System of Trigonomtric Equations [closed]
0
$begingroup$
Please Help to solve the following problem :-
If,
$$sin(x+y)=a$$
$$cos(x-y)=b$$
$$a,bin Bbb R^+$$
Find $tan(2x)$ in terms of a and b.
trigonometry systems-of-equations
edited Dec 9 '18 at 16:44
glowstonetrees
2,368418
asked Dec 9 '18 at 16:27
EtotheipiEtotheipi
344
$endgroup$
closed as off-topic by Nosrati, Jack D'Aurizio, Saad, mrtaurho, Lord_Farin Dec 23 '18 at 11:07
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." – Nosrati, Jack D'Aurizio, Saad, mrtaurho, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
0
$begingroup$
Please Help to solve the following problem :-
If,
$$sin(x+y)=a$$
$$cos(x-y)=b$$
$$a,bin Bbb R^+$$
Find $tan(2x)$ in terms of a and b.
trigonometry systems-of-equations
edited Dec 9 '18 at 16:44
glowstonetrees
2,368418
asked Dec 9 '18 at 16:27
EtotheipiEtotheipi
344
$endgroup$
closed as off-topic by Nosrati, Jack D'Aurizio, Saad, mrtaurho, Lord_Farin Dec 23 '18 at 11:07
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." – Nosrati, Jack D'Aurizio, Saad, mrtaurho, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
$$tan2x=tan(x+y+x-y)=dfrac{tan(x+y)+tan(x-y)}{1-tan(x+y)tan(x-y)}$$
$endgroup$
– Nosrati
Dec 9 '18 at 16:39
$begingroup$
Oh god thats genius, why didnt i think of that!
$endgroup$
– Etotheipi
Dec 9 '18 at 16:42
$begingroup$
@Nosrati,But the question asks for tan(2x) in terms of a and b
$endgroup$
– Dhamnekar Winod
Dec 9 '18 at 16:48
$begingroup$
Write $tan=dfrac{sin}{cos}$.
$endgroup$
– Nosrati
Dec 9 '18 at 16:50
$begingroup$
@DhamnekarWinod You just use $tan x = frac{sin x}{cos x}$ and $cos x = sqrt{1-x^2}$.
$endgroup$
– KM101
Dec 9 '18 at 16:53
add a comment |
0
0
0
$begingroup$
Please Help to solve the following problem :-
If,
$$sin(x+y)=a$$
$$cos(x-y)=b$$
$$a,bin Bbb R^+$$
Find $tan(2x)$ in terms of a and b.
trigonometry systems-of-equations
edited Dec 9 '18 at 16:44
glowstonetrees
2,368418
asked Dec 9 '18 at 16:27
EtotheipiEtotheipi
344
$endgroup$
Please Help to solve the following problem :-
If,
$$sin(x+y)=a$$
$$cos(x-y)=b$$
$$a,bin Bbb R^+$$
Find $tan(2x)$ in terms of a and b.
trigonometry systems-of-equations
trigonometry systems-of-equations
edited Dec 9 '18 at 16:44
glowstonetrees
2,368418
asked Dec 9 '18 at 16:27
EtotheipiEtotheipi
344
edited Dec 9 '18 at 16:44
glowstonetrees
2,368418
asked Dec 9 '18 at 16:27
EtotheipiEtotheipi
344
edited Dec 9 '18 at 16:44
glowstonetrees
2,368418
edited Dec 9 '18 at 16:44
glowstonetrees
2,368418
edited Dec 9 '18 at 16:44
glowstonetrees
2,368418
2,368418
asked Dec 9 '18 at 16:27
EtotheipiEtotheipi
344
asked Dec 9 '18 at 16:27
EtotheipiEtotheipi
344
asked Dec 9 '18 at 16:27
EtotheipiEtotheipi
344
344
closed as off-topic by Nosrati, Jack D'Aurizio, Saad, mrtaurho, Lord_Farin Dec 23 '18 at 11:07
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." – Nosrati, Jack D'Aurizio, Saad, mrtaurho, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Nosrati, Jack D'Aurizio, Saad, mrtaurho, Lord_Farin Dec 23 '18 at 11:07
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." – Nosrati, Jack D'Aurizio, Saad, mrtaurho, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
$$tan2x=tan(x+y+x-y)=dfrac{tan(x+y)+tan(x-y)}{1-tan(x+y)tan(x-y)}$$
$endgroup$
– Nosrati
Dec 9 '18 at 16:39
$begingroup$
Oh god thats genius, why didnt i think of that!
