show this identity with trigometric
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$begingroup$
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
asked Mar 13 at 2:32
function sugfunction sug
2681439
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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
asked Mar 13 at 2:32
function sugfunction sug
2681439
$endgroup$
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
Mar 13 at 2:37
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
asked Mar 13 at 2:32
function sugfunction sug
2681439
$endgroup$
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 Mar 13 at 2:32
function sugfunction sug
2681439
asked Mar 13 at 2:32
function sugfunction sug
2681439
asked Mar 13 at 2:32
function sugfunction sug
2681439
asked Mar 13 at 2:32
function sugfunction sug
2681439
asked Mar 13 at 2:32
function sugfunction sug
2681439
2681439
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
Mar 13 at 2:37
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
Mar 13 at 2:37
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
Mar 13 at 2:37
$begingroup$
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
Mar 13 at 2:37
add a comment |
1 Answer
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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$$
answered Mar 13 at 2:42
BlueBlue
49.6k870158
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$begingroup$
Hint:
$$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$$
answered Mar 13 at 2:42
BlueBlue
49.6k870158
$endgroup$
add a comment |
$begingroup$
Hint:
$$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$$
answered Mar 13 at 2:42
BlueBlue
49.6k870158
$endgroup$
add a comment |
$begingroup$
Hint:
$$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$$
answered Mar 13 at 2:42
BlueBlue
49.6k870158
$endgroup$
Hint:
$$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 Mar 13 at 2:42
BlueBlue
49.6k870158
answered Mar 13 at 2:42
BlueBlue
49.6k870158
answered Mar 13 at 2:42
BlueBlue
49.6k870158
answered Mar 13 at 2:42
BlueBlue
49.6k870158
49.6k870158
add a comment |
add a comment |
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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
Mar 13 at 2:37