If $alpha+beta=90^circ$, then $frac{tan2alpha}{tan2beta} =-1$ [closed]
$begingroup$
Given that $$tan(A+B)= frac{tan A + tan B}{1-tan A tan B}$$
Show that if $alpha+beta=90^circ$, then $frac{tan2alpha}{tan2beta} =-1$.
trigonometry
edited Mar 12 at 22:33
Blue
49.1k870156
asked Mar 12 at 16:32
Nour GNour G
164
$endgroup$
closed as off-topic by Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota Mar 13 at 4:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 2 more comments
$begingroup$
Given that $$tan(A+B)= frac{tan A + tan B}{1-tan A tan B}$$
Show that if $alpha+beta=90^circ$, then $frac{tan2alpha}{tan2beta} =-1$.
trigonometry
edited Mar 12 at 22:33
Blue
49.1k870156
asked Mar 12 at 16:32
Nour GNour G
164
$endgroup$
closed as off-topic by Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota Mar 13 at 4:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
$endgroup$
– saulspatz
Mar 12 at 16:57
1
$begingroup$
You possibly want to add add an edit that the identity is true provided $A+B$ is not equal to (k+0.5)*180 degrees for integer k
$endgroup$
– Martin Hansen
Mar 12 at 16:58
$begingroup$
You don't need the angle addition formula at all, you just need to know that $tan$ is an odd function that is periodic with period $180^circ$.
$endgroup$
– Barry Cipra
Mar 12 at 17:00
$begingroup$
Does your question insist on the identity being used ?
$endgroup$
– Martin Hansen
Mar 12 at 17:03
$begingroup$
Yes they give you the tan(A+B) formula. I am meant to show that if α+β=90 degrees, then tan2α/tan2β=−1
$endgroup$
– Nour G
Mar 12 at 17:07
|
show 2 more comments
$begingroup$
Given that $$tan(A+B)= frac{tan A + tan B}{1-tan A tan B}$$
Show that if $alpha+beta=90^circ$, then $frac{tan2alpha}{tan2beta} =-1$.
trigonometry
edited Mar 12 at 22:33
Blue
49.1k870156
asked Mar 12 at 16:32
Nour GNour G
164
$endgroup$
Given that $$tan(A+B)= frac{tan A + tan B}{1-tan A tan B}$$
Show that if $alpha+beta=90^circ$, then $frac{tan2alpha}{tan2beta} =-1$.
trigonometry
trigonometry
edited Mar 12 at 22:33
Blue
49.1k870156
asked Mar 12 at 16:32
Nour GNour G
164
edited Mar 12 at 22:33
Blue
49.1k870156
asked Mar 12 at 16:32
Nour GNour G
164
edited Mar 12 at 22:33
Blue
49.1k870156
edited Mar 12 at 22:33
Blue
49.1k870156
edited Mar 12 at 22:33
Blue
49.1k870156
49.1k870156
asked Mar 12 at 16:32
Nour GNour G
164
asked Mar 12 at 16:32
Nour GNour G
164
asked Mar 12 at 16:32
Nour GNour G
164
164
closed as off-topic by Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota Mar 13 at 4:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota Mar 13 at 4:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Vinyl_cape_jawa, RRL, Saad, Alex Provost, farruhota
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
$endgroup$
– saulspatz
Mar 12 at 16:57
1
$begingroup$
You possibly want to add add an edit that the identity is true provided $A+B$ is not equal to (k+0.5)*180 degrees for integer k
$endgroup$
– Martin Hansen
Mar 12 at 16:58
$begingroup$
You don't need the angle addition formula at all, you just need to know that $tan$ is an odd function that is periodic with period $180^circ$.
$endgroup$
– Barry Cipra
Mar 12 at 17:00
$begingroup$
Does your question insist on the identity being used ?
