Help prove this basic trig identity please!If $sintheta + sinphi = a$ and $costheta + cosphi = b$, then $sin(theta+phi) = ???$If $ sin alpha + sin beta = a $ and $ cos alpha + cos beta = b $ , then show that $sin(alpha + beta) = frac 2ab a^2 + b^2 $Trig equation help pleaseHelp With Double Angles And Trig Identity ProblemProve Trig IdentityHow to solve this trigonometric identity?Prove the following Trig Identity with reciprocalsHelp needed in verifying a trigonometric identityHow do I prove this seemingly simple trigonometric identityTrig Identity Proof $frac1 + sinthetacostheta + fraccostheta1 - sintheta = 2tanleft(fractheta2 + fracpi4right)$If $x costheta+ysintheta=a$ and $xsintheta-ycostheta=b$, then $tantheta=fracbx+ayax-by$. (Math Olympiad)Precalc Trig Identity, verify: $1 + cos(x) + cos(2x) = frac 12 + fracsin(5x/2)2sin(x/2)$
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Help prove this basic trig identity please!
If $sintheta + sinphi = a$ and $costheta + cosphi = b$, then $sin(theta+phi) = ???$If $ sin alpha + sin beta = a $ and $ cos alpha + cos beta = b $ , then show that $sin(alpha + beta) = frac 2ab a^2 + b^2 $Trig equation help pleaseHelp With Double Angles And Trig Identity ProblemProve Trig IdentityHow to solve this trigonometric identity?Prove the following Trig Identity with reciprocalsHelp needed in verifying a trigonometric identityHow do I prove this seemingly simple trigonometric identityTrig Identity Proof $frac1 + sinthetacostheta + fraccostheta1 - sintheta = 2tanleft(fractheta2 + fracpi4right)$If $x costheta+ysintheta=a$ and $xsintheta-ycostheta=b$, then $tantheta=fracbx+ayax-by$. (Math Olympiad)Precalc Trig Identity, verify: $1 + cos(x) + cos(2x) = frac 12 + fracsin(5x/2)2sin(x/2)$
$begingroup$
I'm really stuck trying to answer this question and have spent endless hours doing so.
If $a=sin(theta)+cos(phi)$ and, $b=cos(theta)+sin(phi)$, prove that $cos(theta-phi)=frac2aba^2+b^2$.
I've tried working LHS to RHS and couldn't get, I've also tried RHS to LHS and still couldn't get it, and advice or help would be much appreciated please.
I've also tried going $ab=...$ and then trying to get it from there, that didn't come to fruition either.
trigonometry
asked 4 hours ago
Avinash ShastriAvinash Shastri
184
New contributor
Avinash Shastri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I'm really stuck trying to answer this question and have spent endless hours doing so.
If $a=sin(theta)+cos(phi)$ and, $b=cos(theta)+sin(phi)$, prove that $cos(theta-phi)=frac2aba^2+b^2$.
I've tried working LHS to RHS and couldn't get, I've also tried RHS to LHS and still couldn't get it, and advice or help would be much appreciated please.
I've also tried going $ab=...$ and then trying to get it from there, that didn't come to fruition either.
trigonometry
asked 4 hours ago
Avinash ShastriAvinash Shastri
184
New contributor
Avinash Shastri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Welcome to Math Stack Exchange. Do you know the formula for cosine of a difference?
$endgroup$
– J. W. Tanner
4 hours ago
$begingroup$
Yes, $cos(theta-phi)=cos(theta)cos(phi)+sin(theta)sin(phi)$, and I went much further but was not able to poduce anything useful in the sense of the proof.
$endgroup$
– Avinash Shastri
4 hours ago
$begingroup$
Set $theta=dfracpi2-psi$ and use math.stackexchange.com/questions/1833153/… or math.stackexchange.com/questions/2021356/…
$endgroup$
– lab bhattacharjee
2 hours ago
add a comment |
$begingroup$
I'm really stuck trying to answer this question and have spent endless hours doing so.
If $a=sin(theta)+cos(phi)$ and, $b=cos(theta)+sin(phi)$, prove that $cos(theta-phi)=frac2aba^2+b^2$.
I've tried working LHS to RHS and couldn't get, I've also tried RHS to LHS and still couldn't get it, and advice or help would be much appreciated please.
I've also tried going $ab=...$ and then trying to get it from there, that didn't come to fruition either.
trigonometry
asked 4 hours ago
Avinash ShastriAvinash Shastri
184
New contributor
Avinash Shastri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm really stuck trying to answer this question and have spent endless hours doing so.
