If $sec theta + tan theta =x$, then find the value of $sin theta$.
$begingroup$
If $sec theta + tan theta =x$, then find the value of $sin theta$.
$$sec theta + tan theta = x$$
$$dfrac {1}{cos theta }+dfrac {sin theta }{cos theta }=x$$
$$dfrac {1+sin theta }{sqrt {1-sin^2 theta }}=x$$
$$1+sin theta =xsqrt {1-sin^2 theta }$$
$$1+2sin theta + sin^2 theta = x^2-x^2 sin^2 theta $$
$$x^2 sin^2 theta + sin^2 theta + 2sin theta = x^2-1$$
$$sin^2 theta (x^2+1) + 2sin theta =x^2-1$$
algebra-precalculus trigonometry
$endgroup$
add a comment |
$begingroup$
If $sec theta + tan theta =x$, then find the value of $sin theta$.
$$sec theta + tan theta = x$$
$$dfrac {1}{cos theta }+dfrac {sin theta }{cos theta }=x$$
$$dfrac {1+sin theta }{sqrt {1-sin^2 theta }}=x$$
$$1+sin theta =xsqrt {1-sin^2 theta }$$
$$1+2sin theta + sin^2 theta = x^2-x^2 sin^2 theta $$
$$x^2 sin^2 theta + sin^2 theta + 2sin theta = x^2-1$$
$$sin^2 theta (x^2+1) + 2sin theta =x^2-1$$
algebra-precalculus trigonometry
$endgroup$
$begingroup$
See also, math.stackexchange.com/questions/2184596/…
$endgroup$
– lab bhattacharjee
May 14 '17 at 14:24
add a comment |
$begingroup$
If $sec theta + tan theta =x$, then find the value of $sin theta$.
$$sec theta + tan theta = x$$
$$dfrac {1}{cos theta }+dfrac {sin theta }{cos theta }=x$$
$$dfrac {1+sin theta }{sqrt {1-sin^2 theta }}=x$$
$$1+sin theta =xsqrt {1-sin^2 theta }$$
$$1+2sin theta + sin^2 theta = x^2-x^2 sin^2 theta $$
$$x^2 sin^2 theta + sin^2 theta + 2sin theta = x^2-1$$
$$sin^2 theta (x^2+1) + 2sin theta =x^2-1$$
algebra-precalculus trigonometry
$endgroup$
If $sec theta + tan theta =x$, then find the value of $sin theta$.
$$sec theta + tan theta = x$$
$$dfrac {1}{cos theta }+dfrac {sin theta }{cos theta }=x$$
$$dfrac {1+sin theta }{sqrt {1-sin^2 theta }}=x$$
$$1+sin theta =xsqrt {1-sin^2 theta }$$
$$1+2sin theta + sin^2 theta = x^2-x^2 sin^2 theta $$
$$x^2 sin^2 theta + sin^2 theta + 2sin theta = x^2-1$$
$$sin^2 theta (x^2+1) + 2sin theta =x^2-1$$
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Dec 28 '18 at 18:16
amWhy
1
1
asked May 14 '17 at 12:55
AryabhattaAryabhatta
371418
371418
$begingroup$
See also, math.stackexchange.com/questions/2184596/…
$endgroup$
– lab bhattacharjee
May 14 '17 at 14:24
add a comment |
$begingroup$
See also, math.stackexchange.com/questions/2184596/…
$endgroup$
– lab bhattacharjee
May 14 '17 at 14:24
$begingroup$
See also, math.stackexchange.com/questions/2184596/…
$endgroup$
– lab bhattacharjee
May 14 '17 at 14:24
$begingroup$
See also, math.stackexchange.com/questions/2184596/…
$endgroup$
– lab bhattacharjee
May 14 '17 at 14:24
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Here is a different approach: Since $1 + tan^2theta = sec^2theta$, we have
$$sec^2theta - tan^2theta = 1$$
Factoring yields
$$(sectheta + tantheta)(sectheta - tantheta) = 1$$
Since we are given that $sectheta + tantheta = x$, we obtain
$$x(sectheta - tantheta) = 1$$
Therefore,
$$sectheta - tantheta = frac{1}{x}$$
This yields the system of equations
begin{align*}
sectheta + tantheta & = x tag{1}\
sectheta - tantheta & = frac{1}{x} tag{2}
end{align*}
Adding equations 1 and 2 and solving for $sectheta$ yields
begin{align*}
2sectheta & = x + frac{1}{x}\
2sectheta & = frac{x^2 + 1}{x}\
sectheta & = frac{x^2 + 1}{2x}
end{align*}
Therefore,
$$costheta = frac{1}{sectheta} = frac{2x}{x^2 + 1}$$
Subtracting equation 2 from equation 1 and solving for $tantheta$ yields
begin{align*}
2tantheta & = x - frac{1}{x}\
2tantheta & = frac{x^2 - 1}{x}\
tantheta & = frac{x^2 - 1}{2x}
end{align*}
Thus,
$$sintheta = tanthetacostheta = frac{x^2 - 1}{2x} cdot frac{2x}{x^2 + 1} = frac{x^2 - 1}{x^2 + 1}$$
$endgroup$
$begingroup$
I didn't get the answer. The answer is $dfrac {1-x^2}{1+x^2}$.
