heureka

if sinθ+cosθ=√2 then θ=°?

$\begin{array}{|rcll|} \hline \text{If }\sin(\theta) + \cos(\theta)=\sqrt{2} \\ \text{ then } \theta=^{\circ}\ ? \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \text{Let } \sin(45^{\circ}) = \cos(45^{\circ}) &=& \frac{\sqrt{2}}{2} \\ \text{then} \sin(45^{\circ})+ \cos(45^{\circ}) &=& 2\cdot \frac{\sqrt{2}}{2} \\ &=& \sqrt{2} \\ \hline \end{array}$

$\theta=45^{\circ}$

Let x and y be integers.

Show that 9x + 5y is divisible by 19 if and only if x + 9y is divisible by 19.

$\begin{array}{lcll} \text{Let } x+9y = n\cdot 19 \qquad n \in Z \\ \end{array}$

$\begin{array}{|rcll|} \hline 9x + 5y \stackrel{?} \equiv 0 \pmod{19} \qquad & | \qquad x = n\cdot 19-9y \\\\ 9\cdot (n\cdot 19-9y) + 5y \stackrel{?} \equiv 0 \pmod{19} \\ 9n\cdot 19-81y + 5y \stackrel{?} \equiv 0 \pmod{19} \\ 9n\cdot 19 - 76y \stackrel{?} \equiv 0 \pmod{19} \qquad & | \qquad 76 = 4\cdot 19 \\ 9n\cdot 19 - 4\cdot 19 y \stackrel{?} \equiv 0 \pmod{19} \\ {\color{red}{19}}\cdot (9n-4y) \equiv 0 \pmod{19} \ \checkmark \\ \hline \end{array}$

What is the tenth term of this function?

f(1) = 30, f(n) = 3 × f(n − 1) − 90

$\begin{array}{|rcll|} \hline f_n =3f_{n-1} - 90 \qquad f_1 = 30 \qquad f_{10} =\ ? \\ \hline \end{array}$

$\small{ \begin{array}{|rcll|} \hline f_1 &=& 30 \\ f_2 &=& 3^1f_1-3^0\cdot 90 \\ f_3 = 3f_2-90 = 3(3f_1-90)-90 &=& 3^2f_1-3^1\cdot90 - 3^0\cdot 90 \\ f_4 = 3f_3-90 = 3(3^2f_1-3\cdot 90 - 90)-90 &=& 3^3f_1-3^2\cdot90- 3^1\cdot90 - 3^0\cdot 90 \\ f_5 = 3f_4-90 = 3(3^3f_1-3^2\cdot 90- 3^1\cdot90 - 3^0\cdot 90)-90 &=& 3^4f_1-3^3\cdot90-3^2\cdot90- 3^1\cdot90 - 3^0\cdot 90 \\ \cdots \\ f_n &=& 3^{n-1}f_1-3^{n-2}\cdot90-3^{n-3}\cdot90-\ldots - 3^0\cdot 90 \\ f_n &=& 3^{n-1}f_1-90\cdot \underbrace{( 3^{n-2} +3^{n-3} +\ldots +1 )}_{=\frac{ 3^{n-1}-1 }{2} }\\ f_n &=& 3^{n-1}f_1-90\cdot \frac{ 3^{n-1}-1 }{2} \\ f_n &=& 3^{n-1}f_1-45\cdot (3^{n-1}-1) \\ \cdots \\ \mathbf{f_n} & \mathbf{=} & \mathbf{-15\cdot(3^{n-1}-3 )} \\ \hline \end{array} }$

$\begin{array}{|rcll|} \hline \mathbf{f_n} & \mathbf{=} & \mathbf{-15\cdot(3^{n-1}-3 )} \\\\ f_{10} & =& -15\cdot (3^9-3) \\ f_{10} & =& -15\cdot (19683-3) \\ f_{10} & =& -15\cdot 19680 \\ \mathbf{f_{10}} & \mathbf{=} & \mathbf{ -295200} \\ \hline \end{array}$