$endgroup$
– Etotheipi
Dec 9 '18 at 16:42
$begingroup$
@Nosrati,But the question asks for tan(2x) in terms of a and b
$endgroup$
– Dhamnekar Winod
Dec 9 '18 at 16:48
$begingroup$
Write $tan=dfrac{sin}{cos}$.
$endgroup$
– Nosrati
Dec 9 '18 at 16:50
$begingroup$
@DhamnekarWinod You just use $tan x = frac{sin x}{cos x}$ and $cos x = sqrt{1-x^2}$.
$endgroup$
– KM101
Dec 9 '18 at 16:53
add a comment |
4
$begingroup$
$$tan2x=tan(x+y+x-y)=dfrac{tan(x+y)+tan(x-y)}{1-tan(x+y)tan(x-y)}$$
$endgroup$
– Nosrati
Dec 9 '18 at 16:39
$begingroup$
Oh god thats genius, why didnt i think of that!
$endgroup$
– Etotheipi
Dec 9 '18 at 16:42
$begingroup$
@Nosrati,But the question asks for tan(2x) in terms of a and b
$endgroup$
– Dhamnekar Winod
Dec 9 '18 at 16:48
$begingroup$
Write $tan=dfrac{sin}{cos}$.
$endgroup$
– Nosrati
Dec 9 '18 at 16:50
$begingroup$
@DhamnekarWinod You just use $tan x = frac{sin x}{cos x}$ and $cos x = sqrt{1-x^2}$.
$endgroup$
– KM101
Dec 9 '18 at 16:53
4
4
$begingroup$
$$tan2x=tan(x+y+x-y)=dfrac{tan(x+y)+tan(x-y)}{1-tan(x+y)tan(x-y)}$$
$endgroup$
– Nosrati
Dec 9 '18 at 16:39
$begingroup$
$$tan2x=tan(x+y+x-y)=dfrac{tan(x+y)+tan(x-y)}{1-tan(x+y)tan(x-y)}$$
$endgroup$
– Nosrati
Dec 9 '18 at 16:39
$begingroup$
Oh god thats genius, why didnt i think of that!
$endgroup$
– Etotheipi
Dec 9 '18 at 16:42
$begingroup$
Oh god thats genius, why didnt i think of that!
$endgroup$
– Etotheipi
Dec 9 '18 at 16:42
$begingroup$
@Nosrati,But the question asks for tan(2x) in terms of a and b
$endgroup$
– Dhamnekar Winod
Dec 9 '18 at 16:48
$begingroup$
@Nosrati,But the question asks for tan(2x) in terms of a and b
$endgroup$
– Dhamnekar Winod
Dec 9 '18 at 16:48
$begingroup$
Write $tan=dfrac{sin}{cos}$.
$endgroup$
– Nosrati
Dec 9 '18 at 16:50
$begingroup$
Write $tan=dfrac{sin}{cos}$.
$endgroup$
– Nosrati
Dec 9 '18 at 16:50
$begingroup$
@DhamnekarWinod You just use $tan x = frac{sin x}{cos x}$ and $cos x = sqrt{1-x^2}$.
$endgroup$
– KM101
Dec 9 '18 at 16:53
$begingroup$
@DhamnekarWinod You just use $tan x = frac{sin x}{cos x}$ and $cos x = sqrt{1-x^2}$.