$endgroup$
– Aryabhatta
May 14 '17 at 13:15
$begingroup$
What did you get for $sectheta$ and $tantheta$?
$endgroup$
– N. F. Taussig
May 14 '17 at 13:19
$begingroup$
I got $sec theta=dfrac {x^2+1}{2x}$ and $tan theta =dfrac {2x-x^2-1}{2x}$.
$endgroup$
– Aryabhatta
May 14 '17 at 13:21
$begingroup$
I agree with your answer for $sectheta$. When I solved for $tantheta$, I obtained $$frac{x^2 - 1}{2x}$$ which led to the answer $$frac{x^2 - 1}{1 + x^2}$$
$endgroup$
– N. F. Taussig
May 14 '17 at 13:26
$begingroup$
I have added the details of my calculations. I checked my answer for the angles $pi/6$, $pi/4$, and $pi/3$. In each case, it gave the correct answer, while the answer you stated gives the wrong sign.
$endgroup$
– N. F. Taussig
May 14 '17 at 13:50
add a comment |
$begingroup$
The equation becomes
$$
1+sintheta=xcostheta
$$
Set $X=costheta$ and $Y=sintheta$, so the equation becomes
$$
begin{cases}
X^2+Y^2=1 \[4px]
1+Y=xX
end{cases}
$$
Note that $xne0$ and substitute $X=x^{-1}(1+Y)$ in the first equation getting
$$
(1+Y)^2+x^2Y^2=x^2
$$
that simplifies to
$$
(1+x^2)Y^2+2Y+1-x^2=0
$$
that yields
$$
Y=-1 qquadtext{or}qquad Y=frac{x^2-1}{x^2+1}
$$
Is $Y=-1$ a solution for the problem?
By the way, you also get $costheta$, since
$$
X=frac{1}{x}(1+Y)=frac{1}{x}frac{x^2+1+x^2-1}{x^2+1}=frac{2x}{x^2+1}
$$
$endgroup$
add a comment |
$begingroup$
From where you are:
You obtained a quadratic function in $sin(theta)$. Perform the substitution $u =sin(theta)$.
We obtain the quadratic (in $u$):
$$(x^2+1)u^2 + 2u - x^2 +1 = 0$$
$$Rightarrow u_{1,2} = frac{- 2 pm sqrt{4 - 4(x^2+1)(1-x^2)}}{2(x^2+1)}$$
$$ = frac{- 2 pm sqrt{4 + 4(x^4 -1)}}{2(x^2+1)}$$
$$ = frac{- 2 pm sqrt{4x^4}}{2(x^2+1)}$$
$$ = frac{- 2 pm 2x^2}{2(x^2+1)}$$
$$ = frac{- 1 pm x^2}{x^2+1}$$
Therefore,
$$sin(theta)_{1,2} = frac{- 1 pm x^2}{x^2+1}$$
One of those solutions will not work out.
This happened because you squared the equation multiple times and we know that $a = b$ is not equivalent with $a^2 = b^2$, so you should fill in both solutions in the original expression and see which one works and which one doesn't.
$endgroup$
$begingroup$
What is discriminant method?
$endgroup$
– Aryabhatta
May 14 '17 at 13:05
$begingroup$
en.wikipedia.org/wiki/Discriminant#Degree_2
$endgroup$
– Math_QED
May 14 '17 at 13:06
$begingroup$
I didn't understand...
