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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.
 11 months ago
 11 months ago
highschoolmom2010 Group Title
Verify that each equation is an identity by showing that each expression on the left simplifies to 1.
 11 months ago
 11 months ago

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highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
how far did you get ?
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
no where im lost
 11 months ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
take left hand side expression, plugin tanA = sinA/cosA
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
im still really confused
 11 months ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
did they teach you ' tan = sin/cos' yet ?
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
nope
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
so im learning it here
 11 months ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
ok, dont learn it here. wait until they start them officially lets do it the known way
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
ok
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
they may want you to use the definitions of sin(A)= opp/adj = a/c and tan(A)= a/b
 11 months ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
dw:1376501404557:dw
 11 months ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
exactly, look at above triangle, first find sinA, and tanA
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
dw:1376501580565:dw
 11 months ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
yes, find tanA also, and plug them in left side of given expression
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
assuming that is alpha....
 11 months ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
uhh this time, thats 'a', sorry
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
they are using lower case for the side and capital for the angle opposite that side
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
\[\tan A=\frac{ a }{ b }\] oh then let me fix sin A
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
\[\sin A=\frac{ a }{ c }\]
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
now what is \[ \frac{1}{\sin A} \]
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
@phi im not sure
 11 months ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
plug them directly, we can simplify after pluggin in may be
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ 1 }{ \frac{ a }{ c} }\]
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
or do what you did , and multiply top and bottom by c/a
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
that ^^kinda confused me
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
\[ \frac{ 1\cdot \frac{c}{a} }{ \frac{ a }{ c} \cdot \frac{c}{a}} \] the idea is you can multiply top and bottom by the same number (or fraction) now simplify
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ \frac{ c }{ a } }{ 1 }\]
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
yes, and anything divided by 1 is itself
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
co the left side =c/a
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
so*
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
right side = b/a then
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
now what is \[ \frac{1}{\sin^2 A} =\frac{1}{\sin A} \cdot \frac{1}{\sin A} \]
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
no clue
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
you know 1/sin A = c/a so you know 1/sin A * 1/sinA = (c/a)*(c/a) \[ \frac{c}{a}\cdot \frac{c}{a}\]
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
which is top times top and bottom times bottom
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
another way to do this is remember \[ \sin^2 A \text{ means } \sin A \cdot \sin A \] you know sin A= a/c so \[ \sin^2 A \text{ means } \sin A \cdot \sin A = \frac{a}{c}\cdot \frac{a}{c}=\frac{a\cdot a}{c\cdot c}=\frac{a^2}{c^2}\]
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
and \[ \frac{1}{\sin^2 A} = \frac{c^2}{a^2} \] using the flip rule
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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)
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
im following and writing too
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
*they mean...
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
so far you have \[ \frac{1}{(\sin A)^2 }= \frac{c^2}{a^2} \] now what is 1/(tan A)^2 ?
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
1 sec ill write it out on paper and then ^^
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ 1 }{ \tan A^2} \]=\[\frac{ b^2 }{ a^2 }\]
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
well i wrote tan A ^2 wrong
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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 rewrite as one fraction ?
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ c^2b^2 }{ a }=1\]
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
* a^2 in the bottom
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
i knew i forgot something
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
now, what is the pythagoras equation for this problem ?
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
what does that mean
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
do you know the pythagorean theorem ?
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
oh yea a^2+b^2=c^2
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
and what do you get if you subtract b^2 from both sides ?
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
c^2/a^2=1b^2
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
no, that is too complicated. start with a^2+b^2=c^2 and write b^2 on both sides and simplify
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
a^2+b^2b^2=c^2b^2 a^2=c^2b^2
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
now use that in your equation \[\frac{ c^2b^2 }{ a^2 }=1 \] I would replace the top
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
in other words c^2 b^2 is the same as a^2 (by pythagoras)
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
oh
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
so what do we do now.....
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
the idea is that c^2b^2 = a^2 so replace c^2b^2 with a^2 in your equation \[ \frac{ c^2b^2 }{ a^2 }=1 \]
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ a^2 }{ a^2 }=1\]
 11 months ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
oh ok
 11 months ago

highschoolmom2010 Group TitleBest ResponseYou've already chosen the best response.1
ty @phi & @ganeshie8
 11 months ago
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