Question

（1）求值：

sin65°+sin15°sin10°

sin25°-cos15°cos80°

；
（2）已知sinθ+2cosθ=0，求

cos2θ-sin2θ

1+cos2θ

的值．

Answer 1

（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°

=

cos15°

sin15°

=cot15°
=

1

tan15°

=

1

tan(45°-30°)

=

1+tan45°tan30°

tan45°-tan30°

=

3+ 3

3- 3

=2+

3

；

（2）由sinθ+2cosθ=0，得sinθ=-2cosθ，又cosθ≠0，則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

．