问题
解答题
（1）求值：
（2）已知sinθ+2cosθ=0，求
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答案
（1）原式=sin(80°-15°)+sin15°sin10° sin(15°+10°)-cos15°cos80°
=sin80°cos15°-cos80°sin15°+sin15°sin10° sin15°cos10°+cos15°sin10°-cos15°cos80°
=sin80°cos15°-sin10°sin15°+sin10°sin15° sin15°cos10°+cos15°sin10°-cos15°sin10°
=
=sin80°cos15° sin15°cos10°
=cos10°cos15° sin15°cos10°
=cot15°cos15° sin15°
=
=1 tan15°
=1 tan(45°-30°)
=1+tan45°tan30° tan45°-tan30°
=2+3+ 3 3- 3
；3
（2）由sinθ+2cosθ=0，得sinθ=-2cosθ，又cosθ≠0，则tanθ=-2，
所以
=cos2θ-sin2θ 1+cos2θ cos2θ-sin2θ-2sinθcosθ sin2θ+2cos2θ
=
=1-tan2θ-2tanθ tan2θ+2
=1-(-2)2-2(-2) (-2)2+2
．1 6