$endgroup$
– KM101
Dec 9 '18 at 16:53
add a comment |
2 Answers
2
active
oldest
votes
1
$begingroup$
$$
x+y = arcsin a\
x-y = arccos b
$$
then
$$
2x = arcsin a + arccos b to tan(2x) = frac{asqrt{1-a^2}+b sqrt{1-b^2}}{b^2-a^2}
$$
NOTE
$$
tan(u+v) = frac{sin (u) cos (v)}{cos (u) cos (v)-sin (u) sin (v)}+frac{cos (u) sin (v)}{cos (u) cos (v)-sin (u) sin (v)}
$$
answered Dec 9 '18 at 17:13
CesareoCesareo
8,8993516
$endgroup$
$begingroup$
,is your answer is equivalent to the answer given by Chris Custer?$sqrt{1-a^2}=cos{(x+y)}$ and$b^2-a^2=cos^2{(x-y)}-sin^2{(x+y)}$ am I right?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:22
$begingroup$
,How did you calculate $tan{(2x)}$? What is the use of your reference note? What are u and v?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:34
$begingroup$
@DhamnekarWinod In the reference note we have a trigonometric identity. Here $u = arcsin a, v=arccos b$. Now we have $sin(arcsin a) = a, sinarccos a = sqrt{1-a^2}$ etc.
$endgroup$
– Cesareo
Dec 10 '18 at 16:54
$begingroup$
@Casareo,$tan{(2x)}=frac{sin{(2x)}}{cos{(2x)}}$ and $sin{(2x)}=ab+sqrt{(1-a^2)}sqrt{(1-b^2)}$ But your numerator is $asqrt{(1-a^2)}+bsqrt{(1-b^2)}$ How is that?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 17:21
$begingroup$
@DhamnekarWinod Please. Refer to me as Cesareo. We have $tan left(sin ^{-1}(a)+cos ^{-1}(b)right) = frac{a b}{sqrt{1-a^2} b-a sqrt{1-b^2}}+frac{sqrt{1-a^2} sqrt{1-b^2}}{sqrt{1-a^2} b-a sqrt{1-b^2}}$ and after simplifivcations we have $frac{asqrt{1-a^2} +b sqrt{1-b^2}}{b^2-a^2}$
$endgroup$
– Cesareo
Dec 10 '18 at 17:38
|
show 1 more comment
2
$begingroup$
$cos2x=cos(x+y+x-y)=cos(x+y)cos(x-y)-sin(x+y)sin(x-y)= asqrt{1-b^2}- bsqrt{1-a^2}$.
Similarly $sin2x=sin(x+y+x-y)=sin(x+y)cos(x-y)+cos(x+y)sin(x-y)=sqrt{1-a^2}sqrt{1-b^2}+ab$.
Thus...?
Note: this works for quadrant $1$. In $2,3$ and $4$ you need to change some signs.
answered Dec 9 '18 at 16:48
Chris CusterChris Custer
13.4k3827
$endgroup$
$begingroup$
You forgot to deal with the signs of the involved square roots. $sin(z)=a$ does not automatically imply $cos(z)=sqrt{1-a^2}$.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 16:50
$begingroup$
well, $a.b$ are infact positive, so maybe we can ignore the signs in square root
$endgroup$
– Etotheipi
Dec 9 '18 at 16:51
$begingroup$
No. The positivity of $sin(x+y)$ does not imply the positivity of $cos(x+y)$.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 16:52
$begingroup$
@Chris Custer, is more simplification of $tan{(2x)}$ possible?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:48
$begingroup$
Nothing really jumps out at me. You could probably get the radicals out of the denominator; but I would just leave it.
