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Expert: - 5/12/2008


Question
QUESTION: Hi, I'm having trouble solving this trig identity,
1-(sin^2xtanx/tanx+1)-(cos^2x/1+tanx)=sinxcosx


ANSWER: Is this really suppose to be
1-(sin^2(x)tan(x)/(tan(x)+1))-(cos^2x/(1+tan(x))=sin(x)cos(x)?
I think so!

Turn tan(x) into sin(x)/cos(x), don't forget parentheses, and combine the fractions.  You will get a sin^3(x)+cos^3(x) on top.
Note the a^3+b^3=(a+b)(a^2-ab+b^2).

Thanks for this and previous questions this year.



---------- FOLLOW-UP ----------

QUESTION: I can't get sin^3(x)+cos^3(x) on top.  It doesn't work.  And once I do get sin^3+cos^3 on top how does that turn into sin(x)cos(x)?

Answer
Take sin^2(x)tan(x)/(tan(x)+1).
It is known that tan(x)=sin(x)/cos(x), so this is
sin^2(x)(sin(x)/cos(x))/(sin(x)/cos(x) + 1).
Multiply top and bottom by cos and note the sin(x)'s combine to give
sin^3(x)/(sin(x)+cos(x)).

Take cos^2x/(1+tan(x)).
Multiply the top and bottom by cos(x).  On top this gives cos^3(x) and on the bottom gives cos(x) + cos(x)tan(x).  Again, it is known that tan(x)=sin(x)/cos(x), so the bottom is cos(x) + sin(x).
Our final fraction is cos^3(x)/(cos(x)+sin(x).

The fractions can be combined (since the denominators are equal, i.e., sin(x)+cos(x)=cos(x)+sin(x)), and both are being subtracted from 1) and as I gave you how to factor a^3+b^3, you can factor sin^3(x)+cos^3(x).

In other words, sin^3(x)+cos^3(x)=
(sin(x)+cos(x))(sin^2(x)-sin(x)cos(x)+cos^2(x)) and there is a sin(x)+cos(x) in the denominator, so the they cancel.

Note that all of this is subtracted from one, so we have
1 - sin^2(x) + sin(x)cos(x) - cos^2(x).  This turns into
1 - (sin^2(x)+cos^2(x)) + sin(x)cos(x).  
Recognize sin^2(x)+cos^2(x)?  All you're left with is the answer.

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