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show this identity with trigometric
A cubed trigonometric identity?Trigonometric Double Angle proofHow to properly use this trigonometric identity?How prove this idenity this $mv-3nu=m-3u$ with unit circleTough Trigonometric Identity ProblemGeometric meaning of a conditional trigonometric identityHow many points to prove a trigonometric identity?Solution for the trigonometric-linear functionHow do I write $csc$ and $tan$ in terms of $sin$ and $cos$?Simplifying a Trigonometric Expression with Sum Difference Identity
$begingroup$
I sent a post earlier. Follow is an original problem. I got an error identity from a previous calculation error. Now there should be no problem.
Problem::
let $x,yin (0,dfracpi2)$. show that
$$dfracsin(x+y)tanx-cos(x+y)sin(x+y)tany-cos(x+y)=dfraccos(2x+y)cosycos(x+2y)cosx$$
This identity comes from the fact that I deal with a geometric problem and use trigonometric functions to calculate an identity that needs to be proved.Thanks
trigonometry
$endgroup$
add a comment |
$begingroup$
I sent a post earlier. Follow is an original problem. I got an error identity from a previous calculation error. Now there should be no problem.
Problem::
let $x,yin (0,dfracpi2)$. show that
$$dfracsin(x+y)tanx-cos(x+y)sin(x+y)tany-cos(x+y)=dfraccos(2x+y)cosycos(x+2y)cosx$$
This identity comes from the fact that I deal with a geometric problem and use trigonometric functions to calculate an identity that needs to be proved.Thanks
trigonometry
$endgroup$
1
$begingroup$
This version seems to be true. :) (Note: You should not delete a question that has received an answer. Doing so is inconsiderate to the answerer who has taken valuable time to respond to your question.)
$endgroup$
– Blue
3 hours ago
add a comment |
$begingroup$
I sent a post earlier. Follow is an original problem. I got an error identity from a previous calculation error. Now there should be no problem.
Problem::
let $x,yin (0,dfracpi2)$. show that
$$dfracsin(x+y)tanx-cos(x+y)sin(x+y)tany-cos(x+y)=dfraccos(2x+y)cosycos(x+2y)cosx$$
This identity comes from the fact that I deal with a geometric problem and use trigonometric functions to calculate an identity that needs to be proved.Thanks
trigonometry
$endgroup$
I sent a post earlier. Follow is an original problem. I got an error identity from a previous calculation error. Now there should be no problem.
Problem::
let $x,yin (0,dfracpi2)$. show that
$$dfracsin(x+y)tanx-cos(x+y)sin(x+y)tany-cos(x+y)=dfraccos(2x+y)cosycos(x+2y)cosx$$
This identity comes from the fact that I deal with a geometric problem and use trigonometric functions to calculate an identity that needs to be proved.Thanks
trigonometry
trigonometry
asked 3 hours ago
function sugfunction sug
3381438
3381438
1
$begingroup$
This version seems to be true. :) (Note: You should not delete a question that has received an answer. Doing so is inconsiderate to the answerer who has taken valuable time to respond to your question.)
$endgroup$
– Blue
3 hours ago
add a comment |
1
$begingroup$
This version seems to be true. :) (Note: You should not delete a question that has received an answer. Doing so is inconsiderate to the answerer who has taken valuable time to respond to your question.)
$endgroup$
– Blue
3 hours ago
1
1
$begingroup$
This version seems to be true. :) (Note: You should not delete a question that has received an answer. Doing so is inconsiderate to the answerer who has taken valuable time to respond to your question.)
$endgroup$
– Blue
3 hours ago
$begingroup$
This version seems to be true. :) (Note: You should not delete a question that has received an answer. Doing so is inconsiderate to the answerer who has taken valuable time to respond to your question.)
$endgroup$
– Blue
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
$$beginalign
sin(x+y)tan x - cos(x+y) &= phantom-frac1cos xleft(;sin(x+y) sin x - cos(x+y)cos x;right) \[4pt]
&= -frac1cos xcosleft((x+y)+xright) \[4pt]
&= -frac1cos xcosleft(2x+yright)
endalign$$
$endgroup$
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Hint:
$$beginalign
sin(x+y)tan x - cos(x+y) &= phantom-frac1cos xleft(;sin(x+y) sin x - cos(x+y)cos x;right) \[4pt]
&= -frac1cos xcosleft((x+y)+xright) \[4pt]
&= -frac1cos xcosleft(2x+yright)
endalign$$
$endgroup$
add a comment |
$begingroup$
Hint:
$$beginalign
sin(x+y)tan x - cos(x+y) &= phantom-frac1cos xleft(;sin(x+y) sin x - cos(x+y)cos x;right) \[4pt]
&= -frac1cos xcosleft((x+y)+xright) \[4pt]
&= -frac1cos xcosleft(2x+yright)
endalign$$
$endgroup$
add a comment |
$begingroup$
Hint:
$$beginalign
sin(x+y)tan x - cos(x+y) &= phantom-frac1cos xleft(;sin(x+y) sin x - cos(x+y)cos x;right) \[4pt]
&= -frac1cos xcosleft((x+y)+xright) \[4pt]
&= -frac1cos xcosleft(2x+yright)
endalign$$
$endgroup$
Hint:
$$beginalign
sin(x+y)tan x - cos(x+y) &= phantom-frac1cos xleft(;sin(x+y) sin x - cos(x+y)cos x;right) \[4pt]
&= -frac1cos xcosleft((x+y)+xright) \[4pt]
&= -frac1cos xcosleft(2x+yright)
endalign$$
answered 3 hours ago
BlueBlue
49k870156
49k870156
add a comment |
add a comment |
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$begingroup$
This version seems to be true. :) (Note: You should not delete a question that has received an answer. Doing so is inconsiderate to the answerer who has taken valuable time to respond to your question.)
$endgroup$
– Blue
3 hours ago