in an arithmatic series x, then 7 then y, then z, their sum equals 12,

what is the first term and common difference

$\begin{array}{|lrcll|} \hline & x = a_1 &=& a_1 + 0\cdot d \\ & 7 = a_2 &=& a_1 + 1\cdot d \\ & y = a_3 &=& a_1 + 2\cdot d \\ & z = a_4 &=& a_1 + 3\cdot d \\\\ (1) & 7 &=& a_1 + d \\ & a_1 &=& 7 - d \\\\ (2)& 12 &=& 4 a_1 + 6d \\ & 12 &=& 4(7-d) + 6d \\ & 12 &=& 28-4d + 6d \\ & 12 &=& 28+2d \quad | \quad : 2\\ & 6 &=& 14 +d \\ & d &=& 6-14\\ & \mathbf{d} & \mathbf{=} & \mathbf{-8} \\\\ & a_1 &=& 7 - d \\ & a_1 &=& 7 - (-8) \\ & a_1 &=& 7+8 \\ & \mathbf{a_1} & \mathbf{=} & \mathbf{15} \\ \hline \end{array}$

The first term is 15 and common difference is -8

$15,\ 7,\ -1,\ -9,$

$x= 15 \\ y=-1 \\ z=-9 \\$

-4x^2+12x-2=p(x)*(2x-1)+3

$\begin{array}{|rcll|} \hline p(x)*(2x-1)+3&=& -4x^2+12x-2 \quad &| \quad - 3 \\ p(x) &=& -4x^2+12x-5 \quad &| \quad :(2x-1)\\ p(x) &=& \frac{-4x^2+12x-5}{2x-1} \\ && -4x^2+12x-5 : 2x-1 = -2x+5 \\ p(x) &=& -2x+5 \\ \hline \end{array}$

Check:

$\begin{array}{|rcll|} \hline && (-2x+5)\cdot (2x-1)+3 \\ &=& -4x^2+2x+10x-5+3 \\ &=& -4x^2+12x-2\ \checkmark \\ \hline \end{array}$

Stimmt diese Ableitung  von
$y = \frac{-2x+1}{x^4-2x^3+x^2} \Rightarrow y=-2x+1 {(x^4-2x^3+x^2)}^{-1}$

Leider ist deine Umformung falsch. Es muss $-2x+1$ in Klammern gesetzt sein.

So wäre deine Umformung richtig: $y=(-2x+1 )\cdot [(x^4-2x^3+x^2)]^{-1}$

$\begin{array}{|rcll|} \hline y &=& (-2x+1 ) \cdot [(x^4-2x^3+x^2)]^{-1} \quad &| \quad x^4-2x^3+x^2 = x^2\cdot(x^2-2x+1) \\\\ y &=& (-2x+1 ) \cdot [x^2\cdot(x^2-2x+1)]^{-1}\quad &| \quad x^2-2x+1 = (x-1)^2 \\\\ y &=& (-2x+1 ) \cdot [x^2\cdot(x-1)^2]^{-1} \\\\ y &=& (-2x+1 ) \cdot x^{-2} \cdot (x-1)^{-2} \\ \hline \end{array}$

Nun kommen wir zur Ableitung.

Unsere Formel dazu lautet allgemein: $y=u \cdot v \cdot w\qquad y'= (u \cdot v \cdot w)' = u'\cdot v \cdot w + u\cdot v' \cdot w +u\cdot v \cdot w'$

$\begin{array}{|rcll|} \hline y &=& \underbrace{(-2x+1 )}_{=u} \cdot \underbrace{x^{-2}}_{=v} \cdot \underbrace{(x-1)^{-2}}_{=w} \\ && u' = -2 \\ && v' = -2\cdot x^{-3} \\ && w' = -2\cdot (x-1)^{-3}\cdot 1 \\\\ y' &=& u'\cdot v \cdot w + u\cdot v' \cdot w +u\cdot v \cdot w' \\\\ y' &=& (-2)\cdot x^{-2} \cdot (x-1)^{-2} \\ && + (-2x+1 ) \cdot (-2\cdot x^{-3}) \cdot (x-1)^{-2} \\ && + (-2x+1 )\cdot x^{-2} \cdot [-2\cdot (x-1)^{-3}] \\\\ y' &=& \frac{-2}{ x^2 \cdot (x-1)^2 } + \frac{ (-2x+1)\cdot(-2)} { x^3 \cdot (x-1)^2 } + \frac{(-2x+1)\cdot (-2)} { x^2\cdot (x-1)^3 } \\\\ y' &=& \frac{-2}{ x^2 \cdot (x-1)^2 } \cdot \frac{x\cdot(x-1)}{x\cdot(x-1)} + \frac{ (-2x+1)\cdot(-2)} { x^3 \cdot (x-1)^2 }\cdot \frac{(x-1)}{(x-1)} + \frac{(-2x+1)\cdot (-2)} { x^2\cdot (x-1)^3 } \cdot \frac{x}{x} \\\\ y' &=& \frac{(-2)\cdot x\cdot(x-1) + (-2x+1)\cdot(-2)\cdot (x-1) + (-2x+1)\cdot (-2)\cdot x } { x^3\cdot (x-1)^3 } \\\\ y' &=& \frac{(-2)\cdot x\cdot(x-1) + (-2x+1)\cdot(-2)\cdot[(x-1) + x] } { x^3\cdot (x-1)^3 } \\\\ y' &=& \frac{(-2)\cdot x\cdot(x-1) + (-2x+1)\cdot(-2)\cdot(2x-1) } { x^3\cdot (x-1)^3 } \\\\ y' &=& \frac{ (-2x)\cdot(x-1) + (4x-2) \cdot(2x-1) } { x^3\cdot (x-1)^3 } \\\\ y' &=& \frac{ -2x^2+2x + 8x^2-4x-4x+2 } { x^3\cdot (x-1)^3 } \\\\ y' &=& \frac{6x^2-6x+2 } { x^3\cdot (x-1)^3 } \\\\ \hline \end{array}$

