,0＜β＜
,0＜k≤
,且sin
=k·cosβ,求证:α+β＜
. 
,0＜β＜,0＜k≤,且sin=k·cosβ,求证:α+β＜. 证明:∵sin=k·cosβ=k·sin(-β)
=2k·sin(-)·cos(-
)
＜2k·sin(
-
)
≤sin(
-
),
又
∈(0,
),
-
∈(0,
),
正弦函数在(0,
)内单调递增,
∴
＜
-
,
即α+β＜
.