$endgroup$
– Martin Hansen
Mar 12 at 17:03
$begingroup$
Yes they give you the tan(A+B) formula. I am meant to show that if α+β=90 degrees, then tan2α/tan2β=−1
$endgroup$
– Nour G
Mar 12 at 17:07
|
show 2 more comments
3
$begingroup$
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
$endgroup$
– saulspatz
Mar 12 at 16:57
1
$begingroup$
You possibly want to add add an edit that the identity is true provided $A+B$ is not equal to (k+0.5)*180 degrees for integer k
$endgroup$
– Martin Hansen
Mar 12 at 16:58
$begingroup$
You don't need the angle addition formula at all, you just need to know that $tan$ is an odd function that is periodic with period $180^circ$.
$endgroup$
– Barry Cipra
Mar 12 at 17:00
$begingroup$
Does your question insist on the identity being used ?
$endgroup$
– Martin Hansen
Mar 12 at 17:03
$begingroup$
Yes they give you the tan(A+B) formula. I am meant to show that if α+β=90 degrees, then tan2α/tan2β=−1
$endgroup$
– Nour G
Mar 12 at 17:07
3
3
$begingroup$
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
$endgroup$
– saulspatz
Mar 12 at 16:57
$begingroup$
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
$endgroup$
– saulspatz
Mar 12 at 16:57
1
1
$begingroup$
You possibly want to add add an edit that the identity is true provided $A+B$ is not equal to (k+0.5)*180 degrees for integer k
$endgroup$
– Martin Hansen
Mar 12 at 16:58
$begingroup$
You possibly want to add add an edit that the identity is true provided $A+B$ is not equal to (k+0.5)*180 degrees for integer k
$endgroup$
– Martin Hansen
Mar 12 at 16:58
$begingroup$
You don't need the angle addition formula at all, you just need to know that $tan$ is an odd function that is periodic with period $180^circ$.
$endgroup$
– Barry Cipra
Mar 12 at 17:00
$begingroup$
You don't need the angle addition formula at all, you just need to know that $tan$ is an odd function that is periodic with period $180^circ$.
$endgroup$
– Barry Cipra
Mar 12 at 17:00
$begingroup$
Does your question insist on the identity being used ?
$endgroup$
– Martin Hansen
Mar 12 at 17:03
$begingroup$
Does your question insist on the identity being used ?
$endgroup$
– Martin Hansen
Mar 12 at 17:03
$begingroup$
Yes they give you the tan(A+B) formula. I am meant to show that if α+β=90 degrees, then tan2α/tan2β=−1
$endgroup$
– Nour G
Mar 12 at 17:07
$begingroup$
Yes they give you the tan(A+B) formula. I am meant to show that if α+β=90 degrees, then tan2α/tan2β=−1
$endgroup$
– Nour G
Mar 12 at 17:07
|
show 2 more comments
5 Answers
5
active
oldest
votes
$begingroup$
Assuming that the question is insisting on the $tan(A+B)$ identity being used;
$$tan(2 alpha)=frac{2tan alpha}{1-tan^2 alpha}$$
and
$$tan(2 beta)=frac{2tan beta}{1-tan^2 beta}$$
Thus
$$frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$
Now draw a right angled triangle with $alpha$ in one corner and $beta$ in the other. Call $x$ the side opposite $alpha$ and $y$ the side opposite $beta$
Observe that
$$tan alpha=frac{x}{y}$$
and
$$tan beta=frac{y}{x}$$
Thus
$$LHS=frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$
$$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}}$$
Multiply both the numerator and the denominator by $x^2y^2$
$$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}} times frac{x^2y^2}{x^2y^2}$$
$$=frac{2xy(x^2-y^2)}{(y^2-x^2)2xy}$$
$$=-1$$
$$=RHS$$
edited Mar 12 at 22:05
Teepeemm
69459
answered Mar 12 at 17:28
Martin HansenMartin Hansen
670114
$endgroup$
1
$begingroup$
Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
$endgroup$
– Nour G
Mar 12 at 17:52
$begingroup$
@Teepeemm Thanks for the suggested edit
$endgroup$
– Martin Hansen
Mar 12 at 22:06
add a comment |
$begingroup$
If
$$2alpha+2beta=180°$$
then
$$tan(2alpha+2beta)=frac{tan2alpha+tan2beta}{1-tan2alphatan2beta}=0$$ and the claim easily follows.
answered Mar 12 at 17:25
Yves DaoustYves Daoust
131k676229
$endgroup$
1
$begingroup$
Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
$endgroup$
– Martin Hansen
Mar 12 at 17:37
1
$begingroup$
Thank you so much! Makes so much sense now.
