2021届高中全程复习构想·数学[新高考]课时作业 22 三角恒等变换

课后必刷题

[保分必刷题]

1．[2018·全国Ⅲ卷]若sin α＝7c1bc20c016ab66f2b43e99fbf038c45.png，则cos 2α＝(　　)

A.d57ff3886ec9281ce84d249dc389a802.png B.777f0540d3ad0400806e73e5d07ea70d.png C．－777f0540d3ad0400806e73e5d07ea70d.png D．－d57ff3886ec9281ce84d249dc389a802.png

答案：B

解析：cos 2α＝1－2sin2α＝1－2×(1c8e4106457a61c77bb8da4f53f588bd.png)2＝f23470e68d2b0ca3d9f10b490017d3fd.png.

2．[2018·全国Ⅲ卷]函数f(x)＝aa69db52ee7d9e0cc884944242a4070f.png的最小正周期为(　　)

A.6e39d14a87b7a35bb9cf5152ecd1ae21.png B.cf2f35d54ae29874f3f2252ef142701d.png C．π D．2π

答案：C

解析：f(x)＝aa69db52ee7d9e0cc884944242a4070f.png＝9ddf7998e56a52fbc7276ed127962dec.png＝fb939e9367fa81c6b90eca6bc76ce963.png＝sin x·cos x＝df4344a8d214cca83c5817f341d32b3d.pngsin 2x.

∴f(x)的最小正周期为T＝π.

3．[2019·山东济南模拟]已知α∈(cf2f35d54ae29874f3f2252ef142701d.png，π)，sin α＝328a3b93f04d7060c617a203f2e833c5.png，则tan(α＋6e39d14a87b7a35bb9cf5152ecd1ae21.png)＝(　　)

A．－07a259d3e3c2ccb207739fa1a11ea8be.png B.07a259d3e3c2ccb207739fa1a11ea8be.png C．－7 D．7

答案：A

解析：因为α∈(cf2f35d54ae29874f3f2252ef142701d.png，π)，sin α＝328a3b93f04d7060c617a203f2e833c5.png，所以cos α＝－95d852bb34370e76df6754a7fe920da2.png＝－353739de3206111130b993a58140aa66.png＝－2e6bc1de54d06d6caa3cab8880a44998.png，tan α＝756570d990d0140de4c2ec7d4132ceef.png＝－b6ad479a47924ebb75f5c54d546eb338.png.所以tan(α＋6e39d14a87b7a35bb9cf5152ecd1ae21.png)＝48b742500835728322d8fa30f184dab9.png＝a4304e8a308379db865f1e42d03f793e.png＝－07a259d3e3c2ccb207739fa1a11ea8be.png，故选A.

4．[2020·山东威海模拟]已知cos(6e39d14a87b7a35bb9cf5152ecd1ae21.png－α)＝2e6bc1de54d06d6caa3cab8880a44998.png，α∈(cf2f35d54ae29874f3f2252ef142701d.png，π)，则sin α－cos α＝(　　)

A.2367094f06c24044bd6ecf658920ba9a.png B．－2367094f06c24044bd6ecf658920ba9a.png C.a0cc1ba1fb35e416a281222af322e436.png D．－a0cc1ba1fb35e416a281222af322e436.png

答案：C

解析：解法一　因为cf2f35d54ae29874f3f2252ef142701d.png<α<π，所以－π<－α<－cf2f35d54ae29874f3f2252ef142701d.png，－aa8ee001c5caf4b01be5866c2c7088f8.png<6e39d14a87b7a35bb9cf5152ecd1ae21.png－α<－6e39d14a87b7a35bb9cf5152ecd1ae21.png.又因为cos(6e39d14a87b7a35bb9cf5152ecd1ae21.png－α)＝2e6bc1de54d06d6caa3cab8880a44998.png>0，则－cf2f35d54ae29874f3f2252ef142701d.png<6e39d14a87b7a35bb9cf5152ecd1ae21.png－α<－6e39d14a87b7a35bb9cf5152ecd1ae21.png，所以sin(6e39d14a87b7a35bb9cf5152ecd1ae21.png－α)＝－328a3b93f04d7060c617a203f2e833c5.png，所以sin α－cos α＝1553867a52c684e18d473467563ea33b.pngsin(α－6e39d14a87b7a35bb9cf5152ecd1ae21.png)＝－1553867a52c684e18d473467563ea33b.pngsin(6e39d14a87b7a35bb9cf5152ecd1ae21.png－α)＝－1553867a52c684e18d473467563ea33b.png×(－328a3b93f04d7060c617a203f2e833c5.png)＝a0cc1ba1fb35e416a281222af322e436.png，故选C.

