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Trig Idetinty proofs
- Thread starter rubeio
- Start date
ChaoticLlama
Junior Member
- Joined
- Dec 11, 2004
- Messages
- 199
So, for the first one
\(\displaystyle \frac{{1 - \cot ^2 x}}{{1 + \cot ^2 x}} + 2\cos ^2 x = 1\)
Clearly the left side is the one you want to tackle here. I Would start by making a common denominator.
\(\displaystyle L.S. = \frac{{1 - \cot ^2 x + 2\cos ^2 x(1 + \cot ^2 x)}}{{1 + \cot ^2 x}}\)
Now, let's just look at the numerator of the left side, and try to manipulate it to show that it equals 1 + cot²x.
\(\displaystyle \begin{array}{l}
L.S. Numerator = 1 - \cot ^2 x + 2\cos ^2 x(1 + \cot ^2 x) \\
= 1 - \cot ^2 x + 2\cos ^2 x + 2\cos ^2 x\cot ^2 x \\
= 1 - \frac{{\cos ^2 x}}{{\sin ^2 x}} + 2\cos ^2 x + \frac{{2\cos ^4 x}}{{\sin ^2 x}} \\
= 1 - \cos ^2 x\left( {\frac{1}{{\sin ^2 x}} - 2 - 2\frac{{\cos ^2 x}}{{\sin ^2 x}}} \right) \\
= 1 - \frac{{\cos ^2 x}}{{\sin ^2 x}}\left( {1 - 2\cos ^2 x - 2\sin ^2 x} \right) \\
= 1 - \cot ^2 x\left( {1 - 2} \right) \\
= 1 + \cot ^2 x \\
\end{array}\)
Therefore, when we look at the whole left side of the equation, we are now left with simply:
\(\displaystyle \begin{array}{l}
L.S. = \frac{{1 + \cot ^2 x}}{{1 + \cot ^2 x}} \\
= 1 \\
= R.S. \\
\end{array}\)
Post if you are having any problems with the second one.
\(\displaystyle \frac{{1 - \cot ^2 x}}{{1 + \cot ^2 x}} + 2\cos ^2 x = 1\)
Clearly the left side is the one you want to tackle here. I Would start by making a common denominator.
\(\displaystyle L.S. = \frac{{1 - \cot ^2 x + 2\cos ^2 x(1 + \cot ^2 x)}}{{1 + \cot ^2 x}}\)
Now, let's just look at the numerator of the left side, and try to manipulate it to show that it equals 1 + cot²x.
\(\displaystyle \begin{array}{l}
L.S. Numerator = 1 - \cot ^2 x + 2\cos ^2 x(1 + \cot ^2 x) \\
= 1 - \cot ^2 x + 2\cos ^2 x + 2\cos ^2 x\cot ^2 x \\
= 1 - \frac{{\cos ^2 x}}{{\sin ^2 x}} + 2\cos ^2 x + \frac{{2\cos ^4 x}}{{\sin ^2 x}} \\
= 1 - \cos ^2 x\left( {\frac{1}{{\sin ^2 x}} - 2 - 2\frac{{\cos ^2 x}}{{\sin ^2 x}}} \right) \\
= 1 - \frac{{\cos ^2 x}}{{\sin ^2 x}}\left( {1 - 2\cos ^2 x - 2\sin ^2 x} \right) \\
= 1 - \cot ^2 x\left( {1 - 2} \right) \\
= 1 + \cot ^2 x \\
\end{array}\)
Therefore, when we look at the whole left side of the equation, we are now left with simply:
\(\displaystyle \begin{array}{l}
L.S. = \frac{{1 + \cot ^2 x}}{{1 + \cot ^2 x}} \\
= 1 \\
= R.S. \\
\end{array}\)
Post if you are having any problems with the second one.
Hello, rubeio!
\(\displaystyle \;\;\)But this time, it works out nicely . . .