$endgroup$
– Chris Custer
Dec 10 '18 at 15:20
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
$$
x+y = arcsin a\
x-y = arccos b
$$
then
$$
2x = arcsin a + arccos b to tan(2x) = frac{asqrt{1-a^2}+b sqrt{1-b^2}}{b^2-a^2}
$$
NOTE
$$
tan(u+v) = frac{sin (u) cos (v)}{cos (u) cos (v)-sin (u) sin (v)}+frac{cos (u) sin (v)}{cos (u) cos (v)-sin (u) sin (v)}
$$
answered Dec 9 '18 at 17:13
CesareoCesareo
8,8993516
$endgroup$
$begingroup$
,is your answer is equivalent to the answer given by Chris Custer?$sqrt{1-a^2}=cos{(x+y)}$ and$b^2-a^2=cos^2{(x-y)}-sin^2{(x+y)}$ am I right?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:22
$begingroup$
,How did you calculate $tan{(2x)}$? What is the use of your reference note? What are u and v?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:34
$begingroup$
@DhamnekarWinod In the reference note we have a trigonometric identity. Here $u = arcsin a, v=arccos b$. Now we have $sin(arcsin a) = a, sinarccos a = sqrt{1-a^2}$ etc.
$endgroup$
– Cesareo
Dec 10 '18 at 16:54
$begingroup$
@Casareo,$tan{(2x)}=frac{sin{(2x)}}{cos{(2x)}}$ and $sin{(2x)}=ab+sqrt{(1-a^2)}sqrt{(1-b^2)}$ But your numerator is $asqrt{(1-a^2)}+bsqrt{(1-b^2)}$ How is that?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 17:21
$begingroup$
@DhamnekarWinod Please. Refer to me as Cesareo. We have $tan left(sin ^{-1}(a)+cos ^{-1}(b)right) = frac{a b}{sqrt{1-a^2} b-a sqrt{1-b^2}}+frac{sqrt{1-a^2} sqrt{1-b^2}}{sqrt{1-a^2} b-a sqrt{1-b^2}}$ and after simplifivcations we have $frac{asqrt{1-a^2} +b sqrt{1-b^2}}{b^2-a^2}$
$endgroup$
– Cesareo
Dec 10 '18 at 17:38
|
show 1 more comment
1
$begingroup$
$$
x+y = arcsin a\
x-y = arccos b
$$
then
$$
2x = arcsin a + arccos b to tan(2x) = frac{asqrt{1-a^2}+b sqrt{1-b^2}}{b^2-a^2}
$$
NOTE
$$
tan(u+v) = frac{sin (u) cos (v)}{cos (u) cos (v)-sin (u) sin (v)}+frac{cos (u) sin (v)}{cos (u) cos (v)-sin (u) sin (v)}
$$
answered Dec 9 '18 at 17:13
CesareoCesareo
8,8993516
$endgroup$
$begingroup$
,is your answer is equivalent to the answer given by Chris Custer?$sqrt{1-a^2}=cos{(x+y)}$ and$b^2-a^2=cos^2{(x-y)}-sin^2{(x+y)}$ am I right?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:22
$begingroup$
,How did you calculate $tan{(2x)}$? What is the use of your reference note? What are u and v?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:34
$begingroup$
@DhamnekarWinod In the reference note we have a trigonometric identity. Here $u = arcsin a, v=arccos b$. Now we have $sin(arcsin a) = a, sinarccos a = sqrt{1-a^2}$ etc.
