第三章 三角恒等变换 章末综合一 学案

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三角恒等变换章末一
知识回顾】
1.两角和、差的正弦、余弦、正切公式



注意：
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2.配方变形
1±sin 2α=sin2α+cos2α±sin 2α=(sin α±cos α)2.
3.因式分解变形
cos 2α=cos2α-sin2α=(cos α-sin α)(cos α+sin α).
4.升幂公式
1+cos 2α=2cos2α;1-cos 2α=2sin2α.
5.降幂公式
cos2α= QUOTE ;sin2α= QUOTE ;
sin αcos α= QUOTE sin 2α;tan2α= QUOTE .
【典型例题】
类型一　运用公式化简求值
例一、:(1)cos 75°;
(2)cos 63°sin 57°+sin 117°sin 33°;
(3)sin 460°sin(-160°)+cos 560°cos(-280°).
(4)cos 105°;
(5)sin 14°cos 16°+sin 76°cos 74°;
(6)sin QUOTE - QUOTE cos QUOTE .
(7) QUOTE ;
(8)tan 20°·tan 30°+tan 30°·tan 40°+tan 40°·tan 20°.
(9)cos QUOTE cos QUOTE =　　　　;
(10) QUOTE -cos215°=　　　　;?
(11) QUOTE =　　　　.?
(12)已知π<α< QUOTE π,化简: QUOTE + QUOTE ;
(13) sin 50°(1+ QUOTE tan 10°).
类型二　由三角函数值求角或范围
例二、(1)已知cos α= QUOTE ,cos(α+β)=- QUOTE ,且0<β<α< QUOTE ,求β的值;



(2)已知sin α+sin β= QUOTE ,求(cos α+cos β)2的取值范围.


类型三　条件求值
例三、(1)已知sin α=- QUOTE ,α∈( QUOTE ,2π),则cos( QUOTE -α)的值为　　　　;




(2)已知α,β∈(0, QUOTE ),且sin α= QUOTE ,cos(α+β)=- QUOTE ,则cos β的值为　　　　.?


例四、已知 QUOTE <β<α< QUOTE π,cos(α-β)= QUOTE ,sin(α+β)=- QUOTE ,求cos 2α,sin2β 的值.




例五、(1)设α为锐角,若cos(α+ QUOTE )= QUOTE ,则sin(2α+ QUOTE )的值为　　　 　;?