$endgroup$
– Aryabhatta
May 14 '17 at 13:14
$begingroup$
If you want to solve a solution in the form $ax^2 + bx + c = 0$, then the solutions are given by $x_{1,2} = frac{-b pm sqrt{b^2 - 4ac}}{2a}$
$endgroup$
– Math_QED
May 14 '17 at 13:21
$begingroup$
Here $x = u, a = (x^2 + 1), b = 2, c = - x^2 + 1$
$endgroup$
– Math_QED
May 14 '17 at 13:22
|
show 1 more comment
$begingroup$
WLOG let $theta=dfracpi2-2yimplies x=csc2y+cot2y=dfrac{1+cos2y}{sin2y}=cot y$
$$sintheta=cos2y=dfrac{1-tan^2y}{1+tan^2y}=dfrac{cot^2y-1}{cot^2y+1}=?$$
$endgroup$
add a comment |
Your Answer
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4 Answers
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active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a different approach: Since $1 + tan^2theta = sec^2theta$, we have
$$sec^2theta - tan^2theta = 1$$
Factoring yields
$$(sectheta + tantheta)(sectheta - tantheta) = 1$$
Since we are given that $sectheta + tantheta = x$, we obtain
$$x(sectheta - tantheta) = 1$$
Therefore,
$$sectheta - tantheta = frac{1}{x}$$
This yields the system of equations
begin{align*}
sectheta + tantheta & = x tag{1}\
sectheta - tantheta & = frac{1}{x} tag{2}
end{align*}
Adding equations 1 and 2 and solving for $sectheta$ yields
begin{align*}
2sectheta & = x + frac{1}{x}\
2sectheta & = frac{x^2 + 1}{x}\
sectheta & = frac{x^2 + 1}{2x}
end{align*}
Therefore,
$$costheta = frac{1}{sectheta} = frac{2x}{x^2 + 1}$$
Subtracting equation 2 from equation 1 and solving for $tantheta$ yields
begin{align*}
2tantheta & = x - frac{1}{x}\
2tantheta & = frac{x^2 - 1}{x}\
tantheta & = frac{x^2 - 1}{2x}
end{align*}
Thus,
$$sintheta = tanthetacostheta = frac{x^2 - 1}{2x} cdot frac{2x}{x^2 + 1} = frac{x^2 - 1}{x^2 + 1}$$
$endgroup$
$begingroup$
I didn't get the answer. The answer is $dfrac {1-x^2}{1+x^2}$.
$endgroup$
– Aryabhatta
May 14 '17 at 13:15
$begingroup$
What did you get for $sectheta$ and $tantheta$?
$endgroup$
– N. F. Taussig
May 14 '17 at 13:19
$begingroup$
I got $sec theta=dfrac {x^2+1}{2x}$ and $tan theta =dfrac {2x-x^2-1}{2x}$.
$endgroup$
– Aryabhatta
May 14 '17 at 13:21
$begingroup$
I agree with your answer for $sectheta$. When I solved for $tantheta$, I obtained $$frac{x^2 - 1}{2x}$$ which led to the answer $$frac{x^2 - 1}{1 + x^2}$$
$endgroup$
– N. F. Taussig
May 14 '17 at 13:26
$begingroup$
I have added the details of my calculations. I checked my answer for the angles $pi/6$, $pi/4$, and $pi/3$. In each case, it gave the correct answer, while the answer you stated gives the wrong sign.
$endgroup$
– N. F. Taussig
May 14 '17 at 13:50
add a comment |
$begingroup$
Here is a different approach: Since $1 + tan^2theta = sec^2theta$, we have
$$sec^2theta - tan^2theta = 1$$
Factoring yields
$$(sectheta + tantheta)(sectheta - tantheta) = 1$$
Since we are given that $sectheta + tantheta = x$, we obtain
$$x(sectheta - tantheta) = 1$$
Therefore,
$$sectheta - tantheta = frac{1}{x}$$
This yields the system of equations
begin{align*}
sectheta + tantheta & = x tag{1}\
sectheta - tantheta & = frac{1}{x} tag{2}
end{align*}
Adding equations 1 and 2 and solving for $sectheta$ yields
begin{align*}
2sectheta & = x + frac{1}{x}\
2sectheta & = frac{x^2 + 1}{x}\
sectheta & = frac{x^2 + 1}{2x}
end{align*}
Therefore,
$$costheta = frac{1}{sectheta} = frac{2x}{x^2 + 1}$$
Subtracting equation 2 from equation 1 and solving for $tantheta$ yields
begin{align*}
2tantheta & = x - frac{1}{x}\
2tantheta & = frac{x^2 - 1}{x}\
tantheta & = frac{x^2 - 1}{2x}
end{align*}
Thus,
$$sintheta = tanthetacostheta = frac{x^2 - 1}{2x} cdot frac{2x}{x^2 + 1} = frac{x^2 - 1}{x^2 + 1}$$
$endgroup$
$begingroup$
I didn't get the answer. The answer is $dfrac {1-x^2}{1+x^2}$.
$endgroup$
– Aryabhatta
May 14 '17 at 13:15
$begingroup$
What did you get for $sectheta$ and $tantheta$?