$endgroup$
– Cesareo
Dec 10 '18 at 16:54
$begingroup$
@Casareo,$tan{(2x)}=frac{sin{(2x)}}{cos{(2x)}}$ and $sin{(2x)}=ab+sqrt{(1-a^2)}sqrt{(1-b^2)}$ But your numerator is $asqrt{(1-a^2)}+bsqrt{(1-b^2)}$ How is that?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 17:21
$begingroup$
@DhamnekarWinod Please. Refer to me as Cesareo. We have $tan left(sin ^{-1}(a)+cos ^{-1}(b)right) = frac{a b}{sqrt{1-a^2} b-a sqrt{1-b^2}}+frac{sqrt{1-a^2} sqrt{1-b^2}}{sqrt{1-a^2} b-a sqrt{1-b^2}}$ and after simplifivcations we have $frac{asqrt{1-a^2} +b sqrt{1-b^2}}{b^2-a^2}$
$endgroup$
– Cesareo
Dec 10 '18 at 17:38
|
show 1 more comment
1
1
1
$begingroup$
$$
x+y = arcsin a\
x-y = arccos b
$$
then
$$
2x = arcsin a + arccos b to tan(2x) = frac{asqrt{1-a^2}+b sqrt{1-b^2}}{b^2-a^2}
$$
NOTE
$$
tan(u+v) = frac{sin (u) cos (v)}{cos (u) cos (v)-sin (u) sin (v)}+frac{cos (u) sin (v)}{cos (u) cos (v)-sin (u) sin (v)}
$$
answered Dec 9 '18 at 17:13
CesareoCesareo
8,8993516
$endgroup$
$$
x+y = arcsin a\
x-y = arccos b
$$
then
$$
2x = arcsin a + arccos b to tan(2x) = frac{asqrt{1-a^2}+b sqrt{1-b^2}}{b^2-a^2}
$$
NOTE
$$
tan(u+v) = frac{sin (u) cos (v)}{cos (u) cos (v)-sin (u) sin (v)}+frac{cos (u) sin (v)}{cos (u) cos (v)-sin (u) sin (v)}
$$
answered Dec 9 '18 at 17:13
CesareoCesareo
8,8993516
edited Dec 9 '18 at 17:19
answered Dec 9 '18 at 17:13
CesareoCesareo
8,8993516
answered Dec 9 '18 at 17:13
CesareoCesareo
8,8993516
answered Dec 9 '18 at 17:13
CesareoCesareo
8,8993516
8,8993516
$begingroup$
,is your answer is equivalent to the answer given by Chris Custer?$sqrt{1-a^2}=cos{(x+y)}$ and$b^2-a^2=cos^2{(x-y)}-sin^2{(x+y)}$ am I right?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:22
$begingroup$
,How did you calculate $tan{(2x)}$? What is the use of your reference note? What are u and v?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:34
$begingroup$
@DhamnekarWinod In the reference note we have a trigonometric identity. Here $u = arcsin a, v=arccos b$. Now we have $sin(arcsin a) = a, sinarccos a = sqrt{1-a^2}$ etc.
$endgroup$
– Cesareo
Dec 10 '18 at 16:54
$begingroup$
@Casareo,$tan{(2x)}=frac{sin{(2x)}}{cos{(2x)}}$ and $sin{(2x)}=ab+sqrt{(1-a^2)}sqrt{(1-b^2)}$ But your numerator is $asqrt{(1-a^2)}+bsqrt{(1-b^2)}$ How is that?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 17:21
$begingroup$
@DhamnekarWinod Please. Refer to me as Cesareo. We have $tan left(sin ^{-1}(a)+cos ^{-1}(b)right) = frac{a b}{sqrt{1-a^2} b-a sqrt{1-b^2}}+frac{sqrt{1-a^2} sqrt{1-b^2}}{sqrt{1-a^2} b-a sqrt{1-b^2}}$ and after simplifivcations we have $frac{asqrt{1-a^2} +b sqrt{1-b^2}}{b^2-a^2}$
$endgroup$
– Cesareo
Dec 10 '18 at 17:38
|
show 1 more comment
$begingroup$
,is your answer is equivalent to the answer given by Chris Custer?$sqrt{1-a^2}=cos{(x+y)}$ and$b^2-a^2=cos^2{(x-y)}-sin^2{(x+y)}$ am I right?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:22
$begingroup$
,How did you calculate $tan{(2x)}$? What is the use of your reference note? What are u and v?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:34
$begingroup$
@DhamnekarWinod In the reference note we have a trigonometric identity. Here $u = arcsin a, v=arccos b$. Now we have $sin(arcsin a) = a, sinarccos a = sqrt{1-a^2}$ etc.