解法二　因为cos(6e39d14a87b7a35bb9cf5152ecd1ae21.png－α)＝2e6bc1de54d06d6caa3cab8880a44998.png，所以cos α＋sin α＝c43d608c785f0fe57aacad740c5977bf.png，等号两边平方得1＋sin 2α＝c3c70df6f664a632c11ba3f9de45a8f0.png，解得sin 2α＝－2367094f06c24044bd6ecf658920ba9a.png，因为α∈(cf2f35d54ae29874f3f2252ef142701d.png，π)，所以sin α－cos α>0，所以sin α－cos α＝d32350e6d7d7546d0b7d06206221236a.png＝f0e6406f048cf9a6beb008a875f25464.png＝be4f12721628463dd827e405e4d733a4.png＝a0cc1ba1fb35e416a281222af322e436.png，故选C.

5．[2020·山东淄博模拟]若sin(cf2f35d54ae29874f3f2252ef142701d.png－α)＝2e6bc1de54d06d6caa3cab8880a44998.png，α∈(0，cf2f35d54ae29874f3f2252ef142701d.png)，则tan 2α＝(　　)

A．－64a2bd4b2d7b67506db81dd08b099d53.png B.003c1a2d00a8d7f1207749755fdc5c69.png C．－003c1a2d00a8d7f1207749755fdc5c69.png D.64a2bd4b2d7b67506db81dd08b099d53.png

答案：A

解析：∵sin(cf2f35d54ae29874f3f2252ef142701d.png－α)＝cos α＝2e6bc1de54d06d6caa3cab8880a44998.png，α∈(0，cf2f35d54ae29874f3f2252ef142701d.png)，∴sin α＝328a3b93f04d7060c617a203f2e833c5.png，∴tan α＝b6ad479a47924ebb75f5c54d546eb338.png，∴tan 2α＝df425ff6955bc76f516757d4c7829973.png＝3af103d797309f0d3cc3bbd008c27f73.png＝－64a2bd4b2d7b67506db81dd08b099d53.png，故选A.

6．若f817f5ec062ea18f76421b70ad0d4ded.png＝df4344a8d214cca83c5817f341d32b3d.png，则sin 2α的值为(　　)

A．－d80dee887b88cc2e849f29df6db3e5a0.png B.d80dee887b88cc2e849f29df6db3e5a0.png C．－49e422ba21459176e081382a4e0cd94f.png D.49e422ba21459176e081382a4e0cd94f.png

答案：B

解析：f817f5ec062ea18f76421b70ad0d4ded.png＝155c1ea632de2cd6aec65701a9ef1249.png

＝1553867a52c684e18d473467563ea33b.png(cos α－sin α)＝df4344a8d214cca83c5817f341d32b3d.png，

即cos α－sin α＝454323c9915a711c22fe0c7f8d2cb103.png，等式两边平方得

cos2α－2sin αcos α＋sin2α＝1－sin 2α＝44bd504c73de9fab21b4573fed2095f1.png，

解得sin 2α＝d80dee887b88cc2e849f29df6db3e5a0.png.

7．化简：f03bba4b3f586e37ef8b729db214a83e.png－4fb6f7db2326eb55b994dfd295ad615f.png＝________.

答案：4

解析：原式＝4cc0b817cfc423342b79b70224c0c7f1.png＝45747bad30c15bc533049cedf66e3e62.png＝4cd802b4c37e2f71ee154c76c8488a2d.png＝b2a33ec236bdb12807100c178541eafe.png＝4.

8．[多填题]若α∈42c56404253a6fd3953459a630f3c03e.png，cos α＝e22dc548c827a07ae42976e6d8caa613.png，则sin α＝________，tan 2α＝________.

答案：6b947573d14816876763af57c7a89b2e.png　4a74c4ef873eb22c5f153063d628cf438.png

解析：若α∈42c56404253a6fd3953459a630f3c03e.png，cos α＝e22dc548c827a07ae42976e6d8caa613.png，则sin α＝090b49c10ef662affe277eca10e7c3aa.png＝6b947573d14816876763af57c7a89b2e.png，

∴tan α＝756570d990d0140de4c2ec7d4132ceef.png＝bb67d19921092521e3c07dba0365b86e.png＝ecae01a1759949383660adc0223f5985.png，∴tan 2α＝df425ff6955bc76f516757d4c7829973.png＝4a74c4ef873eb22c5f153063d628cf438.png，故答案为：6b947573d14816876763af57c7a89b2e.png；4a74c4ef873eb22c5f153063d628cf438.png.