$endgroup$
– N. F. Taussig
May 14 '17 at 13:19
$begingroup$
I got $sec theta=dfrac {x^2+1}{2x}$ and $tan theta =dfrac {2x-x^2-1}{2x}$.
$endgroup$
– Aryabhatta
May 14 '17 at 13:21
$begingroup$
I agree with your answer for $sectheta$. When I solved for $tantheta$, I obtained $$frac{x^2 - 1}{2x}$$ which led to the answer $$frac{x^2 - 1}{1 + x^2}$$
$endgroup$
– N. F. Taussig
May 14 '17 at 13:26
$begingroup$
I have added the details of my calculations. I checked my answer for the angles $pi/6$, $pi/4$, and $pi/3$. In each case, it gave the correct answer, while the answer you stated gives the wrong sign.
$endgroup$
– N. F. Taussig
May 14 '17 at 13:50
add a comment |
$begingroup$
Here is a different approach: Since $1 + tan^2theta = sec^2theta$, we have
$$sec^2theta - tan^2theta = 1$$
Factoring yields
$$(sectheta + tantheta)(sectheta - tantheta) = 1$$
Since we are given that $sectheta + tantheta = x$, we obtain
$$x(sectheta - tantheta) = 1$$
Therefore,
$$sectheta - tantheta = frac{1}{x}$$
This yields the system of equations
begin{align*}
sectheta + tantheta & = x tag{1}\
sectheta - tantheta & = frac{1}{x} tag{2}
end{align*}
Adding equations 1 and 2 and solving for $sectheta$ yields
begin{align*}
2sectheta & = x + frac{1}{x}\
2sectheta & = frac{x^2 + 1}{x}\
sectheta & = frac{x^2 + 1}{2x}
end{align*}
Therefore,
$$costheta = frac{1}{sectheta} = frac{2x}{x^2 + 1}$$
Subtracting equation 2 from equation 1 and solving for $tantheta$ yields
begin{align*}
2tantheta & = x - frac{1}{x}\
2tantheta & = frac{x^2 - 1}{x}\
tantheta & = frac{x^2 - 1}{2x}
end{align*}
Thus,
$$sintheta = tanthetacostheta = frac{x^2 - 1}{2x} cdot frac{2x}{x^2 + 1} = frac{x^2 - 1}{x^2 + 1}$$
$endgroup$
Here is a different approach: Since $1 + tan^2theta = sec^2theta$, we have
$$sec^2theta - tan^2theta = 1$$
Factoring yields
$$(sectheta + tantheta)(sectheta - tantheta) = 1$$
Since we are given that $sectheta + tantheta = x$, we obtain
$$x(sectheta - tantheta) = 1$$
Therefore,
$$sectheta - tantheta = frac{1}{x}$$
This yields the system of equations
begin{align*}
sectheta + tantheta & = x tag{1}\
sectheta - tantheta & = frac{1}{x} tag{2}
end{align*}
Adding equations 1 and 2 and solving for $sectheta$ yields
begin{align*}
2sectheta & = x + frac{1}{x}\
2sectheta & = frac{x^2 + 1}{x}\
sectheta & = frac{x^2 + 1}{2x}
end{align*}
Therefore,
$$costheta = frac{1}{sectheta} = frac{2x}{x^2 + 1}$$
Subtracting equation 2 from equation 1 and solving for $tantheta$ yields
begin{align*}
2tantheta & = x - frac{1}{x}\
2tantheta & = frac{x^2 - 1}{x}\
tantheta & = frac{x^2 - 1}{2x}
end{align*}
Thus,
$$sintheta = tanthetacostheta = frac{x^2 - 1}{2x} cdot frac{2x}{x^2 + 1} = frac{x^2 - 1}{x^2 + 1}$$
edited May 14 '17 at 13:49
answered May 14 '17 at 13:07
N. F. TaussigN. F. Taussig
44.5k93357
44.5k93357
$begingroup$
I didn't get the answer. The answer is $dfrac {1-x^2}{1+x^2}$.
$endgroup$
– Aryabhatta
May 14 '17 at 13:15
$begingroup$
What did you get for $sectheta$ and $tantheta$?
$endgroup$
– N. F. Taussig
May 14 '17 at 13:19
$begingroup$
I got $sec theta=dfrac {x^2+1}{2x}$ and $tan theta =dfrac {2x-x^2-1}{2x}$.
$endgroup$
– Aryabhatta
May 14 '17 at 13:21
$begingroup$
I agree with your answer for $sectheta$. When I solved for $tantheta$, I obtained $$frac{x^2 - 1}{2x}$$ which led to the answer $$frac{x^2 - 1}{1 + x^2}$$
$endgroup$
– N. F. Taussig
May 14 '17 at 13:26
$begingroup$
I have added the details of my calculations. I checked my answer for the angles $pi/6$, $pi/4$, and $pi/3$. In each case, it gave the correct answer, while the answer you stated gives the wrong sign.
