Tiêu đề F. BAI TAP NANG CAO - DA.pdf ✅

( ) 7 3 2cos 3cos 5sin cot 2 2 2cos 3cos 5sin 3 cot 2 2 5cos 5sin cot 2 2 5cos 5cos tan tan B x x x x x x x x x x x x x x x              = − − + − + −             = + + + − + + −             = − − + −         = − + = c) ( ) ( ) 3 C 2sin sin 5 sin cos 2 2 2 2cos sin 4 sin sin 2 x x x x x x x x       = + + − + + + +               = + + − + + + −             2cos sin sin sin ( ) 2 2cos sin cos sin cos 2sin x x x x x x x x x x   = + − − + −     = − − − = −   d) ( ) ( ) ( ) ( ) ( ) ( ) 3 3 D= cos 5 sin tan cot 3 2 2 = cos 4 sin tan cot 2 2 2 = cos +sin tan cot 2 2 = cos +cos cot cot 0 x x x x x x x x x x x x x x x x     − − + + − + −             + − − + + + + − + + −             − + + − + −         − + − =                 Câu 3. Không dùng bảng số và máy tính, rút gọn các biểu thức: a) A=  +  −   tan18 .tan288 sin32 .sin148 sin302 .sin122 . b) 4 4 6 6 1 sin cos 1 sin cos a a B a a + − = − − . Lời giải a) A =  −   −  +   −  −  −   −  tan 90 72 .tan 360 72 sin32 .sin 180 32 sin 360 58 .s ( ) ( ) ( ) ( ) in 180 58 ( ) ( ) 2 2 =  −  +  +  cot 72 . tan 72 sin 32 sin 58 2 2 = − +  +  1 sin 32 cos 32 = − + = 1 1 0.
b) ( )( ) ( )( ) 2 2 2 2 2 2 4 2 2 4 1 sin cos sin cos 1 sin cos sin sin cos cos a a a a B a a a a a a + + − = − + − + ( ) 2 2 2 2 2 2 2 1 sin cos 1 sin cos 3sin cos a a a a a a + − =   − + −   2 2 2 2sin 3sin cos a a a = ( ) 2 2 1 tan 3 = + a . Câu 4. Tính giá trị các biểu thức sau: a) 7 5 7 sin cos9 tan( ) cot 6 4 2 A    = + + − +  b) 1 2sin 2550 cos( 188 ) tan 368 2cos638 cos98 B  −  = +   +  c) 2 2 2 2 C =  +  +  +  sin 25 sin 45 sin 60 sin 65 d) 2 3 5 tan .tan .tan 8 8 8 D    = Lời giải a) Ta có sin cos 4.2 tan cot 3 ( ) 6 4 2 A               = + + + − + + +             1 5 sin cos tan cot 1 1 0 6 4 2 2 2 A     = − + − + = − − − + = −  b) Ta có ( ) ( ) ( ) ( ) 0 0 0 0 0 0 1 2sin 30 7.360 cos(8 180 ) tan 8 360 2cos 90 8 2.360 cos 90 8 B +  +  = + +  − + +  + +  ( ) ( ) ( ) ( ) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2. cos8 1 1 2sin 30 cos8 2 tan8 tan8 2cos 8 90 sin8 2cos 90 8 sin8 1 cos8 1 cos8 0 tan8 2sin8 sin8 tan8 sin8 B − − = + = + = − − − − = − = − = − c) Vì 0 0 0 0 0 25 65 90 sin 65 cos25 + =  = do đó ( ) 2 2 0 2 2 2 2 2 1 sin 25 cos 25 sin 45 sin 60 1 2 2 C     =  + +  +  = + +           Suy ra 7 4 C = . d) 3 5 tan .tan . tan tan 8 8 8 8 D           = − −            