9．[2018·全国Ⅱ卷]已知sin α＋cos β＝1，cos α＋sin β＝0，则sin(α＋β)＝________.

答案：－df4344a8d214cca83c5817f341d32b3d.png

解析：∵sin α＋cos β＝1①

cos α＋sin β＝0②

∴①2＋②2得1＋2(sin αcos β＋cos αsin β)＋1＝1，

∴sin αcos β＋cos αsin β＝－df4344a8d214cca83c5817f341d32b3d.png，

∴sin(α＋β)＝－df4344a8d214cca83c5817f341d32b3d.png.

10．已知函数f(x)＝sin(x＋b0d7892a1bcc5ddedd63a3b4fc04cbdf.png)＋sin(x－b0d7892a1bcc5ddedd63a3b4fc04cbdf.png)＋cos x＋a的最大值为1，

(1)求常数a的值；

(2)求函数f(x)的单调递减区间；

(3)求使f(x)≥0成立的x的取值集合．

解析：f(x)＝9097ad464ca3f4d87bfa261a719ba953.pngsin x＋cos x＋a＝2sin(x＋b0d7892a1bcc5ddedd63a3b4fc04cbdf.png)＋a.

(1)由2＋a＝1得a＝－1.

(2)由cf2f35d54ae29874f3f2252ef142701d.png＋2kπ≤x＋b0d7892a1bcc5ddedd63a3b4fc04cbdf.png≤52b84cc47232834378729c825b2f667b.png＋2kπ，k∈Z，

得5a777e0b4347abb14c3c394ee80f7e68.png＋2kπ≤x≤82be1288fab1939c79019c6c8fd80151.png＋2kπ，k∈Z，

∴f(x)的单调递减区间为[5a777e0b4347abb14c3c394ee80f7e68.png＋2kπ，82be1288fab1939c79019c6c8fd80151.png＋2kπ]，k∈Z.

(3)∵f(x)≥0，即sin(x＋b0d7892a1bcc5ddedd63a3b4fc04cbdf.png)≥df4344a8d214cca83c5817f341d32b3d.png，

∴b0d7892a1bcc5ddedd63a3b4fc04cbdf.png＋2kπ≤x＋b0d7892a1bcc5ddedd63a3b4fc04cbdf.png≤40e0e164a635c508463915d501d1f617.png＋2kπ，k∈Z，

∴2kπ≤x≤6866f5f55bcdb05fc3c4a4512c010e8b.png＋2kπ，k∈Z.

故x的取值集合为{x|2kπ≤x≤6866f5f55bcdb05fc3c4a4512c010e8b.png＋2kπ，k∈Z}．

[提分必刷题]

11．已知cos(α－b0d7892a1bcc5ddedd63a3b4fc04cbdf.png)＋sin α＝abe56650289784aad412d44a29c9d4c9.png，则sin(α＋0993cefd8ea7c93d4e0bcef4184742f4.png)的值是(　　)

A．－f029449ff51d59aa806e6c9c15d08043.png B.f029449ff51d59aa806e6c9c15d08043.png C.328a3b93f04d7060c617a203f2e833c5.png D．－328a3b93f04d7060c617a203f2e833c5.png

答案：D

解析：由cos(α－b0d7892a1bcc5ddedd63a3b4fc04cbdf.png)＋sin α＝abe56650289784aad412d44a29c9d4c9.png，可得b702758df4d9b7bf8fe7a0882928ea08.pngcos α＋df4344a8d214cca83c5817f341d32b3d.pngsin α＋sin α＝abe56650289784aad412d44a29c9d4c9.png，即003c1a2d00a8d7f1207749755fdc5c69.pngsin α＋b702758df4d9b7bf8fe7a0882928ea08.pngcos α＝abe56650289784aad412d44a29c9d4c9.png，所以9097ad464ca3f4d87bfa261a719ba953.pngsin(α＋b0d7892a1bcc5ddedd63a3b4fc04cbdf.png)＝abe56650289784aad412d44a29c9d4c9.png，sin(α＋b0d7892a1bcc5ddedd63a3b4fc04cbdf.png)＝328a3b93f04d7060c617a203f2e833c5.png，所以sin(α＋0993cefd8ea7c93d4e0bcef4184742f4.png)＝－sin(α＋b0d7892a1bcc5ddedd63a3b4fc04cbdf.png)＝－328a3b93f04d7060c617a203f2e833c5.png.