$endgroup$
– Cesareo
Dec 10 '18 at 16:54
$begingroup$
@Casareo,$tan{(2x)}=frac{sin{(2x)}}{cos{(2x)}}$ and $sin{(2x)}=ab+sqrt{(1-a^2)}sqrt{(1-b^2)}$ But your numerator is $asqrt{(1-a^2)}+bsqrt{(1-b^2)}$ How is that?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 17:21
$begingroup$
@DhamnekarWinod Please. Refer to me as Cesareo. We have $tan left(sin ^{-1}(a)+cos ^{-1}(b)right) = frac{a b}{sqrt{1-a^2} b-a sqrt{1-b^2}}+frac{sqrt{1-a^2} sqrt{1-b^2}}{sqrt{1-a^2} b-a sqrt{1-b^2}}$ and after simplifivcations we have $frac{asqrt{1-a^2} +b sqrt{1-b^2}}{b^2-a^2}$
$endgroup$
– Cesareo
Dec 10 '18 at 17:38
$begingroup$
,is your answer is equivalent to the answer given by Chris Custer?$sqrt{1-a^2}=cos{(x+y)}$ and$b^2-a^2=cos^2{(x-y)}-sin^2{(x+y)}$ am I right?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:22
$begingroup$
,is your answer is equivalent to the answer given by Chris Custer?$sqrt{1-a^2}=cos{(x+y)}$ and$b^2-a^2=cos^2{(x-y)}-sin^2{(x+y)}$ am I right?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:22
$begingroup$
,How did you calculate $tan{(2x)}$? What is the use of your reference note? What are u and v?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:34
$begingroup$
,How did you calculate $tan{(2x)}$? What is the use of your reference note? What are u and v?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:34
$begingroup$
@DhamnekarWinod In the reference note we have a trigonometric identity. Here $u = arcsin a, v=arccos b$. Now we have $sin(arcsin a) = a, sinarccos a = sqrt{1-a^2}$ etc.
$endgroup$
– Cesareo
Dec 10 '18 at 16:54
$begingroup$
@DhamnekarWinod In the reference note we have a trigonometric identity. Here $u = arcsin a, v=arccos b$. Now we have $sin(arcsin a) = a, sinarccos a = sqrt{1-a^2}$ etc.
$endgroup$
– Cesareo
Dec 10 '18 at 16:54
$begingroup$
@Casareo,$tan{(2x)}=frac{sin{(2x)}}{cos{(2x)}}$ and $sin{(2x)}=ab+sqrt{(1-a^2)}sqrt{(1-b^2)}$ But your numerator is $asqrt{(1-a^2)}+bsqrt{(1-b^2)}$ How is that?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 17:21
$begingroup$
@Casareo,$tan{(2x)}=frac{sin{(2x)}}{cos{(2x)}}$ and $sin{(2x)}=ab+sqrt{(1-a^2)}sqrt{(1-b^2)}$ But your numerator is $asqrt{(1-a^2)}+bsqrt{(1-b^2)}$ How is that?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 17:21
$begingroup$
@DhamnekarWinod Please. Refer to me as Cesareo. We have $tan left(sin ^{-1}(a)+cos ^{-1}(b)right) = frac{a b}{sqrt{1-a^2} b-a sqrt{1-b^2}}+frac{sqrt{1-a^2} sqrt{1-b^2}}{sqrt{1-a^2} b-a sqrt{1-b^2}}$ and after simplifivcations we have $frac{asqrt{1-a^2} +b sqrt{1-b^2}}{b^2-a^2}$
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– Cesareo
Dec 10 '18 at 17:38
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@DhamnekarWinod Please. Refer to me as Cesareo. We have $tan left(sin ^{-1}(a)+cos ^{-1}(b)right) = frac{a b}{sqrt{1-a^2} b-a sqrt{1-b^2}}+frac{sqrt{1-a^2} sqrt{1-b^2}}{sqrt{1-a^2} b-a sqrt{1-b^2}}$ and after simplifivcations we have $frac{asqrt{1-a^2} +b sqrt{1-b^2}}{b^2-a^2}$
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– Cesareo
Dec 10 '18 at 17:38
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$cos2x=cos(x+y+x-y)=cos(x+y)cos(x-y)-sin(x+y)sin(x-y)= asqrt{1-b^2}- bsqrt{1-a^2}$.
Similarly $sin2x=sin(x+y+x-y)=sin(x+y)cos(x-y)+cos(x+y)sin(x-y)=sqrt{1-a^2}sqrt{1-b^2}+ab$.
