4.2

同角三角函數(shù)的基本關(guān)系

及誘導(dǎo)公式4.2同角三角函數(shù)的基本關(guān)系

及誘導(dǎo)公式-2-知識梳理雙基自測2311.同角三角函數(shù)的基本關(guān)系式(1)平方關(guān)系:sin2α+cos2α=

.

1tanα-2-知識梳理雙基自測2311.同角三角函數(shù)的基本關(guān)系式1-3-知識梳理雙基自測2312.三角函數(shù)的誘導(dǎo)公式

-sinα

-sinαsinαcosαcosα-cosαcosα

-cosαsinα-sinα

tanα-tanα-tanα-3-知識梳理雙基自測2312.三角函數(shù)的誘導(dǎo)公式-sin-4-知識梳理雙基自測2313.特殊角的三角函數(shù)值

0

10

10

-10

1-4-知識梳理雙基自測2313.特殊角的三角函數(shù)值02-5-知識梳理雙基自測34151.下列結(jié)論正確的打“√”,錯誤的打“×”.(1)對任意的角α,β,都有sin2α+cos2β=1.(

)(3)sin(π+α)=-sinα成立的條件是α為銳角.(

)答案答案關(guān)閉(1)×

(2)×

(3)×

(4)×

2-5-知識梳理雙基自測34151.下列結(jié)論正確的打“√”,-6-知識梳理雙基自測23415答案解析解析關(guān)閉答案解析關(guān)閉-6-知識梳理雙基自測23415答案解析解析關(guān)閉答案-7-知識梳理雙基自測23415答案解析解析關(guān)閉答案解析關(guān)閉-7-知識梳理雙基自測23415答案解析解析關(guān)閉答案-8-知識梳理雙基自測23415答案解析解析關(guān)閉答案解析關(guān)閉-8-知識梳理雙基自測23415答案解析解析關(guān)閉答案-9-知識梳理雙基自測234155.(教材習(xí)題改編P22T3)已知tanθ=2,則sinθcosθ=

.

答案解析解析關(guān)閉答案解析關(guān)閉-9-知識梳理雙基自測234155.(教材習(xí)題改編P22T3-10-考點(diǎn)1考點(diǎn)2考點(diǎn)3例1已知α是三角形的內(nèi)角,且sinα+cosα=.(1)求tanα的值;思考同角三角函數(shù)基本關(guān)系式有哪些用途?-10-考點(diǎn)1考點(diǎn)2考點(diǎn)3例1已知α是三角形的內(nèi)角,且sin-11-考點(diǎn)1考點(diǎn)2考點(diǎn)3-11-考點(diǎn)1考點(diǎn)2考點(diǎn)3-12-考點(diǎn)1考點(diǎn)2考點(diǎn)3

-12-考點(diǎn)1考點(diǎn)2考點(diǎn)3

-13-考點(diǎn)1考點(diǎn)2考點(diǎn)3C-13-考點(diǎn)1考點(diǎn)2考點(diǎn)3C-14-考點(diǎn)1考點(diǎn)2考點(diǎn)3-14-考點(diǎn)1考點(diǎn)2考點(diǎn)3-15-考點(diǎn)1考點(diǎn)2考點(diǎn)3例2已知關(guān)于x的方程2x2-(+1)x+m=0的兩根為sinθ和cosθ,且θ∈(0,2π).(2)求m的值;(3)求方程的兩根及此時(shí)θ的值.思考sin

α+cos

α,sin

α-cos

α,sin

αcos

α這三個(gè)式子之間有怎樣的關(guān)系?-15-考點(diǎn)1考點(diǎn)2考點(diǎn)3例2已知關(guān)于x的方程2x2-(-16-考點(diǎn)1考點(diǎn)2考點(diǎn)3-16-考點(diǎn)1考點(diǎn)2考點(diǎn)3-17-考點(diǎn)1考點(diǎn)2考點(diǎn)32.利用上述關(guān)系,對于sin

α+cos

α,sin

α-cos

α,sin

αcos

α這三個(gè)式子,可以知一求二.-17-考點(diǎn)1考點(diǎn)2考點(diǎn)32.利用上述關(guān)系,對于sinα+-18-考點(diǎn)1考點(diǎn)2考點(diǎn)3答案答案關(guān)閉(1)A

(2)C

-18-考點(diǎn)1考點(diǎn)2考點(diǎn)3答案答案關(guān)閉(1)A(2-19-考點(diǎn)1考點(diǎn)2考點(diǎn)3-19-考點(diǎn)1考點(diǎn)2考點(diǎn)3-20-考點(diǎn)1考點(diǎn)2考點(diǎn)3-20-考點(diǎn)1考點(diǎn)2考點(diǎn)3-21-考點(diǎn)1考點(diǎn)2考點(diǎn)3-21-考點(diǎn)1考點(diǎn)2考點(diǎn)3-22-考點(diǎn)1考點(diǎn)2考點(diǎn)3答案答案關(guān)閉-22-考點(diǎn)1考點(diǎn)2考點(diǎn)3答案答案關(guān)閉-23-考點(diǎn)1考點(diǎn)2考點(diǎn)3解析:

(1)原式=-sin

1

200°·cos

1

290°-cos

1

020°sin

1

050°=-sin(3×360°+120°)cos(3×360°+210°)-cos(2×360°+300°)sin(2×360°+330°)=-sin

120°cos

210°-cos

300°sin

330°=-sin(180°-60°)cos(180°+30°)-cos(360°-60°)sin(360°-30°)=sin

60°cos

30°+cos

60°sin

30°-23-考點(diǎn)1考點(diǎn)2考點(diǎn)3解析:(1)原式=-sin1-24-考點(diǎn)1考點(diǎn)2考點(diǎn)3-24-考點(diǎn)1考點(diǎn)2考點(diǎn)3-25-考點(diǎn)1考點(diǎn)2考點(diǎn)3思考觀察題目中的兩角之間有什么關(guān)系?當(dāng)所給兩角互補(bǔ)或互余時(shí),怎樣簡化解題過程?答案答案關(guān)閉-25-考點(diǎn)1考點(diǎn)2考點(diǎn)3思考觀察題目中的兩角之間有什么關(guān)系-26-考點(diǎn)1考點(diǎn)2考點(diǎn)3-26-考點(diǎn)1考點(diǎn)2考點(diǎn)3-27-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得1.利用誘導(dǎo)公式化簡三角函數(shù)的基本思路:(1)分析結(jié)構(gòu)特點(diǎn),選擇恰當(dāng)公式化大角為小角;(2)利用公式化成單角三角函數(shù);(3)整理得最簡形式.2.化簡要求:(1)化簡過程是恒等變形;(2)結(jié)果要求項(xiàng)數(shù)盡可能少,次數(shù)盡可能低,結(jié)構(gòu)盡可能簡單,能求值的要求出值.3.用誘導(dǎo)公式求值時(shí),要善于觀察所給角之間的關(guān)系,利用整體-27-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得1.利用誘導(dǎo)公式化簡三角函-28-考點(diǎn)1考點(diǎn)2考點(diǎn)3A.{1,-1,2,-2} B.{-1,1}C.{2,-2} D.{1,-1,0,2,-2}(2)sin600°+tan240°的值等于

.

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