$endgroup$
– N. F. Taussig
May 14 '17 at 13:50
add a comment |
$begingroup$
I didn't get the answer. The answer is $dfrac {1-x^2}{1+x^2}$.
$endgroup$
– Aryabhatta
May 14 '17 at 13:15
$begingroup$
What did you get for $sectheta$ and $tantheta$?
$endgroup$
– N. F. Taussig
May 14 '17 at 13:19
$begingroup$
I got $sec theta=dfrac {x^2+1}{2x}$ and $tan theta =dfrac {2x-x^2-1}{2x}$.
$endgroup$
– Aryabhatta
May 14 '17 at 13:21
$begingroup$
I agree with your answer for $sectheta$. When I solved for $tantheta$, I obtained $$frac{x^2 - 1}{2x}$$ which led to the answer $$frac{x^2 - 1}{1 + x^2}$$
$endgroup$
– N. F. Taussig
May 14 '17 at 13:26
$begingroup$
I have added the details of my calculations. I checked my answer for the angles $pi/6$, $pi/4$, and $pi/3$. In each case, it gave the correct answer, while the answer you stated gives the wrong sign.
$endgroup$
– N. F. Taussig
May 14 '17 at 13:50
$begingroup$
I didn't get the answer. The answer is $dfrac {1-x^2}{1+x^2}$.
$endgroup$
– Aryabhatta
May 14 '17 at 13:15
$begingroup$
I didn't get the answer. The answer is $dfrac {1-x^2}{1+x^2}$.
$endgroup$
– Aryabhatta
May 14 '17 at 13:15
$begingroup$
What did you get for $sectheta$ and $tantheta$?
$endgroup$
– N. F. Taussig
May 14 '17 at 13:19
$begingroup$
What did you get for $sectheta$ and $tantheta$?
$endgroup$
– N. F. Taussig
May 14 '17 at 13:19
$begingroup$
I got $sec theta=dfrac {x^2+1}{2x}$ and $tan theta =dfrac {2x-x^2-1}{2x}$.
$endgroup$
– Aryabhatta
May 14 '17 at 13:21
$begingroup$
I got $sec theta=dfrac {x^2+1}{2x}$ and $tan theta =dfrac {2x-x^2-1}{2x}$.
$endgroup$
– Aryabhatta
May 14 '17 at 13:21
$begingroup$
I agree with your answer for $sectheta$. When I solved for $tantheta$, I obtained $$frac{x^2 - 1}{2x}$$ which led to the answer $$frac{x^2 - 1}{1 + x^2}$$
$endgroup$
– N. F. Taussig
May 14 '17 at 13:26
$begingroup$
I agree with your answer for $sectheta$. When I solved for $tantheta$, I obtained $$frac{x^2 - 1}{2x}$$ which led to the answer $$frac{x^2 - 1}{1 + x^2}$$
$endgroup$
– N. F. Taussig
May 14 '17 at 13:26
$begingroup$
I have added the details of my calculations. I checked my answer for the angles $pi/6$, $pi/4$, and $pi/3$. In each case, it gave the correct answer, while the answer you stated gives the wrong sign.
$endgroup$
– N. F. Taussig
May 14 '17 at 13:50
$begingroup$
I have added the details of my calculations. I checked my answer for the angles $pi/6$, $pi/4$, and $pi/3$. In each case, it gave the correct answer, while the answer you stated gives the wrong sign.
$endgroup$
– N. F. Taussig
May 14 '17 at 13:50
add a comment |
$begingroup$
The equation becomes
$$
1+sintheta=xcostheta
$$
Set $X=costheta$ and $Y=sintheta$, so the equation becomes
$$
begin{cases}
X^2+Y^2=1 \[4px]
1+Y=xX
end{cases}
$$
Note that $xne0$ and substitute $X=x^{-1}(1+Y)$ in the first equation getting
$$
(1+Y)^2+x^2Y^2=x^2
$$
that simplifies to
$$
(1+x^2)Y^2+2Y+1-x^2=0
$$
that yields
$$
Y=-1 qquadtext{or}qquad Y=frac{x^2-1}{x^2+1}
$$
Is $Y=-1$ a solution for the problem?
