已知tana=2,则sina*cosa=
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解:因为a∈(0,π/2),所以a/2∈(0,π/4),可得sin(a/2)∈(0,√2/2),由二倍角公式可得cosa
=
1
–
2sin
2
(a/2)
=
3/5,移项可得2/5
=
2sin
2
(a/2),因此sin
2
(a/2)
=
1/5,开方可得sin(a/2)
=
√(1/5)
=
√5/5,(负根舍去),所以,sin(a/2)
=
√5/5
;
因为a∈(0,π/2),所以cosa∈(0,1),由三角恒等式可得sin
2
a
+
cos
2
a
=
1,代入可得(1/2)
2
+
cos
2
a
=
1,移项可得cos
2
a
=
1
–
1/4
=
3/4,开方可得cosa
=
√(3/4)
=
√3/2,(负根舍去),所以,cosa
=
√3/2
;
由已知,tana
=2,可知cosa
≠
0,所以(sina
+
cosa)/(sina
–
cosa)
=
(tana
+
1)/(tana
–
1)
=
(2
+
1)/(2
–
1)
=
3
;
而且sinacosa
=
(1/2)sin2a
=
(1/2)*(2tana)/(1
+
tan
2
a)
=
(tana)/(1
+
tan
2
a)
=
2/(1
+
2
2
)
=
2/5
;
因为a∈(π,3π/2),所以a是第三象限角,而且sina
<
0,cosa
<
0,tana
>
0,而题目中已知sina
=
3/5
>
0,与a∈(π,3π/2)相矛盾,所以本题是错题,无解。(检查一下你的题目哪里打错)
=
1
–
2sin
2
(a/2)
=
3/5,移项可得2/5
=
2sin
2
(a/2),因此sin
2
(a/2)
=
1/5,开方可得sin(a/2)
=
√(1/5)
=
√5/5,(负根舍去),所以,sin(a/2)
=
√5/5
;
因为a∈(0,π/2),所以cosa∈(0,1),由三角恒等式可得sin
2
a
+
cos
2
a
=
1,代入可得(1/2)
2
+
cos
2
a
=
1,移项可得cos
2
a
=
1
–
1/4
=
3/4,开方可得cosa
=
√(3/4)
=
√3/2,(负根舍去),所以,cosa
=
√3/2
;
由已知,tana
=2,可知cosa
≠
0,所以(sina
+
cosa)/(sina
–
cosa)
=
(tana
+
1)/(tana
–
1)
=
(2
+
1)/(2
–
1)
=
3
;
而且sinacosa
=
(1/2)sin2a
=
(1/2)*(2tana)/(1
+
tan
2
a)
=
(tana)/(1
+
tan
2
a)
=
2/(1
+
2
2
)
=
2/5
;
因为a∈(π,3π/2),所以a是第三象限角,而且sina
<
0,cosa
<
0,tana
>
0,而题目中已知sina
=
3/5
>
0,与a∈(π,3π/2)相矛盾,所以本题是错题,无解。(检查一下你的题目哪里打错)
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