(2)已知sin( QUOTE -x)= QUOTE ,0



三角恒等变换章末一参考答案
例一、解：(1)cos 75°=cos(120°-45°)
=cos 120°cos 45°+sin 120°sin 45°
=- QUOTE × QUOTE + QUOTE × QUOTE = QUOTE .
(2)原式=cos 63°cos 33°+sin 63°sin 33°=cos(63°-33°)=cos 30°= QUOTE .
(3)原式=-sin 100°sin 160°+cos 200°cos 280°
=-sin 100°sin 20°-cos 20°cos 80°
=-(cos 80°cos 20°+sin 80°sin 20°)
=-cos 60°=- QUOTE .
(4)原式=cos(60°+45°)
=cos 60°cos 45°-sin 60°sin 45°
=×- QUOTE × QUOTE = QUOTE .
(5)原式=sin 14°cos 16°+sin(90°-14°)cos(90°-16°)
=sin 14°cos 16°+cos 14°sin 16°
=sin(14°+16°)=sin 30°= QUOTE .
(6)法一　原式=2( QUOTE sin QUOTE - QUOTE cos QUOTE )
=2(sin QUOTE sin QUOTE -cos QUOTE cos QUOTE )
=-2cos( QUOTE + QUOTE )
=-2cos QUOTE =- QUOTE .
法二　原式=2( QUOTE sin QUOTE - QUOTE cos QUOTE )
=2(cos QUOTE sin QUOTE -sin QUOTE cos QUOTE )
=2sin( QUOTE - QUOTE )
=-2sin QUOTE =- QUOTE .
(7)原式= QUOTE
=tan(45°-75°)
=tan(-30°)=- QUOTE .
(8)原式= QUOTE (tan 20°+tan 40°)+tan 40°·tan 20°
= QUOTE ·tan 60°(1-tan 20°·tan 40°)+tan 40°·tan 20°
=1-tan 20°·tan 40°+tan 20°·tan 40°=1.
(9)原式=cos QUOTE sin QUOTE = QUOTE ×2cos QUOTE sin QUOTE = QUOTE sin QUOTE = QUOTE .
(10)原式= QUOTE (1-2cos215°)=- QUOTE cos 30°=- QUOTE .
(11) 原式= QUOTE =2 QUOTE .
（12)因为π<α< QUOTE π,
所以 QUOTE < QUOTE < QUOTE π,
所以 QUOTE = QUOTE |cos QUOTE |=- QUOTE cos QUOTE ,
QUOTE = QUOTE |sin QUOTE |= QUOTE sin,
所以 QUOTE + QUOTE
= QUOTE + QUOTE
= QUOTE + QUOTE =- QUOTE cos QUOTE .
(13)原式=sin 50°× QUOTE
= QUOTE
= QUOTE
= QUOTE
= QUOTE =1.
例二、解：（1）因为0<β<α< QUOTE ,所以0<α+β<π,
由cos α= QUOTE ,cos(α+β)=- QUOTE ,得sin α= QUOTE ,
sin(α+β)= QUOTE ,
所以cos β=cos[(α+β)-α]
=cos(α+β)cos α+sin(α+β)sin α
=- QUOTE × QUOTE + QUOTE × QUOTE = QUOTE .
所以β= QUOTE .
(2）sin α+sin β= QUOTE ,平方可知,
sin2α+2sin αsin β+sin2β= QUOTE . ①
设cos α+cos β=m,平方可知,
cos2α+2cos αcos β+cos2β=m2. ②
①+②得
sin2α+2sin αsin β+sin2β+cos2α+2cos αcos β+cos2β=m2+ QUOTE ,
整理得m2= QUOTE +2cos(α-β).
又由于cos(α-β)∈[-1,1],
所以m2∈[- QUOTE , QUOTE ],
即得0≤m2≤ QUOTE ,
所以(cos α+cos β)2的取值范围是[0, QUOTE ].
例三、解：(1)因为sin α=- QUOTE ,α∈( QUOTE ,2π),
所以cos α= QUOTE = QUOTE ,
所以cos( QUOTE -α)=cos QUOTE cos α+sin QUOTE sin α
= QUOTE × QUOTE + QUOTE ×(- QUOTE )= QUOTE .
(2)因为α,β∈(0, QUOTE ),
所以α+β∈(0,π),sin(α+β)>0,
sin(α+β)= QUOTE = QUOTE ,
cos α= QUOTE = QUOTE ,
所以cos β=cos(α+β-α)
=cos(α+β)cos α+sin(α+β)sin α
=- QUOTE × QUOTE + QUOTE × QUOTE
= QUOTE .
例四、解：因为 QUOTE <β<α< QUOTE π,
所以- QUOTE π<-β<- QUOTE ,
所以0<α-β< QUOTE ,π<α+β< QUOTE π,
所以sin(α-β)= QUOTE
= QUOTE = QUOTE ,
cos(α+β)=- QUOTE
=- QUOTE
=- QUOTE .
所以cos 2α=cos[(α-β)+(α+β)]
=cos(α-β)cos(α+β)-sin(α-β)sin(α+β)
= QUOTE ×(- QUOTE )- QUOTE ×(- QUOTE )
=- QUOTE .
sin 2β=sin [(α+β)-(α-β)]
=sin(α+β)cos(α-β)-cos(α+β)sin(α-β)
=(- QUOTE )× QUOTE -(- QUOTE )× QUOTE
=- QUOTE + QUOTE =- QUOTE .
例五、解：(1)因为α为锐角,所以α+ QUOTE ∈( QUOTE , QUOTE ).
又因为cos (α+ QUOTE )= QUOTE ,所以sin(α+ QUOTE )= QUOTE ,
所以sin(2α+ QUOTE )=2sin(α+ QUOTE )cos(α+ QUOTE )= QUOTE ,
Cos(2α+ QUOTE )=2cos2(α+ QUOTE )-1= QUOTE ,
所以sin(2α+ QUOTE )
=sin[(2α+ QUOTE )- QUOTE ]
=sin(2α+ QUOTE )cos QUOTE -cos(2α+ QUOTE )sin QUOTE
= QUOTE × QUOTE - QUOTE × QUOTE
= QUOTE .
(2)因为0又因为sin( QUOTE -x)= QUOTE ,
所以cos( QUOTE -x)= QUOTE .
因为cos 2x=sin( QUOTE -2x)
=2sin( QUOTE -x)cos( QUOTE -x)
=2cos[ QUOTE -( QUOTE -x)]cos( QUOTE -x)
=2cos( QUOTE +x)cos( QUOTE -x),
所以 QUOTE =2cos( QUOTE -x)= QUOTE .









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