anonymous
  • anonymous
Simplify the trigonometric expression
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[1\div1+\sin \theta + 1\div 1 - \sin \theta \]
Psymon
  • Psymon
\[\frac{ 1 }{ 1+\sin \theta }+\frac{ 1 }{ 1-\sin \theta } \] Well, you want to make these into one fraction by regular common denominator rules. So if you knowhow to properly combine those into one fraction, things will just kind of simplify together.
anonymous
  • anonymous
well im not asking for an answer but could you help by explaining step by step how to get the answer

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Psymon
  • Psymon
Right. I wasn't sure if youd know how to go about making those into one fraction or not. Well, I want to multiply the two denominators together. Now just like when finding common denominators with all numbers, whatever I multiply into the bottom, I must multiply into the top. So it would kind of look like this: \[\frac{ (1-\sin \theta)1+(1+\sin \theta)1 }{ (1+\sin \theta)(1-\sin \theta) }\] I multiplied the two denominators together. The first fraction on the left was multiplied by 1-sinx on the bottom, so I multiplied it by that on the top. The right fraction was multiplied by 1+sinx on the bottom, so then it was multiplied by 1+sinx on the top. Doing this means they now have a common denominator and can be written as one fraction like above. Now the top can simplify pretty easily: \[\frac{ 2 }{ (1+\sin \theta)(1-\sin \theta) }\] Now you just need to multiply out the bottom. You see what I did up to this point?
anonymous
  • anonymous
okay!:) that helps! would the answer have (sin)? my answers have (cos) (sec) (csc) and (cot)
Psymon
  • Psymon
There would be no sin in your answer. After you multiply the bottom out you can use an identity
anonymous
  • anonymous
well i do kno the answer is 2(???)^2theta
anonymous
  • anonymous
but idk about the identity
Psymon
  • Psymon
Did you multiply the bottomout properly?
anonymous
  • anonymous
yes i did. but i dont understand the identity part
Psymon
  • Psymon
Can I see what you ended up with?
jdoe0001
  • jdoe0001
\(\bf \cfrac{ 1 }{ 1+sin(\theta) }+\cfrac{ 1 }{ 1-sin(\theta) } \implies \cfrac{(1-sin(\theta))+(1+sin(\theta))}{(1+sin(\theta))(1-sin(\theta))}\\ \cfrac{1\cancel{-sin(\theta)}+1\cancel{+sin(\theta)}}{(1+sin(\theta))(1-sin(\theta))} \implies \cfrac{2}{(1+sin(\theta))(1-sin(\theta))}\) as already pointed out by Psymon
jdoe0001
  • jdoe0001
now if you keep in mind that "difference of squares" is => \(\bf (a-b)(a+b) = (a^2-b^2)\) then we could say that \(\bf (1+sin(\theta))(1-sin(\theta)) \implies 1^2-sin^2(\theta) \implies 1-sin^2(\theta)\) thus then \(\bf \cfrac{2}{(1+sin(\theta))(1-sin(\theta))} \implies \cfrac{2}{1-sin^2(\theta)}\\ \textit{recall the trig identity of } \\sin^2(\theta)+cos^2(\theta)=1 \implies cos^2(\theta)= 1-sin^2(\theta)\)
jdoe0001
  • jdoe0001
so, can you get it from there?

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