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受欢迎的三角函数 >

cos(270/2)

解答

cos (\frac{27 0 ^{\circ}}{2})

解答

- \frac{1}{2}

+1

十进制

- 0.70710 \dots

求解步骤

cos (\frac{27 0 ^{\circ}}{2})

使用三角恒等式改写:

- \frac{1 + cos ( 27 0 ^{\circ} )}{2}

cos (\frac{27 0 ^{\circ}}{2})

使用半角公式:

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

使用倍角公式

cos (2 θ) = 2 cos^{2} (θ) - 1

用

\frac{θ}{2}

替代

θ

cos (θ) = 2 cos^{2} (\frac{θ}{2}) - 1

交换两边

2 cos^{2} (\frac{θ}{2}) = 1 + cos (θ)

两边除以

2

cos^{2} (\frac{θ}{2}) = \frac{( 1 + cos ( θ ) )}{2}

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

:

range [0, 9 0^{\circ}] [9 0^{\circ}, 18 0^{\circ}] [18 0^{\circ}, 27 0^{\circ}] [27 0^{\circ}, 36 0^{\circ}] quadrant I II III I V sin positive positive negative negative cos positive negative negative positive

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

= - \frac{1 + cos ( 27 0 ^{\circ} )}{2}

= - \frac{1 + cos ( 27 0 ^{\circ} )}{2}

使用三角恒等式改写:

cos (27 0^{\circ}) = 0

cos (27 0^{\circ})

使用三角恒等式改写:

cos (18 0^{\circ}) cos (9 0^{\circ}) - sin (18 0^{\circ}) sin (9 0^{\circ})

cos (27 0^{\circ})

将

cos (27 0^{\circ})

写为

cos (18 0^{\circ} + 9 0^{\circ})

= cos (18 0^{\circ} + 9 0^{\circ})

使用角和恒等式:

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

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

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

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:

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:

sin (9 0^{\circ}) = 1

sin (9 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}

= 1

= (- 1) \cdot 0 - 0 \cdot 1

化简

= 0

= - \frac{1 + 0}{2}

化简

= - \frac{1}{2}

流行的例子

arctan (49)

tan((25pi)/(12))

tan (\frac{25 π}{12})

(tan^2(-(5pi)/6))/(sec(-(5pi)/6)+1)

\frac{tan ^{2} ( - \frac{5 π}{6} )}{sec ( - \frac{5 π}{6} ) + 1}

cos (2.5 π)

12.4-10(0.5pi-arcsin(0.5)-0.5(1-(0.5)^2)^{1/2})

12.4 - 10 (0.5 π - arcsin (0.5) - 0.5 (1 - (0.5)^{2})^{\frac{1}{2}})