Thus...?
Note: this works for quadrant $1$. In $2,3$ and $4$ you need to change some signs.
answered Dec 9 '18 at 16:48
Chris CusterChris Custer
13.4k3827
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You forgot to deal with the signs of the involved square roots. $sin(z)=a$ does not automatically imply $cos(z)=sqrt{1-a^2}$.
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– Jack D'Aurizio
Dec 9 '18 at 16:50
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well, $a.b$ are infact positive, so maybe we can ignore the signs in square root
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– Etotheipi
Dec 9 '18 at 16:51
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No. The positivity of $sin(x+y)$ does not imply the positivity of $cos(x+y)$.
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– Jack D'Aurizio
Dec 9 '18 at 16:52
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@Chris Custer, is more simplification of $tan{(2x)}$ possible?
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– Dhamnekar Winod
Dec 10 '18 at 12:48
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Nothing really jumps out at me. You could probably get the radicals out of the denominator; but I would just leave it.
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– Chris Custer
Dec 10 '18 at 15:20
add a comment |
2
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$cos2x=cos(x+y+x-y)=cos(x+y)cos(x-y)-sin(x+y)sin(x-y)= asqrt{1-b^2}- bsqrt{1-a^2}$.
Similarly $sin2x=sin(x+y+x-y)=sin(x+y)cos(x-y)+cos(x+y)sin(x-y)=sqrt{1-a^2}sqrt{1-b^2}+ab$.
Thus...?
Note: this works for quadrant $1$. In $2,3$ and $4$ you need to change some signs.
answered Dec 9 '18 at 16:48
Chris CusterChris Custer
13.4k3827
$endgroup$
$begingroup$
You forgot to deal with the signs of the involved square roots. $sin(z)=a$ does not automatically imply $cos(z)=sqrt{1-a^2}$.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 16:50
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well, $a.b$ are infact positive, so maybe we can ignore the signs in square root
$endgroup$
– Etotheipi
Dec 9 '18 at 16:51
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No. The positivity of $sin(x+y)$ does not imply the positivity of $cos(x+y)$.
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– Jack D'Aurizio
Dec 9 '18 at 16:52
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@Chris Custer, is more simplification of $tan{(2x)}$ possible?
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– Dhamnekar Winod
Dec 10 '18 at 12:48
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Nothing really jumps out at me. You could probably get the radicals out of the denominator; but I would just leave it.
$endgroup$
– Chris Custer
Dec 10 '18 at 15:20
add a comment |
2
2
2
$begingroup$
$cos2x=cos(x+y+x-y)=cos(x+y)cos(x-y)-sin(x+y)sin(x-y)= asqrt{1-b^2}- bsqrt{1-a^2}$.
Similarly $sin2x=sin(x+y+x-y)=sin(x+y)cos(x-y)+cos(x+y)sin(x-y)=sqrt{1-a^2}sqrt{1-b^2}+ab$.
Thus...?
Note: this works for quadrant $1$. In $2,3$ and $4$ you need to change some signs.
answered Dec 9 '18 at 16:48
Chris CusterChris Custer
13.4k3827
$endgroup$
$cos2x=cos(x+y+x-y)=cos(x+y)cos(x-y)-sin(x+y)sin(x-y)= asqrt{1-b^2}- bsqrt{1-a^2}$.
Similarly $sin2x=sin(x+y+x-y)=sin(x+y)cos(x-y)+cos(x+y)sin(x-y)=sqrt{1-a^2}sqrt{1-b^2}+ab$.
Thus...?
