highschoolmom2010
Verify that each equation is an identity by showing that each expression on the left simplifies to 1.
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phi
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how far did you get ?
highschoolmom2010
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no where im lost
ganeshie8
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take left hand side expression,
plugin tanA = sinA/cosA
highschoolmom2010
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im still really confused
ganeshie8
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did they teach you ' tan = sin/cos' yet ?
highschoolmom2010
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so im learning it here
ganeshie8
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ok, dont learn it here. wait until they start them officially
lets do it the known way
phi
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they may want you to use the definitions of sin(A)= opp/adj = a/c
and tan(A)= a/b
ganeshie8
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|dw:1376501404557:dw|
ganeshie8
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exactly,
look at above triangle, first find sinA, and tanA
highschoolmom2010
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|dw:1376501580565:dw|
ganeshie8
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yes, find tanA also, and plug them in left side of given expression
highschoolmom2010
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assuming that is alpha....
ganeshie8
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uhh this time, thats 'a', sorry
phi
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they are using lower case for the side and capital for the angle opposite that side
highschoolmom2010
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\[\tan A=\frac{ a }{ b }\]
oh then let me fix sin A
highschoolmom2010
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\[\sin A=\frac{ a }{ c }\]
phi
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now what is
\[ \frac{1}{\sin A} \]
highschoolmom2010
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@phi im not sure
ganeshie8
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plug them directly, we can simplify after pluggin in may be
highschoolmom2010
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\[\frac{ 1 }{ \frac{ a }{ c} }\]
phi
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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
phi
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or do what you did , and multiply top and bottom by c/a
highschoolmom2010
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that ^^kinda confused me
phi
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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
highschoolmom2010
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\[\frac{ \frac{ c }{ a } }{ 1 }\]
phi
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yes, and anything divided by 1 is itself
highschoolmom2010
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co the left side =c/a
phi
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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
highschoolmom2010
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right side = b/a then
phi
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now what is
\[ \frac{1}{\sin^2 A} =\frac{1}{\sin A} \cdot \frac{1}{\sin A} \]
phi
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you know 1/sin A = c/a
so you know 1/sin A * 1/sinA = (c/a)*(c/a)
\[ \frac{c}{a}\cdot \frac{c}{a}\]
phi
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which is top times top and bottom times bottom
phi
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lost ?
phi
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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
\[ \sin^2 A \text{ means } \sin A \cdot \sin A = \frac{a}{c}\cdot \frac{a}{c}=\frac{a\cdot a}{c\cdot c}=\frac{a^2}{c^2}\]
phi
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and
\[ \frac{1}{\sin^2 A} = \frac{c^2}{a^2} \]
using the flip rule
phi
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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)
highschoolmom2010
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im following and writing too
phi
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*they mean...
phi
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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 ?
highschoolmom2010
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1 sec ill write it out on paper and then ^^
highschoolmom2010
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\[\frac{ 1 }{ \tan A^2} \]=\[\frac{ b^2 }{ a^2 }\]
highschoolmom2010
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well i wrote tan A ^2 wrong
phi
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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 ?
highschoolmom2010
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\[\frac{ c^2-b^2 }{ a }=1\]
phi
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* a^2 in the bottom
highschoolmom2010
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i knew i forgot something
phi
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now, what is the pythagoras equation for this problem ?
highschoolmom2010
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what does that mean
phi
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do you know the pythagorean theorem ?
highschoolmom2010
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oh yea
a^2+b^2=c^2
phi
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and what do you get if you subtract b^2 from both sides ?
highschoolmom2010
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c^2/a^2=1-b^2
phi
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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
highschoolmom2010
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a^2+b^2-b^2=c^2-b^2
a^2=c^2-b^2
phi
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now use that in your equation
\[\frac{ c^2-b^2 }{ a^2 }=1 \]
I would replace the top
phi
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in other words c^2 -b^2 is the same as a^2
(by pythagoras)
highschoolmom2010
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so what do we do now.....
phi
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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
\[ \frac{ c^2-b^2 }{ a^2 }=1 \]
highschoolmom2010
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\[\frac{ a^2 }{ a^2 }=1\]
phi
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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
highschoolmom2010
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ty @phi & @ganeshie8