积化和差和差化积记忆口诀及相关练习题
A.sin(A+B)+sin(A-B)=2sin A cos B B.sin(A+B)-sin(A-B)=2cos A sin B C.cos(A+B)+cos(A-B)=2cos A cos B D.cos(A+B)-cos(A-B)=2sin A cos B 2.sin15°sin75°=( )
A.18
B.14
C.12
D .1 3.sin105°+sin15°等于( ) A.32 B.22 C.62 D.64
4.sin37.5°cos7.5°=________.
5.sin70°cos20°-sin10°sin50°的值为( ) A.34 B.32 C.12 D.34
6.cos72°-cos36°的值为( )
A .3-2 3 B.12 C .-12
D .3+2 3 7.在△ABC 中,若sin A sin B =cos 2C
2
,则△ABC 是( ) A .等边三角形 B .等腰三角形 C .不等边三角形 D .直角三角形
8.函数y =sin ?
????x -π6cos x 的最大值为( ) A.12 B.14 C .1 D.22
9.若cos(α+β)cos(α-β)=13
,则cos 2α-sin 2β等于( ) A .-23 B .-13 C.13 D.23
10.函数y =sin ?
????x +π3-sin x (x ∈[0,π2])的值域是( ) A .[-2,2] B.??????-12,32 C.??????12,1 D.????
??12,32 答案
1解析:选D.由两角和与差的正、余弦公式展开左边可知A 、B 、C 正确.
2解析:选B.sin15°sin75°=-12
[cos(15°+75°)-cos(15°-75°)] =-12(cos90°-cos60°)=-12(0-12)=14
. 3解析:选C.sin105°+sin15°=2sin 105°+15°2cos 105°-15°2
=2sin60°cos45°=
62. 答案:2+14=12????22+12=2+14.=12
(sin45°+sin30°) 4解析:sin37.5°cos7.5°=12
[sin(37.5°+7.5°)+sin(37.5°-7.5°)] 5解析:选A.
sin70°cos20°-sin10°sin50°=12(sin90°+sin50°)+12
(cos60°-cos40°) =12+12sin50°+14-12cos40°=34
. 6解析:选C.
原式=-2sin 72°+36°2sin 72°-36°2
=-2sin54°·sin18°=-2cos36°cos72° =-2·sin36°cos36°cos72°sin36°=-sin72°cos72°sin36°=-sin144°2sin36°=-12
,故选C. 7解析:选B.由已知等式得12[cos(A -B )-cos(A +B )]=12
(1+cos C ), 又A +B =π-C .所以cos(A -B )-cos(π-C )=1+cos C .