trig. identity: Verify that 2sinA-cscA= sinA-cotA/cscA
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skeeter
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check the terms in the first equation (see if you copied it correctly) ... it's not an identity.
2tanx/tan(2x) =
since tan(2x) = 2tanx/(1 - tan<sup>2</sup>x) ...
(2tanx)*(1 - tan<sup>2</sup>x)/(2tanx) =
1 - tan<sup>2</sup>x =
since 1 + tan<sup>2</sup>x = sec<sup>2</sup>x ...
1 - (sec<sup>2</sup>x - 1) =
2 - sec<sup>2</sup>x
2tanx/tan(2x) =
since tan(2x) = 2tanx/(1 - tan<sup>2</sup>x) ...
(2tanx)*(1 - tan<sup>2</sup>x)/(2tanx) =
1 - tan<sup>2</sup>x =
since 1 + tan<sup>2</sup>x = sec<sup>2</sup>x ...
1 - (sec<sup>2</sup>x - 1) =
2 - sec<sup>2</sup>x
Re: reply
Hello, aplubin!
\(\displaystyle \csc x\) is negative. .\(\displaystyle x\) is in quadrant 3 or 4.
\(\displaystyle \cos x\) is positive. .\(\displaystyle x\) is in quadrant 1 or 4.
. . Hence: \(\displaystyle x\) is in quadrant 4.
Since \(\displaystyle \csc x\:=\:-\frac{5}{3} \:=\:\frac{hyp}{opp}\),
. . we have: \(\displaystyle \,opp\,=\,-3,\:hyp\,=\,5\) . . . and Pythagorus gives us: \(\displaystyle \,adj\,=\,4\)
Hence: \(\displaystyle \L\:\sin x\,=\,-\frac{3}{5},\:\cos x\,=\,\frac{4}{5}\;\) [1]
\(\displaystyle \cot y\) is positive. .\(\displaystyle y\) is in quadrant 1 or 3.
\(\displaystyle \cos y\) is positive. .\(\displaystyle y\) is in quadrant 1 or 4.
. . Hence: \(\displaystyle \,y\) is in quadrant 1.
Since \(\displaystyle \cot y \:=\:\frac{7}{24}\:=\:\frac{adj}{opp}\)
. . we have: \(\displaystyle \,opp\,=\,24,\:adj\,=\,7\) . . . and Pythagorus gives us: \(\displaystyle \,hyp\,=\,25\)
Hence: \(\displaystyle \L\,\sin y\,=\,\frac{24}{25},\:\cos y\,=\,\frac{7}{25}\;\) [2]
Identity: \(\displaystyle \:\cos(y\,-\,x)\;=\;\cos(y)\cdot\cos(x)\,+\,\sin(x)\cdot\sin(y)\)
We have: \(\displaystyle \:\cos(y\,-\,x)\;=\;\left(\frac{7}{25}\right)\left(\frac{4}{5}\right)\,+\,\left(\frac{24}{25}\right)\left(-\frac{3}{5}\right)\;=\;\frac{28}{125}\,-\,\frac{72}{125}\;=\;\fbox{-\frac{44}{125}}\)
Identity: \(\displaystyle \:\sin(x\,-\,y)\;=\;\sin(x)\cdot\cos(y)\,-\,\sin(y)\cdot\cos(x)\)
We have: \(\displaystyle \:\sin(x\,-\,y) \;=\;\left(-\frac{3}{5}\right)\left(\frac{7}{25}\right)\,-\,\left(\frac{24}{25}\right)\left(\frac{4}{5}\right) \;=\;-\frac{21}{125}\,-\,\frac{96}{125}\;=\;\fbox{-\frac{117}{125}}\)
Hello, aplubin!
\(\displaystyle \csc x\) is negative. .\(\displaystyle x\) is in quadrant 3 or 4.
\(\displaystyle \cos x\) is positive. .\(\displaystyle x\) is in quadrant 1 or 4.
. . Hence: \(\displaystyle x\) is in quadrant 4.
