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

Mathematics
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2 Attachments
  • phi
how far did you get ?
no where im lost

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Other answers:

take left hand side expression, plugin tanA = sinA/cosA
im still really confused
did they teach you ' tan = sin/cos' yet ?
nope
so im learning it here
ok, dont learn it here. wait until they start them officially lets do it the known way
ok
  • phi
they may want you to use the definitions of sin(A)= opp/adj = a/c and tan(A)= a/b
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exactly, look at above triangle, first find sinA, and tanA
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yes, find tanA also, and plug them in left side of given expression
assuming that is alpha....
uhh this time, thats 'a', sorry
  • phi
they are using lower case for the side and capital for the angle opposite that side
\[\tan A=\frac{ a }{ b }\] oh then let me fix sin A
\[\sin A=\frac{ a }{ c }\]
  • phi
now what is \[ \frac{1}{\sin A} \]
@phi im not sure
plug them directly, we can simplify after pluggin in may be
\[\frac{ 1 }{ \frac{ a }{ c} }\]
  • 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
  • phi
or do what you did , and multiply top and bottom by c/a
that ^^kinda confused me
  • 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
\[\frac{ \frac{ c }{ a } }{ 1 }\]
  • phi
yes, and anything divided by 1 is itself
co the left side =c/a
  • 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
so*
right side = b/a then
  • phi
now what is \[ \frac{1}{\sin^2 A} =\frac{1}{\sin A} \cdot \frac{1}{\sin A} \]
no clue
  • phi
you know 1/sin A = c/a so you know 1/sin A * 1/sinA = (c/a)*(c/a) \[ \frac{c}{a}\cdot \frac{c}{a}\]
  • phi
which is top times top and bottom times bottom
  • phi
lost ?
  • 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 \[ \sin^2 A \text{ means } \sin A \cdot \sin A = \frac{a}{c}\cdot \frac{a}{c}=\frac{a\cdot a}{c\cdot c}=\frac{a^2}{c^2}\]
  • phi
and \[ \frac{1}{\sin^2 A} = \frac{c^2}{a^2} \] using the flip rule
  • 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)
im following and writing too
  • phi
*they mean...
  • phi
so far you have \[ \frac{1}{(\sin A)^2 }= \frac{c^2}{a^2} \] now what is 1/(tan A)^2 ?
1 sec ill write it out on paper and then ^^
\[\frac{ 1 }{ \tan A^2} \]=\[\frac{ b^2 }{ a^2 }\]
well i wrote tan A ^2 wrong
  • 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 ?
\[\frac{ c^2-b^2 }{ a }=1\]
  • phi
* a^2 in the bottom
i knew i forgot something
  • phi
now, what is the pythagoras equation for this problem ?
what does that mean
  • phi
do you know the pythagorean theorem ?
oh yea a^2+b^2=c^2
  • phi
and what do you get if you subtract b^2 from both sides ?
c^2/a^2=1-b^2
  • phi
no, that is too complicated. start with a^2+b^2=c^2 and write -b^2 on both sides and simplify
a^2+b^2-b^2=c^2-b^2 a^2=c^2-b^2
  • phi
now use that in your equation \[\frac{ c^2-b^2 }{ a^2 }=1 \] I would replace the top
  • phi
in other words c^2 -b^2 is the same as a^2 (by pythagoras)
oh
so what do we do now.....
  • phi
the idea is that c^2-b^2 = a^2 so replace c^2-b^2 with a^2 in your equation \[ \frac{ c^2-b^2 }{ a^2 }=1 \]
\[\frac{ a^2 }{ a^2 }=1\]
  • 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
oh ok

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