By the way, you also get $costheta$, since
$$
X=frac{1}{x}(1+Y)=frac{1}{x}frac{x^2+1+x^2-1}{x^2+1}=frac{2x}{x^2+1}
$$
$endgroup$
add a comment |
$begingroup$
The equation becomes
$$
1+sintheta=xcostheta
$$
Set $X=costheta$ and $Y=sintheta$, so the equation becomes
$$
begin{cases}
X^2+Y^2=1 \[4px]
1+Y=xX
end{cases}
$$
Note that $xne0$ and substitute $X=x^{-1}(1+Y)$ in the first equation getting
$$
(1+Y)^2+x^2Y^2=x^2
$$
that simplifies to
$$
(1+x^2)Y^2+2Y+1-x^2=0
$$
that yields
$$
Y=-1 qquadtext{or}qquad Y=frac{x^2-1}{x^2+1}
$$
Is $Y=-1$ a solution for the problem?
By the way, you also get $costheta$, since
$$
X=frac{1}{x}(1+Y)=frac{1}{x}frac{x^2+1+x^2-1}{x^2+1}=frac{2x}{x^2+1}
$$
$endgroup$
add a comment |
$begingroup$
The equation becomes
$$
1+sintheta=xcostheta
$$
Set $X=costheta$ and $Y=sintheta$, so the equation becomes
$$
begin{cases}
X^2+Y^2=1 \[4px]
1+Y=xX
end{cases}
$$
Note that $xne0$ and substitute $X=x^{-1}(1+Y)$ in the first equation getting
$$
(1+Y)^2+x^2Y^2=x^2
$$
that simplifies to
$$
(1+x^2)Y^2+2Y+1-x^2=0
$$
that yields
$$
Y=-1 qquadtext{or}qquad Y=frac{x^2-1}{x^2+1}
$$
Is $Y=-1$ a solution for the problem?
By the way, you also get $costheta$, since
$$
X=frac{1}{x}(1+Y)=frac{1}{x}frac{x^2+1+x^2-1}{x^2+1}=frac{2x}{x^2+1}
$$
$endgroup$
The equation becomes
$$
1+sintheta=xcostheta
$$
Set $X=costheta$ and $Y=sintheta$, so the equation becomes
$$
begin{cases}
X^2+Y^2=1 \[4px]
1+Y=xX
end{cases}
$$
Note that $xne0$ and substitute $X=x^{-1}(1+Y)$ in the first equation getting
$$
(1+Y)^2+x^2Y^2=x^2
$$
that simplifies to
$$
(1+x^2)Y^2+2Y+1-x^2=0
$$
that yields
$$
Y=-1 qquadtext{or}qquad Y=frac{x^2-1}{x^2+1}
$$
Is $Y=-1$ a solution for the problem?
By the way, you also get $costheta$, since
$$
X=frac{1}{x}(1+Y)=frac{1}{x}frac{x^2+1+x^2-1}{x^2+1}=frac{2x}{x^2+1}
$$
answered May 14 '17 at 14:00
egregegreg
183k1486205
183k1486205
add a comment |
add a comment |
$begingroup$
From where you are:
You obtained a quadratic function in $sin(theta)$. Perform the substitution $u =sin(theta)$.
We obtain the quadratic (in $u$):
$$(x^2+1)u^2 + 2u - x^2 +1 = 0$$
$$Rightarrow u_{1,2} = frac{- 2 pm sqrt{4 - 4(x^2+1)(1-x^2)}}{2(x^2+1)}$$
$$ = frac{- 2 pm sqrt{4 + 4(x^4 -1)}}{2(x^2+1)}$$
$$ = frac{- 2 pm sqrt{4x^4}}{2(x^2+1)}$$
$$ = frac{- 2 pm 2x^2}{2(x^2+1)}$$
$$ = frac{- 1 pm x^2}{x^2+1}$$
Therefore,
$$sin(theta)_{1,2} = frac{- 1 pm x^2}{x^2+1}$$
One of those solutions will not work out.
This happened because you squared the equation multiple times and we know that $a = b$ is not equivalent with $a^2 = b^2$, so you should fill in both solutions in the original expression and see which one works and which one doesn't.
$endgroup$
$begingroup$
What is discriminant method?
$endgroup$
– Aryabhatta
May 14 '17 at 13:05
$begingroup$
en.wikipedia.org/wiki/Discriminant#Degree_2
$endgroup$
– Math_QED
May 14 '17 at 13:06
$begingroup$
I didn't understand...
$endgroup$
– Aryabhatta
May 14 '17 at 13:14
$begingroup$
If you want to solve a solution in the form $ax^2 + bx + c = 0$, then the solutions are given by $x_{1,2} = frac{-b pm sqrt{b^2 - 4ac}}{2a}$
$endgroup$
– Math_QED
May 14 '17 at 13:21
$begingroup$
Here $x = u, a = (x^2 + 1), b = 2, c = - x^2 + 1$
$endgroup$
– Math_QED
May 14 '17 at 13:22
|
show 1 more comment
$begingroup$
From where you are:
You obtained a quadratic function in $sin(theta)$. Perform the substitution $u =sin(theta)$.
