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show this identity with trigometric
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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 $x,yin (0,dfrac{pi}{2})$. show that
$$dfrac{sin{(x+y)}tan{x}-cos{(x+y)}}{sin{(x+y)}tan{y}-cos{(x+y)}}=dfrac{cos{(2x+y)}cos{y}}{cos{(x+2y)}cos{x}}$$
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 3 hours ago
function sugfunction sug
3381438
$endgroup$
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 $x,yin (0,dfrac{pi}{2})$. show that
$$dfrac{sin{(x+y)}tan{x}-cos{(x+y)}}{sin{(x+y)}tan{y}-cos{(x+y)}}=dfrac{cos{(2x+y)}cos{y}}{cos{(x+2y)}cos{x}}$$
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 3 hours ago
function sugfunction sug
3381438
$endgroup$
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
2 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 $x,yin (0,dfrac{pi}{2})$. show that
$$dfrac{sin{(x+y)}tan{x}-cos{(x+y)}}{sin{(x+y)}tan{y}-cos{(x+y)}}=dfrac{cos{(2x+y)}cos{y}}{cos{(x+2y)}cos{x}}$$
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 3 hours ago
function sugfunction sug
3381438
$endgroup$
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 $x,yin (0,dfrac{pi}{2})$. show that
$$dfrac{sin{(x+y)}tan{x}-cos{(x+y)}}{sin{(x+y)}tan{y}-cos{(x+y)}}=dfrac{cos{(2x+y)}cos{y}}{cos{(x+2y)}cos{x}}$$
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
asked 3 hours ago
function sugfunction sug
3381438
asked 3 hours ago
function sugfunction sug
3381438
asked 3 hours ago
function sugfunction sug
3381438
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
2 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
2 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
2 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
2 hours ago
add a comment |
1 Answer
1
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$begingroup$
Hint:
$$begin{align}
sin(x+y)tan x - cos(x+y) &= phantom{-}frac{1}{cos x}left(;sin(x+y) sin x - cos(x+y)cos x;right) \[4pt]
&= -frac1{cos x}cosleft((x+y)+xright) \[4pt]
&= -frac1{cos x}cosleft(2x+yright)
end{align}$$
answered 2 hours ago
BlueBlue
49k870156
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
Hint:
$$begin{align}
sin(x+y)tan x - cos(x+y) &= phantom{-}frac{1}{cos x}left(;sin(x+y) sin x - cos(x+y)cos x;right) \[4pt]
&= -frac1{cos x}cosleft((x+y)+xright) \[4pt]
&= -frac1{cos x}cosleft(2x+yright)
end{align}$$
answered 2 hours ago
BlueBlue
49k870156
$endgroup$
add a comment |
$begingroup$
Hint:
$$begin{align}
sin(x+y)tan x - cos(x+y) &= phantom{-}frac{1}{cos x}left(;sin(x+y) sin x - cos(x+y)cos x;right) \[4pt]
&= -frac1{cos x}cosleft((x+y)+xright) \[4pt]
&= -frac1{cos x}cosleft(2x+yright)
end{align}$$
answered 2 hours ago
BlueBlue
49k870156
$endgroup$
add a comment |
$begingroup$
Hint:
$$begin{align}
sin(x+y)tan x - cos(x+y) &= phantom{-}frac{1}{cos x}left(;sin(x+y) sin x - cos(x+y)cos x;right) \[4pt]
&= -frac1{cos x}cosleft((x+y)+xright) \[4pt]
&= -frac1{cos x}cosleft(2x+yright)
end{align}$$
answered 2 hours ago
BlueBlue
49k870156
$endgroup$
Hint:
$$begin{align}
sin(x+y)tan x - cos(x+y) &= phantom{-}frac{1}{cos x}left(;sin(x+y) sin x - cos(x+y)cos x;right) \[4pt]
&= -frac1{cos x}cosleft((x+y)+xright) \[4pt]
&= -frac1{cos x}cosleft(2x+yright)
end{align}$$
answered 2 hours ago
BlueBlue
49k870156
answered 2 hours ago
BlueBlue
49k870156
answered 2 hours ago
BlueBlue
49k870156
answered 2 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
2 hours ago