If $a=sin(theta)+cos(phi)$ and, $b=cos(theta)+sin(phi)$, prove that $cos(theta-phi)=frac2aba^2+b^2$.
I've tried working LHS to RHS and couldn't get, I've also tried RHS to LHS and still couldn't get it, and advice or help would be much appreciated please.
I've also tried going $ab=...$ and then trying to get it from there, that didn't come to fruition either.
trigonometry
trigonometry
asked 4 hours ago
Avinash ShastriAvinash Shastri
184
New contributor
Avinash Shastri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Avinash ShastriAvinash Shastri
184
New contributor
Avinash Shastri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Avinash ShastriAvinash Shastri
184
New contributor
Avinash Shastri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Avinash ShastriAvinash Shastri
184
asked 4 hours ago
Avinash ShastriAvinash Shastri
184
184
New contributor
Avinash Shastri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Avinash Shastri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Avinash Shastri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Welcome to Math Stack Exchange. Do you know the formula for cosine of a difference?
$endgroup$
– J. W. Tanner
4 hours ago
$begingroup$
Yes, $cos(theta-phi)=cos(theta)cos(phi)+sin(theta)sin(phi)$, and I went much further but was not able to poduce anything useful in the sense of the proof.
$endgroup$
– Avinash Shastri
4 hours ago
$begingroup$
Set $theta=dfracpi2-psi$ and use math.stackexchange.com/questions/1833153/… or math.stackexchange.com/questions/2021356/…
$endgroup$
– lab bhattacharjee
2 hours ago
add a comment |
$begingroup$
Welcome to Math Stack Exchange. Do you know the formula for cosine of a difference?
$endgroup$
– J. W. Tanner
4 hours ago
$begingroup$
Yes, $cos(theta-phi)=cos(theta)cos(phi)+sin(theta)sin(phi)$, and I went much further but was not able to poduce anything useful in the sense of the proof.
$endgroup$
– Avinash Shastri
4 hours ago
$begingroup$
Set $theta=dfracpi2-psi$ and use math.stackexchange.com/questions/1833153/… or math.stackexchange.com/questions/2021356/…
$endgroup$
– lab bhattacharjee
2 hours ago
$begingroup$
Welcome to Math Stack Exchange. Do you know the formula for cosine of a difference?
$endgroup$
– J. W. Tanner
4 hours ago
$begingroup$
Welcome to Math Stack Exchange. Do you know the formula for cosine of a difference?
$endgroup$
– J. W. Tanner
4 hours ago
$begingroup$
Yes, $cos(theta-phi)=cos(theta)cos(phi)+sin(theta)sin(phi)$, and I went much further but was not able to poduce anything useful in the sense of the proof.
$endgroup$
– Avinash Shastri
4 hours ago
$begingroup$
Yes, $cos(theta-phi)=cos(theta)cos(phi)+sin(theta)sin(phi)$, and I went much further but was not able to poduce anything useful in the sense of the proof.
$endgroup$
– Avinash Shastri
4 hours ago
$begingroup$
Set $theta=dfracpi2-psi$ and use math.stackexchange.com/questions/1833153/… or math.stackexchange.com/questions/2021356/…
$endgroup$
– lab bhattacharjee
2 hours ago
$begingroup$
Set $theta=dfracpi2-psi$ and use math.stackexchange.com/questions/1833153/… or math.stackexchange.com/questions/2021356/…
$endgroup$
– lab bhattacharjee
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$(i).a=sin(theta)+cos(phi)$$
$$(ii).b=cos(theta)+sin(phi)$$
$$(i)^2+(ii)^2=2+2sin(theta +phi)$$so
$$sin(theta+phi) =(a^2+b^2)over 2-1$$.
$$(i)*(ii)=sin(2theta)+sin(2phi) over 2+cos(theta-phi)=sin(theta+phi)cos(theta-phi)+cos(theta-phi)$$
so$$cos(theta-phi)=abover 1+sin(theta+phi)=2abover a^2+b^2$$
answered 3 hours ago
StAKmodStAKmod
406110
$endgroup$
$begingroup$
Wow, thank you, you've made it seem so simple. Was the path intuitive or how did you decide to choose this path?