$endgroup$
– Nour G
Mar 12 at 17:53
add a comment |
$begingroup$
Hint: if $alpha+beta=90^circ$ then $2alpha=180^circ-2beta$.
Use that $sin 2 alpha=sin(180^circ-2beta)=sin 2beta$, $cos 2 alpha=cos(180^circ-2beta)=-cos 2beta$
Alternatively, $tan(2alpha+2beta)=0 rightarrow tan(2alpha)+tan(2beta)=0 $ (thanks, @BarryCipra)
answered Mar 12 at 17:01
VasyaVasya
4,1081618
$endgroup$
add a comment |
$begingroup$
$alpha+beta=90^circ$
$implies2alpha+2beta=180^circ$
$implies2beta=180 ^circ-2alpha$
Now
$$frac{tan2alpha}{tan2beta} =frac{tan2alpha}{tan(180 ^circ-2alpha)} =frac{tan2alpha}{-tan2alpha} = -1= RHS $$
answered Mar 13 at 3:34
saket kumarsaket kumar
143113
$endgroup$
add a comment |
$begingroup$
$$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$
answered Mar 12 at 17:31
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming that the question is insisting on the $tan(A+B)$ identity being used;
$$tan(2 alpha)=frac{2tan alpha}{1-tan^2 alpha}$$
and
$$tan(2 beta)=frac{2tan beta}{1-tan^2 beta}$$
Thus
$$frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$
Now draw a right angled triangle with $alpha$ in one corner and $beta$ in the other. Call $x$ the side opposite $alpha$ and $y$ the side opposite $beta$
Observe that
$$tan alpha=frac{x}{y}$$
and
$$tan beta=frac{y}{x}$$
Thus
$$LHS=frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$
$$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}}$$
Multiply both the numerator and the denominator by $x^2y^2$
$$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}} times frac{x^2y^2}{x^2y^2}$$
$$=frac{2xy(x^2-y^2)}{(y^2-x^2)2xy}$$
$$=-1$$
$$=RHS$$
edited Mar 12 at 22:05
Teepeemm
69459
answered Mar 12 at 17:28
Martin HansenMartin Hansen
670114
$endgroup$
1
$begingroup$
Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
$endgroup$
– Nour G
Mar 12 at 17:52
$begingroup$
@Teepeemm Thanks for the suggested edit
$endgroup$
– Martin Hansen
Mar 12 at 22:06
add a comment |
$begingroup$
Assuming that the question is insisting on the $tan(A+B)$ identity being used;
$$tan(2 alpha)=frac{2tan alpha}{1-tan^2 alpha}$$
and
$$tan(2 beta)=frac{2tan beta}{1-tan^2 beta}$$
Thus
$$frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$
Now draw a right angled triangle with $alpha$ in one corner and $beta$ in the other. Call $x$ the side opposite $alpha$ and $y$ the side opposite $beta$
Observe that
$$tan alpha=frac{x}{y}$$
and
$$tan beta=frac{y}{x}$$
Thus
$$LHS=frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$
$$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}}$$
Multiply both the numerator and the denominator by $x^2y^2$
$$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}} times frac{x^2y^2}{x^2y^2}$$
$$=frac{2xy(x^2-y^2)}{(y^2-x^2)2xy}$$
$$=-1$$
$$=RHS$$
edited Mar 12 at 22:05
Teepeemm
69459
answered Mar 12 at 17:28
Martin HansenMartin Hansen
670114
$endgroup$
1
$begingroup$
Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
$endgroup$
– Nour G
Mar 12 at 17:52
$begingroup$
@Teepeemm Thanks for the suggested edit
$endgroup$
– Martin Hansen
Mar 12 at 22:06
add a comment |
$begingroup$
Assuming that the question is insisting on the $tan(A+B)$ identity being used;
$$tan(2 alpha)=frac{2tan alpha}{1-tan^2 alpha}$$
and
$$tan(2 beta)=frac{2tan beta}{1-tan^2 beta}$$
