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highschoolmom2010 Group Title

Verify that each equation is an identity by showing that each expression on the left simplifies to 1.

  • one year ago
  • one year ago

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  1. highschoolmom2010 Group Title
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    • one year ago
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  2. phi Group Title
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    how far did you get ?

    • one year ago
  3. highschoolmom2010 Group Title
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    no where im lost

    • one year ago
  4. ganeshie8 Group Title
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    take left hand side expression, plugin tanA = sinA/cosA

    • one year ago
  5. highschoolmom2010 Group Title
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    im still really confused

    • one year ago
  6. ganeshie8 Group Title
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    did they teach you ' tan = sin/cos' yet ?

    • one year ago
  7. highschoolmom2010 Group Title
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    nope

    • one year ago
  8. highschoolmom2010 Group Title
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    so im learning it here

    • one year ago
  9. ganeshie8 Group Title
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    ok, dont learn it here. wait until they start them officially lets do it the known way

    • one year ago
  10. highschoolmom2010 Group Title
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    ok

    • one year ago
  11. phi Group Title
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    they may want you to use the definitions of sin(A)= opp/adj = a/c and tan(A)= a/b

    • one year ago
  12. ganeshie8 Group Title
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    |dw:1376501404557:dw|

    • one year ago
  13. ganeshie8 Group Title
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    exactly, look at above triangle, first find sinA, and tanA

    • one year ago
  14. highschoolmom2010 Group Title
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    |dw:1376501580565:dw|

    • one year ago
  15. ganeshie8 Group Title
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    yes, find tanA also, and plug them in left side of given expression

    • one year ago
  16. highschoolmom2010 Group Title
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    assuming that is alpha....

    • one year ago
  17. ganeshie8 Group Title
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    uhh this time, thats 'a', sorry

    • one year ago
  18. phi Group Title
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    they are using lower case for the side and capital for the angle opposite that side

    • one year ago
  19. highschoolmom2010 Group Title
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    \[\tan A=\frac{ a }{ b }\] oh then let me fix sin A

    • one year ago
  20. highschoolmom2010 Group Title
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    \[\sin A=\frac{ a }{ c }\]

    • one year ago
  21. phi Group Title
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    now what is \[ \frac{1}{\sin A} \]

    • one year ago
  22. highschoolmom2010 Group Title
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    @phi im not sure

    • one year ago
  23. ganeshie8 Group Title
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    plug them directly, we can simplify after pluggin in may be

    • one year ago
  24. highschoolmom2010 Group Title
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    \[\frac{ 1 }{ \frac{ a }{ c} }\]

    • one year ago
  25. phi Group Title
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    the short way is "flip" both sides of sin A = a/c the long way is go through these steps multiply both sides by c: c sin A = a divide both sides by a (c sin A)/a = 1 divide both sides by sin A c/a = 1/ sin A

    • one year ago
  26. phi Group Title
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    or do what you did , and multiply top and bottom by c/a

    • one year ago
  27. highschoolmom2010 Group Title
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    that ^^kinda confused me

    • one year ago
  28. phi Group Title
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    \[ \frac{ 1\cdot \frac{c}{a} }{ \frac{ a }{ c} \cdot \frac{c}{a}} \] the idea is you can multiply top and bottom by the same number (or fraction) now simplify

    • one year ago
  29. highschoolmom2010 Group Title
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    \[\frac{ \frac{ c }{ a } }{ 1 }\]

    • one year ago
  30. phi Group Title
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    yes, and anything divided by 1 is itself

    • one year ago
  31. highschoolmom2010 Group Title
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    co the left side =c/a

    • one year ago
  32. phi Group Title
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    yes. (and remember the short cut) if you have a/b = c/d then b/a = d/c (flip both sides) in this case sin A = a/c 1/sinA = c/a

    • one year ago
  33. highschoolmom2010 Group Title
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    so*

    • one year ago
  34. highschoolmom2010 Group Title
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    right side = b/a then

    • one year ago
  35. phi Group Title
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    now what is \[ \frac{1}{\sin^2 A} =\frac{1}{\sin A} \cdot \frac{1}{\sin A} \]

    • one year ago
  36. highschoolmom2010 Group Title
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    no clue

    • one year ago
  37. phi Group Title
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    you know 1/sin A = c/a so you know 1/sin A * 1/sinA = (c/a)*(c/a) \[ \frac{c}{a}\cdot \frac{c}{a}\]

    • one year ago
  38. phi Group Title
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    which is top times top and bottom times bottom

    • one year ago
  39. phi Group Title
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    lost ?

