Arthur D. answered • 04/25/17
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(sin/[1-csc])-(sin/[1+csc])=-2cot^2
rewrite with a common denominator
[(sin)(1+csc)-(sin)(1-csc)]/(1-csc)(1+csc)=
{(sin)([1+csc]-(sin)[1-csc]}/(1-csc^2)=
factor out sin in the numerator
(sin)(1+csc-[1-csc])/(1-csc^2)=
(sin)(1+csc-1+csc)/(1-csc^2)=
(sin)(2csc)/(1-csc^2)=
1+cot^2=csc^2 is a trig identity which yields -cot^2=1-csc^2
(sin)(2csc)/(-cot^2)=
csc=1/sin is a trig identity
(sin)(2)(1/sin)/(-cot^2)=
2/(-cot2)=
-2/cot2 is what I got instead of -2cot2
is there a typo ?
Kenneth S.
Arthur, please do not write trig functions without arguments!
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04/25/17
Arthur D.
tutor
Kenneth, you're right. I should write the arguments. I got in the habit of not writing the arguments when I had to use a symbol such as "θ" and got tired of using the special character symbol to access it.
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04/25/17
Kenneth S.
Here's a good trick, Arthur. Click on the capital omega ikon and select θ.
Then highlight this character in the answer or comment text box.and do CTRL C, thus making a copy of it.
Then you can type sin, do control V, thus producing sin θ ... and you can drop θ repeatedly whenever needed for more θ characters, at will, avoiding calling on the images through the Omega dropdown method over & over.
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04/25/17
Kemal G.
04/25/17