Thus
$$frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$
Now draw a right angled triangle with $alpha$ in one corner and $beta$ in the other. Call $x$ the side opposite $alpha$ and $y$ the side opposite $beta$
Observe that
$$tan alpha=frac{x}{y}$$
and
$$tan beta=frac{y}{x}$$
Thus
$$LHS=frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$
$$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}}$$
Multiply both the numerator and the denominator by $x^2y^2$
$$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}} times frac{x^2y^2}{x^2y^2}$$
$$=frac{2xy(x^2-y^2)}{(y^2-x^2)2xy}$$
$$=-1$$
$$=RHS$$
edited Mar 12 at 22:05
Teepeemm
69459
answered Mar 12 at 17:28
Martin HansenMartin Hansen
670114
$endgroup$
Assuming that the question is insisting on the $tan(A+B)$ identity being used;
$$tan(2 alpha)=frac{2tan alpha}{1-tan^2 alpha}$$
and
$$tan(2 beta)=frac{2tan beta}{1-tan^2 beta}$$
Thus
$$frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$
Now draw a right angled triangle with $alpha$ in one corner and $beta$ in the other. Call $x$ the side opposite $alpha$ and $y$ the side opposite $beta$
Observe that
$$tan alpha=frac{x}{y}$$
and
$$tan beta=frac{y}{x}$$
Thus
$$LHS=frac{tan(2 alpha)}{tan(2 beta)}=frac{2tan alpha(1-tan^2 beta)}{(1-tan^2 alpha) 2tan beta}$$
$$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}}$$
Multiply both the numerator and the denominator by $x^2y^2$
$$=frac{frac{2x}{y}big(1-frac{y^2}{x^2}big)}{big(1-frac{x^2}{y^2}big) frac{2y}{x}} times frac{x^2y^2}{x^2y^2}$$
$$=frac{2xy(x^2-y^2)}{(y^2-x^2)2xy}$$
$$=-1$$
$$=RHS$$
edited Mar 12 at 22:05
Teepeemm
69459
answered Mar 12 at 17:28
Martin HansenMartin Hansen
670114
edited Mar 12 at 22:05
Teepeemm
69459
edited Mar 12 at 22:05
Teepeemm
69459
edited Mar 12 at 22:05
Teepeemm
69459
69459
answered Mar 12 at 17:28
Martin HansenMartin Hansen
670114
answered Mar 12 at 17:28
Martin HansenMartin Hansen
670114
answered Mar 12 at 17:28
Martin HansenMartin Hansen
670114
670114
1
$begingroup$
Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
$endgroup$
– Nour G
Mar 12 at 17:52
$begingroup$
@Teepeemm Thanks for the suggested edit
$endgroup$
– Martin Hansen
Mar 12 at 22:06
add a comment |
1
$begingroup$
Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
$endgroup$
– Nour G
Mar 12 at 17:52
$begingroup$
@Teepeemm Thanks for the suggested edit
$endgroup$
– Martin Hansen
Mar 12 at 22:06
1
1
$begingroup$
Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
$endgroup$
– Nour G
Mar 12 at 17:52
$begingroup$
Thank you Martin! That's a brilliant answer. I would've never come to think of doing it this way.
$endgroup$
– Nour G
Mar 12 at 17:52
$begingroup$
@Teepeemm Thanks for the suggested edit
$endgroup$
– Martin Hansen
Mar 12 at 22:06
$begingroup$
@Teepeemm Thanks for the suggested edit
$endgroup$
– Martin Hansen
Mar 12 at 22:06
add a comment |
$begingroup$
If
$$2alpha+2beta=180°$$
then
$$tan(2alpha+2beta)=frac{tan2alpha+tan2beta}{1-tan2alphatan2beta}=0$$ and the claim easily follows.
answered Mar 12 at 17:25
Yves DaoustYves Daoust
131k676229
$endgroup$
1
$begingroup$
Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
$endgroup$
– Martin Hansen
Mar 12 at 17:37
1
$begingroup$
Thank you so much! Makes so much sense now.
