三角恒等变换1函数y=sin ^4+cos ^2的最小正周期是2已知在三角形ABC中,3sinA+4cosB=6,4si
1个回答

1、先化简函数,再根据公式即可求出周期:

y

=(sin²x)²+cos²x

=[(1-cos2x)/2]²+(1+cos2x)/2

=(3+cos²2x)/4

=[3+(1+cos4x)/2]/4

=(1/8)*(7+cos4x)

∴周期T=2π/4=π/2,

2、两式平方后相加即可:

(3sinA+4cosB)²+(4sinB+3cosA)²

=(9sin²A+9cos²A)+(16cos²B+16sin²B)+(24sinAcosB+24sinBcosA)

=9+16+24sin(A+B)

=25+24sin[π-(A+B)]

=25+24sinC

=6²+1²

=37

∴sinC=1/2

∴∠C=π/6或5π/6

3、(sin65°+sin15°sin10°)/(sin25°-cos15°cos80°)

=(cos25°+sin15°cos10°)/[sin(15°+10°)-cos15°sin10°]

=[cos(15°+10°)+sin15°cos10°]/[(sin15°cos10°+sin10°cos15°)-cos15°sin10°]

=[(cos15°cos10°-sin15°sin10°)+sin15°sin10°]/(sin15°cos10°)

=(cos15°cos10°)/(sin15°cos10°)

=cot15°

4、乘以cos6°再计算:

sin6°sin42°sin66°sin78°

=cos6°sin6°sin42°sin66°sin78°/cos6°

=(1/2)sin12°cos12°cos24°cos48°/cos6°

=(1/4)sin24°cos24°cos48°/cos6°

=(1/8)sin48°cos48°/cos6°

=(1/16)sin96°/cos6°

=(1/16)sin84°/sin84°

=1/16

5、sin²20°+cos²50°+sin20°cos50°

=(1-cos40°)/2+(1+cos100°)/2+(1/2)*(sin70°-sin30°)

=(1-cos40°)/2+(1-cos80°)/2+(1/2)*(cos20°-1/2)

=1-(1/2)*(cos40°+cos80°)+(1/2)*(cos20°-1/2)

=1-(cos60°cos20°)+(1/2)cos20°-1/4

=1-(1/2)cos20°+(1/2)cos20°-1/4

=3/4

6、作和后,真数部分乘以sin(π/9)再算:

log2 cosπ/9+ log2 cos2π/9+ log2 cos4π/9

=log2 [(cosπ/9)*(cos2π/9)*(cos4π/9)]

=log2 [(sinπ/9)*(cosπ/9)*(cos2π/9)*(cos4π/9)/(sinπ/9)]

=log2 [(1/8)*sin(8π/9)/(sinπ/9)]

=log2 (1/8)

=-3

7、(1+tanA)(1+tanB)

=1+tanA+tanB+tanAtanB

=1+tan(A+B)(1-tanAtanB)+tanAtanB

=2