Can anybody derive this expression:

arctan(a1/b1) + arctan(a2/b2) = arctan[a1b2 + a2b1] / [b1b2 - a1a2],

IF

-Pi/2 < arctan(a1/b1) + arctan(a2/b2) < Pi/2 {nl} From these two? 

Sin(a + b) = SinaCosb + CosaSinb, and

Cos(a + b) = CosaCosb - SinaSinb

$\begin{array}{|lrcll|} \hline \text{Let} & a &=& \arctan(\frac{a_1}{b_1}) \\ \text{Let} & b &=& \arctan(\frac{a_2}{b_2}) \\ \hline \end{array}$

$\small{ \begin{array}{rcll} \sin(a + b) &=& \sin(a) \cdot \cos(b) + \cos(a) \cdot \sin(b) \quad |\quad a = \arctan(\frac{a_1}{b_1}) \quad b = \arctan(\frac{a_2}{b_2})\\\\ \sin(\arctan(\frac{a_1}{b_1}) + \arctan(\frac{a_2}{b_2})) &=& \sin(\arctan(\frac{a_1}{b_1})) \cdot \cos(\arctan(\frac{a_2}{b_2})) + \cos(\arctan(\frac{a_1}{b_1})) \cdot \sin(\arctan(\frac{a_2}{b_2})) \\\\ \hline \\ \cos(a + b) &=& \cos(a) \cdot \cos(b) - \sin(a) \cdot \sin(b) \quad |\quad a = \arctan(\frac{a_1}{b_1}) \quad b = \arctan(\frac{a_2}{b_2})\\\\ \cos(\arctan(\frac{a_1}{b_1}) + \arctan(\frac{a_2}{b_2})) &=& \cos(\arctan(\frac{a_1}{b_1})) \cdot \cos(\arctan(\frac{a_2}{b_2})) - \sin(\arctan(\frac{a_1}{b_1})) \cdot \sin(\arctan(\frac{a_2}{b_2})) \\\\ \end{array} }$