    • one year ago
  40. phi Group Title
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    another way to do this is remember \[ \sin^2 A \text{ means } \sin A \cdot \sin A \] you know sin A= a/c so \[ \sin^2 A \text{ means } \sin A \cdot \sin A = \frac{a}{c}\cdot \frac{a}{c}=\frac{a\cdot a}{c\cdot c}=\frac{a^2}{c^2}\]

    • one year ago
  41. phi Group Title
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    and \[ \frac{1}{\sin^2 A} = \frac{c^2}{a^2} \] using the flip rule

    • one year ago
  42. phi Group Title
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    I should mention that \[ \left(\sin A \right)^2 \text{ is often written as }\sin^2 A \] then mean the same thing. (sin A) times itself: (sin A)(sin A)

    • one year ago
  43. highschoolmom2010 Group Title
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    im following and writing too

    • one year ago
  44. phi Group Title
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    *they mean...

    • one year ago
  45. phi Group Title
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    so far you have \[ \frac{1}{(\sin A)^2 }= \frac{c^2}{a^2} \] now what is 1/(tan A)^2 ?

    • one year ago
  46. highschoolmom2010 Group Title
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    1 sec ill write it out on paper and then ^^

    • one year ago
  47. highschoolmom2010 Group Title
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    \[\frac{ 1 }{ \tan A^2} \]=\[\frac{ b^2 }{ a^2 }\]

    • one year ago
  48. highschoolmom2010 Group Title
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    well i wrote tan A ^2 wrong

    • one year ago
  49. phi Group Title
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    yes, so the equation is \[ \frac{c^2}{a^2}- \frac{b^2}{a^2} = 1 \] re: tan A^2 yes, you should put parens in (tan A)^2. but we know what you mean now, the fractions on the left side have a common denominator can you re-write as one fraction ?

    • one year ago
  50. highschoolmom2010 Group Title
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    \[\frac{ c^2-b^2 }{ a }=1\]

    • one year ago
  51. phi Group Title
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    * a^2 in the bottom

    • one year ago
  52. highschoolmom2010 Group Title
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    i knew i forgot something

    • one year ago
  53. phi Group Title
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    now, what is the pythagoras equation for this problem ?

    • one year ago
  54. highschoolmom2010 Group Title
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    what does that mean

    • one year ago
  55. phi Group Title
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    do you know the pythagorean theorem ?

    • one year ago
  56. highschoolmom2010 Group Title
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    oh yea a^2+b^2=c^2

    • one year ago
  57. phi Group Title
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    and what do you get if you subtract b^2 from both sides ?

    • one year ago
  58. highschoolmom2010 Group Title
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    c^2/a^2=1-b^2

    • one year ago
  59. phi Group Title
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    no, that is too complicated. start with a^2+b^2=c^2 and write -b^2 on both sides and simplify

    • one year ago
  60. highschoolmom2010 Group Title
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    a^2+b^2-b^2=c^2-b^2 a^2=c^2-b^2

    • one year ago
  61. phi Group Title
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    now use that in your equation \[\frac{ c^2-b^2 }{ a^2 }=1 \] I would replace the top

    • one year ago
  62. phi Group Title
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    in other words c^2 -b^2 is the same as a^2 (by pythagoras)

    • one year ago
  63. highschoolmom2010 Group Title
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    oh

    • one year ago
  64. highschoolmom2010 Group Title
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    so what do we do now.....

    • one year ago
  65. phi Group Title
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    the idea is that c^2-b^2 = a^2 so replace c^2-b^2 with a^2 in your equation \[ \frac{ c^2-b^2 }{ a^2 }=1 \]

    • one year ago
  66. highschoolmom2010 Group Title
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    \[\frac{ a^2 }{ a^2 }=1\]

    • one year ago
  67. phi Group Title
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    and something divided by itself is 1 i.e. a^2 divided by itself is 1 you get 1=1 and you are done. You showed both sides are equal

    • one year ago
  68. highschoolmom2010 Group Title
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    oh ok

    • one year ago
  69. highschoolmom2010 Group Title
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    ty @phi & @ganeshie8

    • one year ago
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