【正文】
)( ???C? ? ?????? s i ns i nc o sc o sc o s ??? 簡(jiǎn)記: )( ???C將 替換為 ? ??解: )3045c o s (15c o s??? ??42621222322 ??????)6045co s (1 0 5co s ??? ??46223222122 ?????????? 30s i n45s i n30c os45c os ?????? 60s i n45s i n60c os45c os ??例 2. 已知 , , , , 求 的值 . 32s in ??)2( ??? ,? 43c o s ??? )23( ??? ,?? ??? ?c o s 兩角和與差的正弦、余弦、正切 解:由 , 得 32s in ?? )2( ??? ,?35321s i n1c o s 22 ?????????????? ??又由 , 得 , 43c o s ??? )23( ??? ,?47431c o s1s i n 22 ???????? ??????? ??? ? ?????? s i ns i nc o sc o sc o s ??????????? ???????? ???47324335127253 ?? 兩角和與差的正弦、余弦、正切 ???? 55c os10c os35c os80c os ?( 3) 解: ???? 20s i n80s i n20c os80c os ?( 1) ? ??? 2080c o s ??2160c o s ?? ??? 15s i n15c os 22 ?( 2) ???? 15s i n15s i n15c os15c os ??例 3. 不查表,求下列各式的值: ???? 20s i n80s i n20c os80c os ?( 1) ?? 15s i n15c os 22 ?( 2) ? ? 2 330c o s1515c o s ???? ??????? 55c os10c os35c osc os ?( 3) ???? 35s i n80s i n35c os80c os ??? ? ???? ??? 45co s3580co s221, c o s c o s2? ? ? ???1已 知 s i n s i n = 2??求 c o s ( ) 的 值引申: 練習(xí) ① ???? 14c os44s i n14s i n44c os ??② 已知 , , ? ?131230c o s ?? ?? ?? 9030 ?? ? ?cos四 . 小結(jié) COS( ? +? ) =COS? COS? – sin? sin? COS (? –? ) =COS? COS ? + sin?sin? 3。 公式中的運(yùn)算符號(hào) 2。公式中三角符號(hào)的順序 CCSS 兩角和與差的正弦、余弦、正切 例 4.證明公式: ( 1) ; ??? s in2c o s ??????? ???? c o s2s in ??????? ?( 2) ? ????C證明:( 1)利用 可得 ?????? s i n2s i nc o s2c o s2c o s ???????? ???? s i ns i n1c o s0 ?????∴ ??? s i n2c o s ??????? ?( 2)因?yàn)樯鲜街? 為任意角,故可將 換成 , )2( ?? ?? ? 即 ??? c o s2s in ??????? ?就得 ??????