$\begin{array}{|rcll|} \hline \tan(a+b) &=& \frac{\sin(a+b)}{\cos(a+b)} \quad |\quad a = \arctan(\frac{a_1}{b_1}) \quad b = \arctan(\frac{a_2}{b_2})\\\\ \tan(\arctan(\frac{a_1}{b_1})+\arctan(\frac{a_2}{b_2})) &=& \frac{\sin(\arctan(\frac{a_1}{b_1})+\arctan(\frac{a_2}{b_2}))}{\cos(\arctan(\frac{a_1}{b_1})+\arctan(\frac{a_2}{b_2}))} \\ \tan(\arctan(\frac{a_1}{b_1})+\arctan(\frac{a_2}{b_2})) &=& \frac { \sin(\arctan(\frac{a_1}{b_1})) \cdot \cos(\arctan(\frac{a_2}{b_2})) + \cos(\arctan(\frac{a_1}{b_1})) \cdot \sin(\arctan(\frac{a_2}{b_2})) } { \cos(\arctan(\frac{a_1}{b_1})) \cdot \cos(\arctan(\frac{a_2}{b_2})) - \sin(\arctan(\frac{a_1}{b_1})) \cdot \sin(\arctan(\frac{a_2}{b_2})) } \\ \tan(\arctan(\frac{a_1}{b_1})+\arctan(\frac{a_2}{b_2})) &=& \frac { \frac{ \sin(\arctan(\frac{a_1}{b_1})) \cdot \cos(\arctan(\frac{a_2}{b_2})) + \cos(\arctan(\frac{a_1}{b_1})) \cdot \sin(\arctan(\frac{a_2}{b_2})) } {\cos(\arctan(\frac{a_1}{b_1})) \cdot \cos(\arctan(\frac{a_2}{b_2}))} } { \frac{ \cos(\arctan(\frac{a_1}{b_1})) \cdot \cos(\arctan(\frac{a_2}{b_2})) - \sin(\arctan(\frac{a_1}{b_1})) \cdot \sin(\arctan(\frac{a_2}{b_2})) } {\cos(\arctan(\frac{a_1}{b_1})) \cdot \cos(\arctan(\frac{a_2}{b_2}))} }\\ \tan(\arctan(\frac{a_1}{b_1})+\arctan(\frac{a_2}{b_2})) &=& \frac { \frac{ \sin(\arctan(\frac{a_1}{b_1})) \cdot \cos(\arctan(\frac{a_2}{b_2}))}{\cos(\arctan(\frac{a_1}{b_1})) \cdot \cos(\arctan(\frac{a_2}{b_2}))} + \frac{\cos(\arctan(\frac{a_1}{b_1})) \cdot \sin(\arctan(\frac{a_2}{b_2})) } {\cos(\arctan(\frac{a_1}{b_1})) \cdot \cos(\arctan(\frac{a_2}{b_2}))} } { \frac{ \cos(\arctan(\frac{a_1}{b_1})) \cdot \cos(\arctan(\frac{a_2}{b_2}))}{\cos(\arctan(\frac{a_1}{b_1})) \cdot \cos(\arctan(\frac{a_2}{b_2}))} - \frac{\sin(\arctan(\frac{a_1}{b_1})) \cdot \sin(\arctan(\frac{a_2}{b_2})) } {\cos(\arctan(\frac{a_1}{b_1})) \cdot \cos(\arctan(\frac{a_2}{b_2}))} }\\ \tan(\arctan(\frac{a_1}{b_1})+\arctan(\frac{a_2}{b_2})) &=& \frac { \frac{ \sin(\arctan(\frac{a_1}{b_1}))} {\cos(\arctan(\frac{a_1}{b_1})) } + \frac{\sin(\arctan(\frac{a_2}{b_2})) } {\cos(\arctan(\frac{a_2}{b_2}))} } { 1 - \frac{\sin(\arctan(\frac{a_1}{b_1})) \cdot \sin(\arctan(\frac{a_2}{b_2})) } {\cos(\arctan(\frac{a_1}{b_1})) \cdot \cos(\arctan(\frac{a_2}{b_2}))} }\\ \tan(\arctan(\frac{a_1}{b_1})+\arctan(\frac{a_2}{b_2})) &=& \frac { \tan(\arctan(\frac{a_1}{b_1})) + \tan(\arctan(\frac{a_2}{b_2})) } { 1 - \tan(\arctan(\frac{a_1}{b_1})) \cdot \tan(\arctan(\frac{a_2}{b_2})) }\\ \tan(\arctan(\frac{a_1}{b_1})+\arctan(\frac{a_2}{b_2})) &=& \frac { \frac{a_1}{b_1} + \frac{a_2}{b_2} } { 1 - \frac{a_1}{b_1} \cdot \frac{a_2}{b_2} }\\ \tan(\arctan(\frac{a_1}{b_1})+\arctan(\frac{a_2}{b_2})) &=& \frac { \frac{ a_1b_2+a_2b_1}{b_1b_2} } { \frac{ b_1b_2-a_1a_2}{b_1b_2} }\\ \tan(\arctan(\frac{a_1}{b_1})+\arctan(\frac{a_2}{b_2})) &=& \frac{ a_1b_2+a_2b_1 }{ b_1b_2-a_1a_2 } \quad | \quad \arctan() \text{ both sides }\\\\ \mathbf{ \arctan(\frac{a_1}{b_1})+\arctan(\frac{a_2}{b_2}) } & \mathbf{=} & \mathbf{ \arctan(\frac{ a_1b_2+a_2b_1 }{ b_1b_2-a_1a_2 }) } \\ \hline \end{array}$

 how many numbers between 2000 and 5000 can be made
 from the digits 1, 2, 4, 5, 7, and 8 if each digit is used only once?

