(1)已知点P(-3,-4)是角a终边上的一点,求cos(2a+π/4)的值
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解
r=√(-3)²+(-4)²=√25=5
∴sina=y/r=-4/5
∴cosa=x/r=-3/5
∴sin2a=2sinacosa=2×(-4/5)(-3/5)=24/25
∴cos2a=2cos²a-1=2×(-3/5)²-1=18/25-1=-7/25
∴cos(2a+π/4)
=√2/2cos2a-√2/2sin2a
=√2/2(-7/25-24/25)
=√2/2×(-31/25)
=-31√2/25
tana=3
∴tan(a-60)
=(tana-tan60)/(1+tanatan60)
=(3-√3)/(1+3√3)
=(3-√3)(1-3√3)/(1-27)
=(3-9√3-√3+9)/(-26)
=(12-10√3)/(-26)
=(5√3-6)/13
tan2a=(2tana)/(1-tan²a)=6/(1-9)=-3/4
tan(2a+45)
=(tan2a+1)/(1-tan2a)
=(-3/4+1)/(1+3/4)
=1/4×(4/7)
=1/7
r=√(-3)²+(-4)²=√25=5
∴sina=y/r=-4/5
∴cosa=x/r=-3/5
∴sin2a=2sinacosa=2×(-4/5)(-3/5)=24/25
∴cos2a=2cos²a-1=2×(-3/5)²-1=18/25-1=-7/25
∴cos(2a+π/4)
=√2/2cos2a-√2/2sin2a
=√2/2(-7/25-24/25)
=√2/2×(-31/25)
=-31√2/25
tana=3
∴tan(a-60)
=(tana-tan60)/(1+tanatan60)
=(3-√3)/(1+3√3)
=(3-√3)(1-3√3)/(1-27)
=(3-9√3-√3+9)/(-26)
=(12-10√3)/(-26)
=(5√3-6)/13
tan2a=(2tana)/(1-tan²a)=6/(1-9)=-3/4
tan(2a+45)
=(tan2a+1)/(1-tan2a)
=(-3/4+1)/(1+3/4)
=1/4×(4/7)
=1/7
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