We obtain the quadratic (in $u$):
$$(x^2+1)u^2 + 2u - x^2 +1 = 0$$
$$Rightarrow u_{1,2} = frac{- 2 pm sqrt{4 - 4(x^2+1)(1-x^2)}}{2(x^2+1)}$$
$$ = frac{- 2 pm sqrt{4 + 4(x^4 -1)}}{2(x^2+1)}$$
$$ = frac{- 2 pm sqrt{4x^4}}{2(x^2+1)}$$
$$ = frac{- 2 pm 2x^2}{2(x^2+1)}$$
$$ = frac{- 1 pm x^2}{x^2+1}$$
Therefore,
$$sin(theta)_{1,2} = frac{- 1 pm x^2}{x^2+1}$$
One of those solutions will not work out.
This happened because you squared the equation multiple times and we know that $a = b$ is not equivalent with $a^2 = b^2$, so you should fill in both solutions in the original expression and see which one works and which one doesn't.
$endgroup$
$begingroup$
What is discriminant method?
$endgroup$
– Aryabhatta
May 14 '17 at 13:05
$begingroup$
en.wikipedia.org/wiki/Discriminant#Degree_2
$endgroup$
– Math_QED
May 14 '17 at 13:06
$begingroup$
I didn't understand...
$endgroup$
– Aryabhatta
May 14 '17 at 13:14
$begingroup$
If you want to solve a solution in the form $ax^2 + bx + c = 0$, then the solutions are given by $x_{1,2} = frac{-b pm sqrt{b^2 - 4ac}}{2a}$
$endgroup$
– Math_QED
May 14 '17 at 13:21
$begingroup$
Here $x = u, a = (x^2 + 1), b = 2, c = - x^2 + 1$
$endgroup$
– Math_QED
May 14 '17 at 13:22
|
show 1 more comment
$begingroup$
From where you are:
You obtained a quadratic function in $sin(theta)$. Perform the substitution $u =sin(theta)$.
We obtain the quadratic (in $u$):
$$(x^2+1)u^2 + 2u - x^2 +1 = 0$$
$$Rightarrow u_{1,2} = frac{- 2 pm sqrt{4 - 4(x^2+1)(1-x^2)}}{2(x^2+1)}$$
$$ = frac{- 2 pm sqrt{4 + 4(x^4 -1)}}{2(x^2+1)}$$
$$ = frac{- 2 pm sqrt{4x^4}}{2(x^2+1)}$$
$$ = frac{- 2 pm 2x^2}{2(x^2+1)}$$
$$ = frac{- 1 pm x^2}{x^2+1}$$
Therefore,
$$sin(theta)_{1,2} = frac{- 1 pm x^2}{x^2+1}$$
One of those solutions will not work out.
This happened because you squared the equation multiple times and we know that $a = b$ is not equivalent with $a^2 = b^2$, so you should fill in both solutions in the original expression and see which one works and which one doesn't.
$endgroup$
From where you are:
You obtained a quadratic function in $sin(theta)$. Perform the substitution $u =sin(theta)$.
We obtain the quadratic (in $u$):
$$(x^2+1)u^2 + 2u - x^2 +1 = 0$$
$$Rightarrow u_{1,2} = frac{- 2 pm sqrt{4 - 4(x^2+1)(1-x^2)}}{2(x^2+1)}$$
$$ = frac{- 2 pm sqrt{4 + 4(x^4 -1)}}{2(x^2+1)}$$
$$ = frac{- 2 pm sqrt{4x^4}}{2(x^2+1)}$$
$$ = frac{- 2 pm 2x^2}{2(x^2+1)}$$
$$ = frac{- 1 pm x^2}{x^2+1}$$
Therefore,
$$sin(theta)_{1,2} = frac{- 1 pm x^2}{x^2+1}$$
One of those solutions will not work out.
This happened because you squared the equation multiple times and we know that $a = b$ is not equivalent with $a^2 = b^2$, so you should fill in both solutions in the original expression and see which one works and which one doesn't.
edited May 14 '17 at 18:24
answered May 14 '17 at 13:01
Math_QEDMath_QED
7,64531452
7,64531452
$begingroup$
What is discriminant method?
$endgroup$
– Aryabhatta
May 14 '17 at 13:05
$begingroup$
en.wikipedia.org/wiki/Discriminant#Degree_2
$endgroup$
– Math_QED
May 14 '17 at 13:06
$begingroup$
I didn't understand...