$\begin{array}{|rcll|} \hline 2xxx &=& \frac{5!}{(5-3)!} = \frac{5!}{2!} = 60 \\ 4xxx &=& \frac{5!}{(5-3)!} = \frac{5!}{2!} = 60 \\ \hline \end{array}$

1. 2145

2. 2147

3. 2148

4. 2154

5. 2157

6. 2158

7. 2174

8. 2175

9. 2178

10. 2184

11. 2185

12. 2187

13. 2415

14. 2417

15. 2418

16. 2451

17. 2457

18. 2458

19. 2471

20. 2475

21. 2478

22. 2481

23. 2485

24. 2487

25. 2514

26. 2517

27. 2518

28. 2541

29. 2547

30. 2548

31. 2571

32. 2574

33. 2578

34. 2581

35. 2584

36. 2587

37. 2714

38. 2715

39. 2718

40. 2741

41. 2745

42. 2748

43. 2751

44. 2754

45. 2758

46. 2781

47. 2784

48. 2785

49. 2814

50. 2815

51. 2817

52. 2841

53. 2845

54. 2847

55. 2851

56. 2854

57. 2857

58. 2871

59. 2874

60. 2875

61. 4125

62. 4127

63. 4128

64. 4152

65. 4157

66. 4158

67. 4172

68. 4175

69. 4178

70. 4182

71. 4185

72. 4187

73. 4215

74. 4217

75. 4218

76. 4251

77. 4257

78. 4258

79. 4271

80. 4275

81. 4278

82. 4281

83. 4285

84. 4287

85. 4512

86. 4517

87. 4518

88. 4521

89. 4527

90. 4528

91. 4571

92. 4572

93. 4578

94. 4581

95. 4582

96. 4587

97. 4712

98. 4715

99. 4718

100. 4721

101. 4725

102. 4728

103. 4751

104. 4752

105. 4758

106. 4781

107. 4782

108. 4785

109. 4812

110. 4815

111. 4817

112. 4821

113. 4825

114. 4827

115. 4851

116. 4852

117. 4857

118. 4871

119. 4872

120. 4875

120 numbers between 2000 and 5000 can be made.

Fill in the other coordinate for the line

3x−5y=2 

(2,y )

$\begin{array}{|rcll|} \hline 3x-5y &=& 2 \quad & | \quad x = 2\\ 3\cdot 2-5y &=& 2 \\ 6 - 5y &=& 2 \quad & | \quad \cdot (-1) \\ -6 + 5y &=& -2 \quad & | \quad +6 \\ 5y &=& -2 +6 \\ 5y &=& 4 \quad & | \quad :5 \\ y &=& \frac{4} {5} \\ \hline \end{array}$

$(2,y ) = (2,\frac{4} {5} )$

For a circle with a diameter of 6 meters,

what is the measurement of a central angle (in degrees) subtended by an arc with a length of 5.2 π meters?

Let radius r = 3 meters

Let arc = $5.2\pi$ meters

Let angle subtended by the arc at the centre = $\alpha$

$\begin{array}{|rcll|} \hline \text{Formula: } ~ arc &=& 2\cdot \pi \cdot r \cdot \frac{\alpha}{360^{\circ}} \\\\ 5.2\pi\ m &=& 2\pi \cdot 3\ m \cdot \frac{\alpha}{360^{\circ}} \\ 5.2\pi &=& 2\pi \cdot 3 \cdot \frac{\alpha}{360^{\circ}} \\ 5.2\pi &=& 6\pi \frac{\alpha}{360^{\circ}} \quad & | \quad :\pi \\ 5.2 &=& 6 \frac{\alpha}{360^{\circ}} \quad & | \quad \cdot 360^{\circ} \\ 5.2 \cdot 360^{\circ} &=& 6 \alpha \quad & | \quad :6 \\ 5.2 \cdot 60^{\circ} &=& \alpha \\ 312^{\circ} &=& \alpha \\ \hline \end{array}$

 The measurement of a central angle (in degrees) is $312^{\circ}$