12．[多选题][2020·临沂一中新高考备考监测联考]已知函数f(x)＝1bcb2cbb9fe068e5487bae020d7ce7d6.png，则(　　)

A．f(x)的最小正周期为π

B．f(x)的最大值为2

C．f(x)的值域为(－2,2)

D．f(x)的图象关于(－e4ca3275ee4385ac97d838fdbe9e589a.png，0)对称

答案：ACD

解析：∵f(x)＝1bcb2cbb9fe068e5487bae020d7ce7d6.png＝a4facb53bcc60b34e6e858ab7664e738.png

＝－2sin(2x＋b0d7892a1bcc5ddedd63a3b4fc04cbdf.png)(cos(2x＋b0d7892a1bcc5ddedd63a3b4fc04cbdf.png)≠0)，

当且仅当cos(2x＋b0d7892a1bcc5ddedd63a3b4fc04cbdf.png)＝0时，|sin(2x＋b0d7892a1bcc5ddedd63a3b4fc04cbdf.png)|＝1，

∴f(x)的值域为(－2,2)，f(x)的最小正周期为π，图象关于(－e4ca3275ee4385ac97d838fdbe9e589a.png，0)对称．

13．已知cos(6e39d14a87b7a35bb9cf5152ecd1ae21.png＋θ)cos(6e39d14a87b7a35bb9cf5152ecd1ae21.png－θ)＝70e7efdd0b858341812e625a071abd09.png，则sin4θ＋cos4θ的值为________．

答案：9d7e450dafeb7e55b99416b11c0d9ce2.png

解析：因为cos(6e39d14a87b7a35bb9cf5152ecd1ae21.png＋θ)cos(6e39d14a87b7a35bb9cf5152ecd1ae21.png－θ)

＝(193acac34cd52a51c1973c3ce22b6172.pngcos θ－193acac34cd52a51c1973c3ce22b6172.pngsin θ)(193acac34cd52a51c1973c3ce22b6172.pngcos θ＋193acac34cd52a51c1973c3ce22b6172.pngsin θ)

＝df4344a8d214cca83c5817f341d32b3d.png(cos2θ－sin2θ)＝66cb1286d2ae3e9092235381221e59d2.pngcos 2θ＝70e7efdd0b858341812e625a071abd09.png.

所以cos 2θ＝df4344a8d214cca83c5817f341d32b3d.png.

故sin4θ＋cos4θ＝(b86b8c8ba370dbf5d792494399cb3705.png)2＋(39c97bd38dc28b9c36a82eb7bca665fc.png)2＝79c9652d1db2bc2c380fbc5d0e3be7bf.png＋75c32518f5bcf0a525f2f2ebc1051265.png＝9d7e450dafeb7e55b99416b11c0d9ce2.png.

14．[2020·山东师大附中月考]已知函数f(x)＝2sin(7c1bc20c016ab66f2b43e99fbf038c45.pngx－b0d7892a1bcc5ddedd63a3b4fc04cbdf.png)，x∈R.

(1)求f(fc1b1c86323f41fe78750967e1c742e8.png)的值；

(2)设α，β∈[0，cf2f35d54ae29874f3f2252ef142701d.png]，f(3α＋cf2f35d54ae29874f3f2252ef142701d.png)＝87cbd2891f09ea2e497f70ca24aa07c7.png，f(3β＋2π)＝df26f4ade16a8d5ebf2906ee86758da3.png，求cos(α＋β)的值．

解析：(1)f(fc1b1c86323f41fe78750967e1c742e8.png)＝2sin(7c1bc20c016ab66f2b43e99fbf038c45.png×9d3355dd2ffe42827c14804d953fb335.pngπ－b0d7892a1bcc5ddedd63a3b4fc04cbdf.png)＝2sin 6e39d14a87b7a35bb9cf5152ecd1ae21.png＝1553867a52c684e18d473467563ea33b.png.