Note: this works for quadrant $1$. In $2,3$ and $4$ you need to change some signs.
answered Dec 9 '18 at 16:48
Chris CusterChris Custer
13.4k3827
edited Dec 9 '18 at 17:23
answered Dec 9 '18 at 16:48
Chris CusterChris Custer
13.4k3827
answered Dec 9 '18 at 16:48
Chris CusterChris Custer
13.4k3827
answered Dec 9 '18 at 16:48
Chris CusterChris Custer
13.4k3827
13.4k3827
$begingroup$
You forgot to deal with the signs of the involved square roots. $sin(z)=a$ does not automatically imply $cos(z)=sqrt{1-a^2}$.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 16:50
$begingroup$
well, $a.b$ are infact positive, so maybe we can ignore the signs in square root
$endgroup$
– Etotheipi
Dec 9 '18 at 16:51
$begingroup$
No. The positivity of $sin(x+y)$ does not imply the positivity of $cos(x+y)$.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 16:52
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@Chris Custer, is more simplification of $tan{(2x)}$ possible?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:48
$begingroup$
Nothing really jumps out at me. You could probably get the radicals out of the denominator; but I would just leave it.
$endgroup$
– Chris Custer
Dec 10 '18 at 15:20
add a comment |
$begingroup$
You forgot to deal with the signs of the involved square roots. $sin(z)=a$ does not automatically imply $cos(z)=sqrt{1-a^2}$.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 16:50
$begingroup$
well, $a.b$ are infact positive, so maybe we can ignore the signs in square root
$endgroup$
– Etotheipi
Dec 9 '18 at 16:51
$begingroup$
No. The positivity of $sin(x+y)$ does not imply the positivity of $cos(x+y)$.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 16:52
$begingroup$
@Chris Custer, is more simplification of $tan{(2x)}$ possible?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:48
$begingroup$
Nothing really jumps out at me. You could probably get the radicals out of the denominator; but I would just leave it.
$endgroup$
– Chris Custer
Dec 10 '18 at 15:20
$begingroup$
You forgot to deal with the signs of the involved square roots. $sin(z)=a$ does not automatically imply $cos(z)=sqrt{1-a^2}$.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 16:50
$begingroup$
You forgot to deal with the signs of the involved square roots. $sin(z)=a$ does not automatically imply $cos(z)=sqrt{1-a^2}$.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 16:50
$begingroup$
well, $a.b$ are infact positive, so maybe we can ignore the signs in square root
$endgroup$
– Etotheipi
Dec 9 '18 at 16:51
$begingroup$
well, $a.b$ are infact positive, so maybe we can ignore the signs in square root
$endgroup$
– Etotheipi
Dec 9 '18 at 16:51
$begingroup$
No. The positivity of $sin(x+y)$ does not imply the positivity of $cos(x+y)$.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 16:52
$begingroup$
No. The positivity of $sin(x+y)$ does not imply the positivity of $cos(x+y)$.
$endgroup$
– Jack D'Aurizio
Dec 9 '18 at 16:52
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@Chris Custer, is more simplification of $tan{(2x)}$ possible?
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– Dhamnekar Winod
Dec 10 '18 at 12:48
$begingroup$
@Chris Custer, is more simplification of $tan{(2x)}$ possible?
$endgroup$
– Dhamnekar Winod
Dec 10 '18 at 12:48
$begingroup$
Nothing really jumps out at me. You could probably get the radicals out of the denominator; but I would just leave it.
$endgroup$
– Chris Custer
Dec 10 '18 at 15:20
$begingroup$
Nothing really jumps out at me. You could probably get the radicals out of the denominator; but I would just leave it.
$endgroup$
– Chris Custer
Dec 10 '18 at 15:20
add a comment |
4
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$$tan2x=tan(x+y+x-y)=dfrac{tan(x+y)+tan(x-y)}{1-tan(x+y)tan(x-y)}$$
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– Nosrati
Dec 9 '18 at 16:39
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Oh god thats genius, why didnt i think of that!
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– Etotheipi
Dec 9 '18 at 16:42
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@Nosrati,But the question asks for tan(2x) in terms of a and b
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– Dhamnekar Winod
Dec 9 '18 at 16:48
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Write $tan=dfrac{sin}{cos}$.
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– Nosrati
Dec 9 '18 at 16:50
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@DhamnekarWinod You just use $tan x = frac{sin x}{cos x}$ and $cos x = sqrt{1-x^2}$.
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– KM101
Dec 9 '18 at 16:53