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受欢迎的三角函数 >

sin(75)sin(15)

解答

sin (7 5^{\circ}) sin (1 5^{\circ})

解答

\frac{1}{4}

+1

十进制

0.25

求解步骤

sin (7 5^{\circ}) sin (1 5^{\circ})

使用三角恒等式改写:

\frac{cos ( 6 0 ^{\circ} ) - cos ( 9 0 ^{\circ} )}{2}

sin (7 5^{\circ}) sin (1 5^{\circ})

使用积化和差恒等式:

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

= \frac{1}{2} (cos (7 5^{\circ} - 1 5^{\circ}) - cos (7 5^{\circ} + 1 5^{\circ}))

化简

= \frac{cos ( 6 0 ^{\circ} ) - cos ( 9 0 ^{\circ} )}{2}

= \frac{cos ( 6 0 ^{\circ} ) - cos ( 9 0 ^{\circ} )}{2}

使用以下普通恒等式:

cos (6 0^{\circ}) = \frac{1}{2}

cos (6 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{1}{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{1}{2} - 0}{2}

化简

\frac{\frac{1}{2} - 0}{2} : \frac{1}{4}

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

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

= \frac{\frac{1}{2}}{2}

使用分式法则:

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

= \frac{1}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{1}{4}

= \frac{1}{4}

流行的例子

(tan(74)-tan(14))/(1+tan(74)tan(14))

\frac{tan ( 7 4 ^{\circ} ) - tan ( 1 4 ^{\circ} )}{1 + tan ( 7 4 ^{\circ} ) tan ( 1 4 ^{\circ} )}

7 sin (3 5^{\circ})

sin (\frac{π}{20})

cos (\frac{π}{3} - \frac{π}{4})

cos(arctan(24/7))

cos (arctan (\frac{24}{7}))