$endgroup$
– Nour G
Mar 12 at 17:53
add a comment |
$begingroup$
If
$$2alpha+2beta=180°$$
then
$$tan(2alpha+2beta)=frac{tan2alpha+tan2beta}{1-tan2alphatan2beta}=0$$ and the claim easily follows.
answered Mar 12 at 17:25
Yves DaoustYves Daoust
131k676229
$endgroup$
1
$begingroup$
Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
$endgroup$
– Martin Hansen
Mar 12 at 17:37
1
$begingroup$
Thank you so much! Makes so much sense now.
$endgroup$
– Nour G
Mar 12 at 17:53
add a comment |
$begingroup$
If
$$2alpha+2beta=180°$$
then
$$tan(2alpha+2beta)=frac{tan2alpha+tan2beta}{1-tan2alphatan2beta}=0$$ and the claim easily follows.
answered Mar 12 at 17:25
Yves DaoustYves Daoust
131k676229
$endgroup$
If
$$2alpha+2beta=180°$$
then
$$tan(2alpha+2beta)=frac{tan2alpha+tan2beta}{1-tan2alphatan2beta}=0$$ and the claim easily follows.
answered Mar 12 at 17:25
Yves DaoustYves Daoust
131k676229
answered Mar 12 at 17:25
Yves DaoustYves Daoust
131k676229
answered Mar 12 at 17:25
Yves DaoustYves Daoust
131k676229
answered Mar 12 at 17:25
Yves DaoustYves Daoust
131k676229
131k676229
1
$begingroup$
Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
$endgroup$
– Martin Hansen
Mar 12 at 17:37
1
$begingroup$
Thank you so much! Makes so much sense now.
$endgroup$
– Nour G
Mar 12 at 17:53
add a comment |
1
$begingroup$
Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
$endgroup$
– Martin Hansen
Mar 12 at 17:37
1
$begingroup$
Thank you so much! Makes so much sense now.
$endgroup$
– Nour G
Mar 12 at 17:53
1
1
$begingroup$
Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
$endgroup$
– Martin Hansen
Mar 12 at 17:37
$begingroup$
Ah yes, that a clever answer, and shorter than mine and it used the identity provided. It gets my up-vote !
$endgroup$
– Martin Hansen
Mar 12 at 17:37
1
1
$begingroup$
Thank you so much! Makes so much sense now.
$endgroup$
– Nour G
Mar 12 at 17:53
$begingroup$
Thank you so much! Makes so much sense now.
$endgroup$
– Nour G
Mar 12 at 17:53
add a comment |
$begingroup$
Hint: if $alpha+beta=90^circ$ then $2alpha=180^circ-2beta$.
Use that $sin 2 alpha=sin(180^circ-2beta)=sin 2beta$, $cos 2 alpha=cos(180^circ-2beta)=-cos 2beta$
Alternatively, $tan(2alpha+2beta)=0 rightarrow tan(2alpha)+tan(2beta)=0 $ (thanks, @BarryCipra)
answered Mar 12 at 17:01
VasyaVasya
4,1081618
$endgroup$
add a comment |
$begingroup$
Hint: if $alpha+beta=90^circ$ then $2alpha=180^circ-2beta$.
Use that $sin 2 alpha=sin(180^circ-2beta)=sin 2beta$, $cos 2 alpha=cos(180^circ-2beta)=-cos 2beta$
Alternatively, $tan(2alpha+2beta)=0 rightarrow tan(2alpha)+tan(2beta)=0 $ (thanks, @BarryCipra)
answered Mar 12 at 17:01
VasyaVasya
4,1081618
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Hint: if $alpha+beta=90^circ$ then $2alpha=180^circ-2beta$.
Use that $sin 2 alpha=sin(180^circ-2beta)=sin 2beta$, $cos 2 alpha=cos(180^circ-2beta)=-cos 2beta$
Alternatively, $tan(2alpha+2beta)=0 rightarrow tan(2alpha)+tan(2beta)=0 $ (thanks, @BarryCipra)
answered Mar 12 at 17:01
VasyaVasya
4,1081618
$endgroup$
Hint: if $alpha+beta=90^circ$ then $2alpha=180^circ-2beta$.
