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受欢迎的三角函数 >

(sin(570)+2cos(1170))/(cos(1980))

解答

\frac{sin ( 57 0 ^{\circ} ) + 2 cos ( 117 0 ^{\circ} )}{cos ( 198 0 ^{\circ} )}

解答

\frac{1}{2}

+1

十进制

0.5

求解步骤

\frac{sin ( 57 0 ^{\circ} ) + 2 cos ( 117 0 ^{\circ} )}{cos ( 198 0 ^{\circ} )}

化简

= - \frac{- 2 cos ( 117 0 ^{\circ} ) - sin ( 57 0 ^{\circ} )}{cos ( 198 0 ^{\circ} )}

cos (117 0^{\circ}) = cos (9 0^{\circ})

cos (117 0^{\circ})

将

117 0^{\circ}

改写为

36 0^{\circ} \cdot 3 + 9 0^{\circ}

= cos (36 0^{\circ} 3 + 9 0^{\circ})

使用周期

cos

:

cos (x + 36 0^{\circ} \cdot k) = cos (x)

cos (36 0^{\circ} \cdot 3 + 9 0^{\circ}) = cos (9 0^{\circ})

= cos (9 0^{\circ})

= - \frac{- 2 cos ( 9 0 ^{\circ} ) - sin ( 57 0 ^{\circ} )}{cos ( 198 0 ^{\circ} )}

sin (57 0^{\circ}) = sin (21 0^{\circ})

sin (57 0^{\circ})

将

57 0^{\circ}

改写为

36 0^{\circ} + 21 0^{\circ}

= sin (36 0^{\circ} + 21 0^{\circ})

使用周期

sin

:

sin (x + 36 0^{\circ}) = sin (x)

sin (36 0^{\circ} + 21 0^{\circ}) = sin (21 0^{\circ})

= sin (21 0^{\circ})

= - \frac{- 2 cos ( 9 0 ^{\circ} ) - sin ( 21 0 ^{\circ} )}{cos ( 198 0 ^{\circ} )}

cos (198 0^{\circ}) = cos (18 0^{\circ})

cos (198 0^{\circ})

将

198 0^{\circ}

改写为

36 0^{\circ} \cdot 5 + 18 0^{\circ}

= cos (36 0^{\circ} 5 + 18 0^{\circ})

使用周期

cos

:

cos (x + 36 0^{\circ} \cdot k) = cos (x)

cos (36 0^{\circ} \cdot 5 + 18 0^{\circ}) = cos (18 0^{\circ})

= cos (18 0^{\circ})

= - \frac{- 2 cos ( 9 0 ^{\circ} ) - sin ( 21 0 ^{\circ} )}{cos ( 18 0 ^{\circ} )}

使用以下普通恒等式:

cos (9 0^{\circ}) = 0

cos (9 0^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 0

使用三角恒等式改写:

sin (21 0^{\circ}) = - \frac{1}{2}

sin (21 0^{\circ})

使用三角恒等式改写:

sin (18 0^{\circ}) cos (3 0^{\circ}) + cos (18 0^{\circ}) sin (3 0^{\circ})

sin (21 0^{\circ})

将

sin (21 0^{\circ})

写为

sin (18 0^{\circ} + 3 0^{\circ})

= sin (18 0^{\circ} + 3 0^{\circ})

使用角和恒等式:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (18 0^{\circ}) cos (3 0^{\circ}) + cos (18 0^{\circ}) sin (3 0^{\circ})

= sin (18 0^{\circ}) cos (3 0^{\circ}) + cos (18 0^{\circ}) sin (3 0^{\circ})

使用以下普通恒等式:

sin (18 0^{\circ}) = 0

sin (18 0^{\circ})

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

使用以下普通恒等式:

cos (3 0^{\circ}) = \frac{3}{2}

cos (3 0^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{3}{2}

使用以下普通恒等式:

cos (18 0^{\circ}) = (- 1)

cos (18 0^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= (- 1)

使用以下普通恒等式:

sin (3 0^{\circ}) = \frac{1}{2}

sin (3 0^{\circ})

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

= 0 \cdot \frac{3}{2} + (- 1) \frac{1}{2}

化简

= - \frac{1}{2}

使用以下普通恒等式:

cos (18 0^{\circ}) = (- 1)

cos (18 0^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= (- 1)

= - \frac{- 2 \cdot 0 - ( - \frac{1}{2} )}{- 1}

化简

= \frac{1}{2}

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