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受欢迎的三角函数 >

cos(67.5)cos(22.5)

解答

cos (67. 5^{\circ}) cos (22. 5^{\circ})

解答

\frac{2}{4}

+1

十进制

0.35355 \dots

求解步骤

cos (67. 5^{\circ}) cos (22. 5^{\circ})

使用三角恒等式改写:

\frac{cos ( 4 5 ^{\circ} ) + cos ( 9 0 ^{\circ} )}{2}

cos (67. 5^{\circ}) cos (22. 5^{\circ})

使用积化和差恒等式:

cos (s) cos (t) = \frac{1}{2} (cos (s - t) + cos (s + t))

= \frac{1}{2} (cos (67. 5^{\circ} - 22. 5^{\circ}) + cos (67. 5^{\circ} + 22. 5^{\circ}))

化简

= \frac{cos ( 4 5 ^{\circ} ) + cos ( 9 0 ^{\circ} )}{2}

= \frac{cos ( 4 5 ^{\circ} ) + cos ( 9 0 ^{\circ} )}{2}

使用以下普通恒等式:

cos (4 5^{\circ}) = \frac{2}{2}

cos (4 5^{\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{2}{2}

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

= \frac{\frac{2}{2} + 0}{2}

化简

\frac{\frac{2}{2} + 0}{2} : \frac{2}{4}

\frac{\frac{2}{2} + 0}{2}

\frac{2}{2} + 0 = \frac{2}{2}

= \frac{\frac{2}{2}}{2}

使用分式法则:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{2}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

= \frac{2}{4}

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6sqrt(2)cos((7pi)/4)

62 cos (\frac{7 π}{4})

cot(arccsc(sqrt(10)))

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