Use that $sin 2 alpha=sin(180^circ-2beta)=sin 2beta$, $cos 2 alpha=cos(180^circ-2beta)=-cos 2beta$
Alternatively, $tan(2alpha+2beta)=0 rightarrow tan(2alpha)+tan(2beta)=0 $ (thanks, @BarryCipra)
answered Mar 12 at 17:01
VasyaVasya
4,1081618
edited Mar 12 at 17:17
answered Mar 12 at 17:01
VasyaVasya
4,1081618
answered Mar 12 at 17:01
VasyaVasya
4,1081618
answered Mar 12 at 17:01
VasyaVasya
4,1081618
4,1081618
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$alpha+beta=90^circ$
$implies2alpha+2beta=180^circ$
$implies2beta=180 ^circ-2alpha$
Now
$$frac{tan2alpha}{tan2beta} =frac{tan2alpha}{tan(180 ^circ-2alpha)} =frac{tan2alpha}{-tan2alpha} = -1= RHS $$
answered Mar 13 at 3:34
saket kumarsaket kumar
143113
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add a comment |
$begingroup$
$alpha+beta=90^circ$
$implies2alpha+2beta=180^circ$
$implies2beta=180 ^circ-2alpha$
Now
$$frac{tan2alpha}{tan2beta} =frac{tan2alpha}{tan(180 ^circ-2alpha)} =frac{tan2alpha}{-tan2alpha} = -1= RHS $$
answered Mar 13 at 3:34
saket kumarsaket kumar
143113
$endgroup$
add a comment |
$begingroup$
$alpha+beta=90^circ$
$implies2alpha+2beta=180^circ$
$implies2beta=180 ^circ-2alpha$
Now
$$frac{tan2alpha}{tan2beta} =frac{tan2alpha}{tan(180 ^circ-2alpha)} =frac{tan2alpha}{-tan2alpha} = -1= RHS $$
answered Mar 13 at 3:34
saket kumarsaket kumar
143113
$endgroup$
$alpha+beta=90^circ$
$implies2alpha+2beta=180^circ$
$implies2beta=180 ^circ-2alpha$
Now
$$frac{tan2alpha}{tan2beta} =frac{tan2alpha}{tan(180 ^circ-2alpha)} =frac{tan2alpha}{-tan2alpha} = -1= RHS $$
answered Mar 13 at 3:34
saket kumarsaket kumar
143113
edited Mar 13 at 3:46
answered Mar 13 at 3:34
saket kumarsaket kumar
143113
answered Mar 13 at 3:34
saket kumarsaket kumar
143113
answered Mar 13 at 3:34
saket kumarsaket kumar
143113
143113
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$$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$
answered Mar 12 at 17:31
Maria MazurMaria Mazur
47.9k1260120
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$begingroup$
$$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$
answered Mar 12 at 17:31
Maria MazurMaria Mazur
47.9k1260120
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add a comment |
$begingroup$
$$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$
answered Mar 12 at 17:31
Maria MazurMaria Mazur
47.9k1260120
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$$tan 2alpha = tan (180 -2beta) = tan (-2beta ) = -tan (2beta )$$
answered Mar 12 at 17:31
Maria MazurMaria Mazur
47.9k1260120
answered Mar 12 at 17:31
Maria MazurMaria Mazur
47.9k1260120
answered Mar 12 at 17:31
Maria MazurMaria Mazur
47.9k1260120
answered Mar 12 at 17:31
Maria MazurMaria Mazur
47.9k1260120
47.9k1260120
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add a comment |
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Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments.
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– saulspatz
Mar 12 at 16:57
1
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You possibly want to add add an edit that the identity is true provided $A+B$ is not equal to (k+0.5)*180 degrees for integer k
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– Martin Hansen
Mar 12 at 16:58
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You don't need the angle addition formula at all, you just need to know that $tan$ is an odd function that is periodic with period $180^circ$.
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– Barry Cipra
Mar 12 at 17:00
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Does your question insist on the identity being used ?
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– Martin Hansen
Mar 12 at 17:03
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Yes they give you the tan(A+B) formula. I am meant to show that if α+β=90 degrees, then tan2α/tan2β=−1
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– Nour G
Mar 12 at 17:07