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受欢迎的三角函数 >

证明 sin^2(90-b)+cos^2(b-450)=1

解答

证明

sin^{2} (9 0^{\circ} - b) + cos^{2} (b - 45 0^{\circ}) = 1

解答

真

求解步骤

sin^{2} (9 0^{\circ} - b) + cos^{2} (b - 45 0^{\circ}) = 1

调整左侧

sin^{2} (9 0^{\circ} - b) + cos^{2} (b - 45 0^{\circ})

使用三角恒等式改写

cos (b - 45 0^{\circ})

使用角差恒等式:

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

= cos (b) cos (45 0^{\circ}) + sin (b) sin (45 0^{\circ})

化简

cos (b) cos (45 0^{\circ}) + sin (b) sin (45 0^{\circ}) : sin (b)

cos (b) cos (45 0^{\circ}) + sin (b) sin (45 0^{\circ})

cos (b) cos (45 0^{\circ}) = 0

cos (b) cos (45 0^{\circ})

cos (45 0^{\circ}) = 0

cos (45 0^{\circ})

cos (45 0^{\circ}) = cos (9 0^{\circ})

cos (45 0^{\circ})

将

45 0^{\circ}

改写为

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

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

使用周期

cos

:

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

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

= cos (9 0^{\circ})

= cos (9 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

= 0

= 0 \cdot cos (b)

使用法则

0 \cdot a = 0

= 0

sin (b) sin (45 0^{\circ}) = sin (b)

sin (b) sin (45 0^{\circ})

sin (45 0^{\circ}) = 1

sin (45 0^{\circ})

sin (45 0^{\circ}) = sin (9 0^{\circ})

sin (45 0^{\circ})

将

45 0^{\circ}

改写为

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

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

使用周期

sin

:

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

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

= sin (9 0^{\circ})

= sin (9 0^{\circ})

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

= 1 \cdot sin (b)

乘以:

sin (b) \cdot 1 = sin (b)

= sin (b)

= 0 + sin (b)

0 + sin (b) = sin (b)

= sin (b)

= sin (b)

= sin^{2} (9 0^{\circ} - b) + (sin (b))^{2}

使用三角恒等式改写

sin (9 0^{\circ} - b)

使用角差恒等式:

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

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

化简

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

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

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

sin (9 0^{\circ}) cos (b)

化简

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

sin (9 0^{\circ})

使用以下普通恒等式:

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

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 cos (b)

乘以:

1 \cdot cos (b) = cos (b)

= cos (b)

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

cos (9 0^{\circ}) sin (b)

化简

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

cos (9 0^{\circ})

使用以下普通恒等式:

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

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

= 0 \cdot sin (b)

使用法则

0 \cdot a = 0

= 0

= cos (b) - 0

cos (b) - 0 = cos (b)

= cos (b)

= cos (b)

= (cos (b))^{2} + (sin (b))^{2}

去除括号:

(a) = a

= cos^{2} (b) + sin^{2} (b)

使用三角恒等式改写

cos^{2} (b) + sin^{2} (b)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

= 1

= 1

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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