$endgroup$
– Avinash Shastri
3 hours ago
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
$$(i).a=sin(theta)+cos(phi)$$
$$(ii).b=cos(theta)+sin(phi)$$
$$(i)^2+(ii)^2=2+2sin(theta +phi)$$so
$$sin(theta+phi) =(a^2+b^2)over 2-1$$.
$$(i)*(ii)=sin(2theta)+sin(2phi) over 2+cos(theta-phi)=sin(theta+phi)cos(theta-phi)+cos(theta-phi)$$
so$$cos(theta-phi)=abover 1+sin(theta+phi)=2abover a^2+b^2$$
answered 3 hours ago
StAKmodStAKmod
406110
$endgroup$
$begingroup$
Wow, thank you, you've made it seem so simple. Was the path intuitive or how did you decide to choose this path?
$endgroup$
– Avinash Shastri
3 hours ago
add a comment |
$begingroup$
$$(i).a=sin(theta)+cos(phi)$$
$$(ii).b=cos(theta)+sin(phi)$$
$$(i)^2+(ii)^2=2+2sin(theta +phi)$$so
$$sin(theta+phi) =(a^2+b^2)over 2-1$$.
$$(i)*(ii)=sin(2theta)+sin(2phi) over 2+cos(theta-phi)=sin(theta+phi)cos(theta-phi)+cos(theta-phi)$$
so$$cos(theta-phi)=abover 1+sin(theta+phi)=2abover a^2+b^2$$
answered 3 hours ago
StAKmodStAKmod
406110
$endgroup$
$begingroup$
Wow, thank you, you've made it seem so simple. Was the path intuitive or how did you decide to choose this path?
$endgroup$
– Avinash Shastri
3 hours ago
add a comment |
$begingroup$
$$(i).a=sin(theta)+cos(phi)$$
$$(ii).b=cos(theta)+sin(phi)$$
$$(i)^2+(ii)^2=2+2sin(theta +phi)$$so
$$sin(theta+phi) =(a^2+b^2)over 2-1$$.
$$(i)*(ii)=sin(2theta)+sin(2phi) over 2+cos(theta-phi)=sin(theta+phi)cos(theta-phi)+cos(theta-phi)$$
so$$cos(theta-phi)=abover 1+sin(theta+phi)=2abover a^2+b^2$$
answered 3 hours ago
StAKmodStAKmod
406110
$endgroup$
$$(i).a=sin(theta)+cos(phi)$$
$$(ii).b=cos(theta)+sin(phi)$$
$$(i)^2+(ii)^2=2+2sin(theta +phi)$$so
$$sin(theta+phi) =(a^2+b^2)over 2-1$$.
$$(i)*(ii)=sin(2theta)+sin(2phi) over 2+cos(theta-phi)=sin(theta+phi)cos(theta-phi)+cos(theta-phi)$$
so$$cos(theta-phi)=abover 1+sin(theta+phi)=2abover a^2+b^2$$
answered 3 hours ago
StAKmodStAKmod
406110
answered 3 hours ago
StAKmodStAKmod
406110
answered 3 hours ago
StAKmodStAKmod
406110
answered 3 hours ago
StAKmodStAKmod
406110
406110
$begingroup$
Wow, thank you, you've made it seem so simple. Was the path intuitive or how did you decide to choose this path?
$endgroup$
– Avinash Shastri
3 hours ago
add a comment |
$begingroup$
Wow, thank you, you've made it seem so simple. Was the path intuitive or how did you decide to choose this path?
$endgroup$
– Avinash Shastri
3 hours ago
$begingroup$
Wow, thank you, you've made it seem so simple. Was the path intuitive or how did you decide to choose this path?
$endgroup$
– Avinash Shastri
3 hours ago
$begingroup$
Wow, thank you, you've made it seem so simple. Was the path intuitive or how did you decide to choose this path?
$endgroup$
– Avinash Shastri
3 hours ago
add a comment |
Avinash Shastri is a new contributor. Be nice, and check out our Code of Conduct.
Avinash Shastri is a new contributor. Be nice, and check out our Code of Conduct.
Avinash Shastri is a new contributor. Be nice, and check out our Code of Conduct.
Avinash Shastri is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Welcome to Math Stack Exchange. Do you know the formula for cosine of a difference?
$endgroup$
– J. W. Tanner
4 hours ago
$begingroup$
Yes, $cos(theta-phi)=cos(theta)cos(phi)+sin(theta)sin(phi)$, and I went much further but was not able to poduce anything useful in the sense of the proof.
$endgroup$
– Avinash Shastri
4 hours ago
$begingroup$
Set $theta=dfracpi2-psi$ and use math.stackexchange.com/questions/1833153/… or math.stackexchange.com/questions/2021356/…
$endgroup$
– lab bhattacharjee
2 hours ago