## highschoolmom2010 3 years ago Verify that each equation is an identity by showing that each expression on the left simplifies to 1.

1. highschoolmom2010

2. phi

how far did you get ?

3. highschoolmom2010

no where im lost

4. ganeshie8

take left hand side expression, plugin tanA = sinA/cosA

5. highschoolmom2010

im still really confused

6. ganeshie8

did they teach you ' tan = sin/cos' yet ?

7. highschoolmom2010

nope

8. highschoolmom2010

so im learning it here

9. ganeshie8

ok, dont learn it here. wait until they start them officially lets do it the known way

10. highschoolmom2010

ok

11. phi

they may want you to use the definitions of sin(A)= opp/adj = a/c and tan(A)= a/b

12. ganeshie8

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13. ganeshie8

exactly, look at above triangle, first find sinA, and tanA

14. highschoolmom2010

|dw:1376501580565:dw|

15. ganeshie8

yes, find tanA also, and plug them in left side of given expression

16. highschoolmom2010

assuming that is alpha....

17. ganeshie8

uhh this time, thats 'a', sorry

18. phi

they are using lower case for the side and capital for the angle opposite that side

19. highschoolmom2010

$\tan A=\frac{ a }{ b }$ oh then let me fix sin A

20. highschoolmom2010

$\sin A=\frac{ a }{ c }$

21. phi

now what is $\frac{1}{\sin A}$

22. highschoolmom2010

@phi im not sure

23. ganeshie8

plug them directly, we can simplify after pluggin in may be

24. highschoolmom2010

$\frac{ 1 }{ \frac{ a }{ c} }$

25. phi

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

26. phi

or do what you did , and multiply top and bottom by c/a

27. highschoolmom2010

that ^^kinda confused me

28. phi

$\frac{ 1\cdot \frac{c}{a} }{ \frac{ a }{ c} \cdot \frac{c}{a}}$ the idea is you can multiply top and bottom by the same number (or fraction) now simplify

29. highschoolmom2010

$\frac{ \frac{ c }{ a } }{ 1 }$

30. phi

yes, and anything divided by 1 is itself

31. highschoolmom2010

co the left side =c/a

32. phi

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

33. highschoolmom2010

so*

34. highschoolmom2010

right side = b/a then

35. phi

now what is $\frac{1}{\sin^2 A} =\frac{1}{\sin A} \cdot \frac{1}{\sin A}$

36. highschoolmom2010

no clue

37. phi

you know 1/sin A = c/a so you know 1/sin A * 1/sinA = (c/a)*(c/a) $\frac{c}{a}\cdot \frac{c}{a}$

38. phi

which is top times top and bottom times bottom

39. phi

lost ?

40. phi

another way to do this is remember $\sin^2 A \text{ means } \sin A \cdot \sin A$ you know sin A= a/c so $\sin^2 A \text{ means } \sin A \cdot \sin A = \frac{a}{c}\cdot \frac{a}{c}=\frac{a\cdot a}{c\cdot c}=\frac{a^2}{c^2}$

41. phi

and $\frac{1}{\sin^2 A} = \frac{c^2}{a^2}$ using the flip rule

42. phi

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)

43. highschoolmom2010

im following and writing too

44. phi

*they mean...

45. phi

so far you have $\frac{1}{(\sin A)^2 }= \frac{c^2}{a^2}$ now what is 1/(tan A)^2 ?

46. highschoolmom2010

1 sec ill write it out on paper and then ^^

47. highschoolmom2010

$\frac{ 1 }{ \tan A^2}$=$\frac{ b^2 }{ a^2 }$

48. highschoolmom2010

well i wrote tan A ^2 wrong

49. phi

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 ?

50. highschoolmom2010

$\frac{ c^2-b^2 }{ a }=1$

51. phi

* a^2 in the bottom

52. highschoolmom2010

i knew i forgot something

53. phi

now, what is the pythagoras equation for this problem ?

54. highschoolmom2010

what does that mean

55. phi

do you know the pythagorean theorem ?

56. highschoolmom2010

oh yea a^2+b^2=c^2

57. phi

and what do you get if you subtract b^2 from both sides ?

58. highschoolmom2010

c^2/a^2=1-b^2

59. phi

no, that is too complicated. start with a^2+b^2=c^2 and write -b^2 on both sides and simplify

60. highschoolmom2010

a^2+b^2-b^2=c^2-b^2 a^2=c^2-b^2

61. phi

now use that in your equation $\frac{ c^2-b^2 }{ a^2 }=1$ I would replace the top

62. phi

in other words c^2 -b^2 is the same as a^2 (by pythagoras)

63. highschoolmom2010

oh

64. highschoolmom2010

so what do we do now.....

65. phi

the idea is that c^2-b^2 = a^2 so replace c^2-b^2 with a^2 in your equation $\frac{ c^2-b^2 }{ a^2 }=1$

66. highschoolmom2010

$\frac{ a^2 }{ a^2 }=1$

67. phi

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

68. highschoolmom2010

oh ok

69. highschoolmom2010

ty @phi & @ganeshie8