2(cos30度)(cos150度)

=-2[ (根號3)/2 ][ -(根號3)/2 ]

=-2(-3/4)

=3/2

(2)(cosA)^3+(cosB)^3

=(cosA+cosB)^3-3(cosA)(cosB)(cosA+cosB)

=0-3(cos30度)(cos150度)(0)

=0

已知SinA+SinB=1，CosA+CosB=0，求
(1)Cos^2 A+Cos^2 B=?
Sol
1=Sin^2 A+Cos^2 A
=(1－SinB)^2+(－CosB)^2
=1－2SinB+1
SinB=1/2
SinA=1/2
(1) Cos^2 A+Cos^2 B
=2－Sin^2 A－Sin^2 B
=2－1/4－1/4
=3/2
(2) Cos^3 A+Cos^3 B=?
Sol
Cos^3 A+Cos^3 B
=(CosA+CosB)(Cos^2 A－CosACosB+Cos^2 B)
=0*(Cos^2 A－CosACosB+Cos^2 B)
=0