We have: \(\displaystyle \L\,\frac{1\,-\,\frac{\cos^2x}{\sin^2x}}{1\,+\,\frac{\cos^2x}{\sin^2x}}\)\(\displaystyle \,+\,2\cdot\cos^2x\)
Multiply top and bottom of the fraction by \(\displaystyle \sin^2x:\)
\(\displaystyle \L\;\;\frac{\sin^2x\left(1\,-\,\frac{\cos^2x}{\sin^2x}\right)}{\sin^2x\left(1\,+\,\frac{\cos^2x}{\sin^2x}\right)}\)\(\displaystyle \,+\,2\cdot\cos^2x \;= \;\L\frac{\sin^2x\,-\,\cos^2x}{\sin^2x\,+\,\cos^2x}\)\(\displaystyle \,+\,2\cos^2x\)
Since \(\displaystyle \,sin^2x\,+\,\cos^2x\:=\:1\), we have: \(\displaystyle \,\sin^2x\,-\,\cos^2x\,+\,2\cdot\cos^2x \;=\;\sin^2x\,+\,\cos^2x\;=\;1\)
We have: \(\displaystyle \L\,\frac{\sin^2x\,-\,\frac{\sin x}{\cos x}}{\cos^2x\,-\,\frac{\cos x}{\sin x}}\)
Factor: \(\displaystyle \L\,\frac{\sin x\left(\sin x\,-\,\frac{1}{\cos x}\right)}{\cos x\left(\cos x\,-\,\frac{1}{\sin x}\right)} \;= \;\frac{\sin x\left(\frac{\sout{\sin x\cdot\cos x\,-\,1}}{\cos x}\right)}{\cos x\left(\frac{\sout{\sin x\cdot\cos x\,-\,1}}{\sin x}\right)} \;= \;\frac{\sin x\cdot\frac{1}{\cos x}}{\cos x\cdot\frac{1}{\sin x}}\)
\(\displaystyle \L\;\;\;=\;\frac{\sin^2x}{\cos^2x}\;=\;\left(\frac{\sin x}{\cos x}\right)^2\)\(\displaystyle \;= \;\tan^2x\)
I usually don't recommend changin everything to sines and cosines; there's usually a better way.\(\displaystyle \L\frac{1\,-\,\cot^2x}{1\,+\,\cot^2x}\)\(\displaystyle \,+\,2\cdot\cos^2x\:=\:1\)
\(\displaystyle \;\;\)But this time, it works out nicely . . .
We have: \(\displaystyle \L\,\frac{1\,-\,\frac{\cos^2x}{\sin^2x}}{1\,+\,\frac{\cos^2x}{\sin^2x}}\)\(\displaystyle \,+\,2\cdot\cos^2x\)
Multiply top and bottom of the fraction by \(\displaystyle \sin^2x:\)
\(\displaystyle \L\;\;\frac{\sin^2x\left(1\,-\,\frac{\cos^2x}{\sin^2x}\right)}{\sin^2x\left(1\,+\,\frac{\cos^2x}{\sin^2x}\right)}\)\(\displaystyle \,+\,2\cdot\cos^2x \;= \;\L\frac{\sin^2x\,-\,\cos^2x}{\sin^2x\,+\,\cos^2x}\)\(\displaystyle \,+\,2\cos^2x\)
Since \(\displaystyle \,sin^2x\,+\,\cos^2x\:=\:1\), we have: \(\displaystyle \,\sin^2x\,-\,\cos^2x\,+\,2\cdot\cos^2x \;=\;\sin^2x\,+\,\cos^2x\;=\;1\)
There are several approaches to this one . . . none of them are pleasant.\(\displaystyle \L\frac{\sin^2x\,-\,\tan x}{\cos^2x\,-\,\cot x}\)\(\displaystyle \;=\;\tan^2x\)
We have: \(\displaystyle \L\,\frac{\sin^2x\,-\,\frac{\sin x}{\cos x}}{\cos^2x\,-\,\frac{\cos x}{\sin x}}\)
Factor: \(\displaystyle \L\,\frac{\sin x\left(\sin x\,-\,\frac{1}{\cos x}\right)}{\cos x\left(\cos x\,-\,\frac{1}{\sin x}\right)} \;= \;\frac{\sin x\left(\frac{\sout{\sin x\cdot\cos x\,-\,1}}{\cos x}\right)}{\cos x\left(\frac{\sout{\sin x\cdot\cos x\,-\,1}}{\sin x}\right)} \;= \;\frac{\sin x\cdot\frac{1}{\cos x}}{\cos x\cdot\frac{1}{\sin x}}\)
\(\displaystyle \L\;\;\;=\;\frac{\sin^2x}{\cos^2x}\;=\;\left(\frac{\sin x}{\cos x}\right)^2\)\(\displaystyle \;= \;\tan^2x\)