$endgroup$
– Aryabhatta
May 14 '17 at 13:14
$begingroup$
If you want to solve a solution in the form $ax^2 + bx + c = 0$, then the solutions are given by $x_{1,2} = frac{-b pm sqrt{b^2 - 4ac}}{2a}$
$endgroup$
– Math_QED
May 14 '17 at 13:21
$begingroup$
Here $x = u, a = (x^2 + 1), b = 2, c = - x^2 + 1$
$endgroup$
– Math_QED
May 14 '17 at 13:22
|
show 1 more comment
$begingroup$
What is discriminant method?
$endgroup$
– Aryabhatta
May 14 '17 at 13:05
$begingroup$
en.wikipedia.org/wiki/Discriminant#Degree_2
$endgroup$
– Math_QED
May 14 '17 at 13:06
$begingroup$
I didn't understand...
$endgroup$
– Aryabhatta
May 14 '17 at 13:14
$begingroup$
If you want to solve a solution in the form $ax^2 + bx + c = 0$, then the solutions are given by $x_{1,2} = frac{-b pm sqrt{b^2 - 4ac}}{2a}$
$endgroup$
– Math_QED
May 14 '17 at 13:21
$begingroup$
Here $x = u, a = (x^2 + 1), b = 2, c = - x^2 + 1$
$endgroup$
– Math_QED
May 14 '17 at 13:22
$begingroup$
What is discriminant method?
$endgroup$
– Aryabhatta
May 14 '17 at 13:05
$begingroup$
What is discriminant method?
$endgroup$
– Aryabhatta
May 14 '17 at 13:05
$begingroup$
en.wikipedia.org/wiki/Discriminant#Degree_2
$endgroup$
– Math_QED
May 14 '17 at 13:06
$begingroup$
en.wikipedia.org/wiki/Discriminant#Degree_2
$endgroup$
– Math_QED
May 14 '17 at 13:06
$begingroup$
I didn't understand...
$endgroup$
– Aryabhatta
May 14 '17 at 13:14
$begingroup$
I didn't understand...
$endgroup$
– Aryabhatta
May 14 '17 at 13:14
$begingroup$
If you want to solve a solution in the form $ax^2 + bx + c = 0$, then the solutions are given by $x_{1,2} = frac{-b pm sqrt{b^2 - 4ac}}{2a}$
$endgroup$
– Math_QED
May 14 '17 at 13:21
$begingroup$
If you want to solve a solution in the form $ax^2 + bx + c = 0$, then the solutions are given by $x_{1,2} = frac{-b pm sqrt{b^2 - 4ac}}{2a}$
$endgroup$
– Math_QED
May 14 '17 at 13:21
$begingroup$
Here $x = u, a = (x^2 + 1), b = 2, c = - x^2 + 1$
$endgroup$
– Math_QED
May 14 '17 at 13:22
$begingroup$
Here $x = u, a = (x^2 + 1), b = 2, c = - x^2 + 1$
$endgroup$
– Math_QED
May 14 '17 at 13:22
|
show 1 more comment
$begingroup$
WLOG let $theta=dfracpi2-2yimplies x=csc2y+cot2y=dfrac{1+cos2y}{sin2y}=cot y$
$$sintheta=cos2y=dfrac{1-tan^2y}{1+tan^2y}=dfrac{cot^2y-1}{cot^2y+1}=?$$
$endgroup$
add a comment |
$begingroup$
WLOG let $theta=dfracpi2-2yimplies x=csc2y+cot2y=dfrac{1+cos2y}{sin2y}=cot y$
$$sintheta=cos2y=dfrac{1-tan^2y}{1+tan^2y}=dfrac{cot^2y-1}{cot^2y+1}=?$$
$endgroup$
add a comment |
$begingroup$
WLOG let $theta=dfracpi2-2yimplies x=csc2y+cot2y=dfrac{1+cos2y}{sin2y}=cot y$
$$sintheta=cos2y=dfrac{1-tan^2y}{1+tan^2y}=dfrac{cot^2y-1}{cot^2y+1}=?$$
$endgroup$
WLOG let $theta=dfracpi2-2yimplies x=csc2y+cot2y=dfrac{1+cos2y}{sin2y}=cot y$
$$sintheta=cos2y=dfrac{1-tan^2y}{1+tan^2y}=dfrac{cot^2y-1}{cot^2y+1}=?$$
answered May 14 '17 at 14:27
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
add a comment |
add a comment |
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See also, math.stackexchange.com/questions/2184596/…
$endgroup$
– lab bhattacharjee
May 14 '17 at 14:24