Chứng minh rằng
a) 1−cos2a+sin2a1+cos2a+sin2a=tana
b) cota+tana1+tan2tana=2cot2a
c) √2−sina−cosasina−cosa=−tan(a2−π8)
d) cos2a−cos3a−cos4a+cos5a=−4sina2sinacos7a2
Gợi ý làm bài
a)
1−cos2a+sin2a1+cos2a+sin2a=2sin2a+2sinacosa1+2cos2a−1+2sinacosa
=2sina(sina+cosa)2cosa(sina+cosa)=tana
b) cota+tana1+tan2atana=1tana+tana1+2tana1−tan2a
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=1+tan2atana:1−tan2a+2tan2a1−tan2a
=1−tan2atana=2cot2a
c) √2−sina−cosasina−cosa=√2−√2sin(a+π4)√2sin(a−π4)
=1−sin(a+π4)sin(a−π4)=sinπ2−sin(a+π4)sin(a−π4)
=cos(a2+3π8)sin(π8−a2)2sin(a2−π8)cos(a2−π8)=sin(−a2+π8)sin(π8−a2)sin(a2−π8)sin(a2−π8)
=−sin(a2−π8)cos(a2−π8)=−tan(a2−π8)
d)
cos2a−cos3a−cos4a+cos5a=(cos2a−cos4a)+(cos5a−cos3a)
=−2sin3asin(−a)−2sin4asina=2sina(sin3a−sin4a)
=4sinacos7a2sin(−a2)=−4sina2sinacos7a2