(2)∵87cbd2891f09ea2e497f70ca24aa07c7.png＝f(3α＋cf2f35d54ae29874f3f2252ef142701d.png)＝2sin(7c1bc20c016ab66f2b43e99fbf038c45.png×(3α＋cf2f35d54ae29874f3f2252ef142701d.png)－b0d7892a1bcc5ddedd63a3b4fc04cbdf.png)＝2sin α，

df26f4ade16a8d5ebf2906ee86758da3.png＝f(3β＋2π)＝2sin(7c1bc20c016ab66f2b43e99fbf038c45.png×(3β＋2π)－b0d7892a1bcc5ddedd63a3b4fc04cbdf.png)＝2sin(β＋cf2f35d54ae29874f3f2252ef142701d.png)＝2cos β，

∴sin α＝22f56af511cb1b10a2f7582b6ee59667.png，cos β＝2e6bc1de54d06d6caa3cab8880a44998.png，又∵α，β∈[0，cf2f35d54ae29874f3f2252ef142701d.png]，

∴cos α＝95d852bb34370e76df6754a7fe920da2.png＝bfb2e49030c2b6f975fc3f37baf2d769.png＝00f8e28a5d34ae1525e4942bf3d9a8b4.png，

sin β＝dced0911efafa720b6cf94a5dbd420c1.png＝97fa581eff04cea2d3cef0a84e8266a6.png＝328a3b93f04d7060c617a203f2e833c5.png，

故cos(α＋β)＝cos αcos β－sin αsin β＝2e6bc1de54d06d6caa3cab8880a44998.png×00f8e28a5d34ae1525e4942bf3d9a8b4.png－22f56af511cb1b10a2f7582b6ee59667.png×328a3b93f04d7060c617a203f2e833c5.png＝7548fffd239386d1fd71aa327bc10106.png.

[优生必刷题]

15．[清华大学自招]sin410°＋sin450°＋sin470°＝________.

答案：2a7d339d997db8878e40e681801fb525.png

解析：sin410°＋sin450°＋sin470°＝cos480°＋cos440°＋cos420°＝(e790aa95930cac41134771c018d77b11.png)2＋(a1244c52556e847de128a2ff648c692f.png)2＋(9e2a8e02bde8a1744cb1f6ad7efb6a4b.png)2＝0495c943adeef64d4e2ed7db21a0c83e.png(3＋2cos 160°＋2cos 80°＋2cos 40°＋cos2160°＋cos280°＋cos240°)＝42f93fa0b2b03732fd53e94cbbfa8840.png＋df4344a8d214cca83c5817f341d32b3d.png(－cos 20°＋cos 80°＋cos 40°)＋70e7efdd0b858341812e625a071abd09.png(232852c97e7a5faf052ec7002fbdb928.png＋131af3fdf14389281a51a37d2adce5a1.png＋bb7ccd089b6bf3e4740843e3d590c7af.png)＝2a7d339d997db8878e40e681801fb525.png＋9d7e450dafeb7e55b99416b11c0d9ce2.png(cos 80°＋cos 40°－cos 20°)＝2a7d339d997db8878e40e681801fb525.png＋9d7e450dafeb7e55b99416b11c0d9ce2.png[cos(60°＋20°)＋cos(60°－20°)－cos 20°]＝2a7d339d997db8878e40e681801fb525.png＋9d7e450dafeb7e55b99416b11c0d9ce2.png(2cos 60°cos 20°－cos 20°)＝2a7d339d997db8878e40e681801fb525.png.

16．若θ为任意角，求32cos6θ－cos 6θ－6cos 4θ－15cos 2θ的值．

解析：解法一　根据二倍角和三倍角公式知：32cos6θ－cos 6θ－6cos 4θ－15cos 2θ＝32cos6θ－(2cos23θ－1)－6(2cos22θ－1)－15(2cos2θ－1)＝32cos6θ－[2(4cos3θ－3cos θ)2－1]－6[2(2cos2θ－1)2－1]－15(2cos2θ－1)＝32cos6θ－(32cos6θ－48cos4θ＋18cos2θ－1)－(48cos4θ－48cos2θ＋6)－(30cos2θ－15)＝10.

解法二　由cos2θ＝39c97bd38dc28b9c36a82eb7bca665fc.png，cos 6θ＝4cos32θ－3cos 2θ，cos 4θ＝2cos22θ－1，知待求式中的每一项均可用cos 2θ表示，令cos 2θ＝a，则32cos6θ－cos 6θ－6cos 4θ－15cos 2θ＝32(8332cf4fe9af042a002e20aca7f43749.png)3－(4a3－3a)－6(2a2－1)－15a＝10.

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