Since \(\displaystyle \csc x\:=\:-\frac{5}{3} \:=\:\frac{hyp}{opp}\),
. . we have: \(\displaystyle \,opp\,=\,-3,\:hyp\,=\,5\) . . . and Pythagorus gives us: \(\displaystyle \,adj\,=\,4\)
Hence: \(\displaystyle \L\:\sin x\,=\,-\frac{3}{5},\:\cos x\,=\,\frac{4}{5}\;\) [1]
\(\displaystyle \cot y\) is positive. .\(\displaystyle y\) is in quadrant 1 or 3.
\(\displaystyle \cos y\) is positive. .\(\displaystyle y\) is in quadrant 1 or 4.
. . Hence: \(\displaystyle \,y\) is in quadrant 1.
Since \(\displaystyle \cot y \:=\:\frac{7}{24}\:=\:\frac{adj}{opp}\)
. . we have: \(\displaystyle \,opp\,=\,24,\:adj\,=\,7\) . . . and Pythagorus gives us: \(\displaystyle \,hyp\,=\,25\)
Hence: \(\displaystyle \L\,\sin y\,=\,\frac{24}{25},\:\cos y\,=\,\frac{7}{25}\;\) [2]
Identity: \(\displaystyle \:\cos(y\,-\,x)\;=\;\cos(y)\cdot\cos(x)\,+\,\sin(x)\cdot\sin(y)\)
We have: \(\displaystyle \:\cos(y\,-\,x)\;=\;\left(\frac{7}{25}\right)\left(\frac{4}{5}\right)\,+\,\left(\frac{24}{25}\right)\left(-\frac{3}{5}\right)\;=\;\frac{28}{125}\,-\,\frac{72}{125}\;=\;\fbox{-\frac{44}{125}}\)
Identity: \(\displaystyle \:\sin(x\,-\,y)\;=\;\sin(x)\cdot\cos(y)\,-\,\sin(y)\cdot\cos(x)\)
We have: \(\displaystyle \:\sin(x\,-\,y) \;=\;\left(-\frac{3}{5}\right)\left(\frac{7}{25}\right)\,-\,\left(\frac{24}{25}\right)\left(\frac{4}{5}\right) \;=\;-\frac{21}{125}\,-\,\frac{96}{125}\;=\;\fbox{-\frac{117}{125}}\)
skeeter
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Re: reply
as I said before, this is not an identity ... let A = pi/6, the two sides of the equation are not equal.
2sin(pi/6) - csc(pi/6) = 2(1/2) - 2 = -1
sin(pi/6) - [cot(pi/6)/csc(pi/6)] = 1/2 - [sqrt(3)/2] = [1 - sqrt(3)]/2
aplubin said:
as I said before, this is not an identity ... let A = pi/6, the two sides of the equation are not equal.
2sin(pi/6) - csc(pi/6) = 2(1/2) - 2 = -1
sin(pi/6) - [cot(pi/6)/csc(pi/6)] = 1/2 - [sqrt(3)/2] = [1 - sqrt(3)]/2
skeeter
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\(\displaystyle \L \sin{x} - \frac{\cot{x}}{\sec{x}} =\)
\(\displaystyle \L \sin{x} - \frac{\cos^2{x}}{\sin{x}} =\)
\(\displaystyle \L \frac{\sin^2{x} - \cos^2{x}}{\sin{x}} =\)
\(\displaystyle \L \frac{\sin^2{x} - (1 - \sin^2{x})}{\sin{x}} =\)
\(\displaystyle \L \frac{2\sin^2{x} - 1}{\sin{x}} =\)
\(\displaystyle \L \frac{2\sin^2{x}}{\sin{x}} - \frac{1}{\sin{x}} =\)
\(\displaystyle \L 2\sin{x} - \csc{x}\)
\(\displaystyle \L \sin{x} - \frac{\cos^2{x}}{\sin{x}} =\)
\(\displaystyle \L \frac{\sin^2{x} - \cos^2{x}}{\sin{x}} =\)
\(\displaystyle \L \frac{\sin^2{x} - (1 - \sin^2{x})}{\sin{x}} =\)
\(\displaystyle \L \frac{2\sin^2{x} - 1}{\sin{x}} =\)
\(\displaystyle \L \frac{2\sin^2{x}}{\sin{x}} - \frac{1}{\sin{x}} =\)
\(\displaystyle \L 2\sin{x} - \csc{x}\)