Is there a way to find $cos(a+b)$ and $cos(a-b)$ in terms of $tan a$ and $tan b$?
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I know $tan(a)$ and $tan(b)$. Is there a way that I can find $cos x$ and $cos y$ in term of $tan a$ and $tan b$, where $x=a+b$ and $y=a-b$?
Thanks.
trigonometry
edited Nov 27 at 4:53
Blue
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asked Nov 27 at 4:50
Asghar Razzaqi
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I know $tan(a)$ and $tan(b)$. Is there a way that I can find $cos x$ and $cos y$ in term of $tan a$ and $tan b$, where $x=a+b$ and $y=a-b$?
Thanks.
trigonometry
edited Nov 27 at 4:53
Blue
47.2k870149
asked Nov 27 at 4:50
Asghar Razzaqi
31
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I know $tan(a)$ and $tan(b)$. Is there a way that I can find $cos x$ and $cos y$ in term of $tan a$ and $tan b$, where $x=a+b$ and $y=a-b$?
Thanks.
trigonometry
edited Nov 27 at 4:53
Blue
47.2k870149
asked Nov 27 at 4:50
Asghar Razzaqi
31
I know $tan(a)$ and $tan(b)$. Is there a way that I can find $cos x$ and $cos y$ in term of $tan a$ and $tan b$, where $x=a+b$ and $y=a-b$?
Thanks.
trigonometry
trigonometry
edited Nov 27 at 4:53
Blue
47.2k870149
asked Nov 27 at 4:50
Asghar Razzaqi
31
edited Nov 27 at 4:53
Blue
47.2k870149
asked Nov 27 at 4:50
Asghar Razzaqi
31
edited Nov 27 at 4:53
Blue
47.2k870149
edited Nov 27 at 4:53
Blue
47.2k870149
edited Nov 27 at 4:53
Blue
47.2k870149
47.2k870149
asked Nov 27 at 4:50
Asghar Razzaqi
31
asked Nov 27 at 4:50
Asghar Razzaqi
31
asked Nov 27 at 4:50
Asghar Razzaqi
31
31
add a comment |
add a comment |
2 Answers
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oldest
votes
up vote
3
down vote
accepted
$cos(a+b)=cos acos b- sin a sin b$
Multiply by $frac{cos acos b}{cos acos b}$
You'll get $$cos(a+b)=cos acos b- (cos acos b)(tan a tan b)$$
Now substitute instead of $cos a=sqrt{frac{1}{1+tan^2 a}}$
And you'll get that $$cos(a+b)=sqrt{frac{1}{(1+tan^2 a)(1+tan^2 b)}} (1-tan a tan b)$$
It is a bit complicated as you see, you can try and do some simplifications, and there might be another easier method to do that, but thats what came into my mind now.
answered Nov 27 at 5:06
Fareed AF
41711
add a comment |
up vote
0
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Strictly speaking, no. As $tan(x)=tan(x+kpi)$ for every integer $k$, you can only determine $cos(a+b)$ and $cos(a-b)$ up to a sign.
answered Nov 27 at 5:51
William McGonagall
1337
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$cos(a+b)=cos acos b- sin a sin b$
Multiply by $frac{cos acos b}{cos acos b}$
You'll get $$cos(a+b)=cos acos b- (cos acos b)(tan a tan b)$$
Now substitute instead of $cos a=sqrt{frac{1}{1+tan^2 a}}$
And you'll get that $$cos(a+b)=sqrt{frac{1}{(1+tan^2 a)(1+tan^2 b)}} (1-tan a tan b)$$
It is a bit complicated as you see, you can try and do some simplifications, and there might be another easier method to do that, but thats what came into my mind now.
answered Nov 27 at 5:06
Fareed AF
41711
add a comment |
up vote
3
down vote
accepted
$cos(a+b)=cos acos b- sin a sin b$
Multiply by $frac{cos acos b}{cos acos b}$
You'll get $$cos(a+b)=cos acos b- (cos acos b)(tan a tan b)$$
Now substitute instead of $cos a=sqrt{frac{1}{1+tan^2 a}}$
And you'll get that $$cos(a+b)=sqrt{frac{1}{(1+tan^2 a)(1+tan^2 b)}} (1-tan a tan b)$$
It is a bit complicated as you see, you can try and do some simplifications, and there might be another easier method to do that, but thats what came into my mind now.
answered Nov 27 at 5:06
Fareed AF
41711
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$cos(a+b)=cos acos b- sin a sin b$
Multiply by $frac{cos acos b}{cos acos b}$
You'll get $$cos(a+b)=cos acos b- (cos acos b)(tan a tan b)$$
Now substitute instead of $cos a=sqrt{frac{1}{1+tan^2 a}}$
And you'll get that $$cos(a+b)=sqrt{frac{1}{(1+tan^2 a)(1+tan^2 b)}} (1-tan a tan b)$$
It is a bit complicated as you see, you can try and do some simplifications, and there might be another easier method to do that, but thats what came into my mind now.
answered Nov 27 at 5:06
Fareed AF
41711
$cos(a+b)=cos acos b- sin a sin b$
Multiply by $frac{cos acos b}{cos acos b}$
You'll get $$cos(a+b)=cos acos b- (cos acos b)(tan a tan b)$$
Now substitute instead of $cos a=sqrt{frac{1}{1+tan^2 a}}$
And you'll get that $$cos(a+b)=sqrt{frac{1}{(1+tan^2 a)(1+tan^2 b)}} (1-tan a tan b)$$
It is a bit complicated as you see, you can try and do some simplifications, and there might be another easier method to do that, but thats what came into my mind now.
answered Nov 27 at 5:06
Fareed AF
41711
edited Nov 27 at 5:11
answered Nov 27 at 5:06
Fareed AF
41711
answered Nov 27 at 5:06
Fareed AF
41711
answered Nov 27 at 5:06
Fareed AF
41711
41711
add a comment |
add a comment |
up vote
0
down vote
Strictly speaking, no. As $tan(x)=tan(x+kpi)$ for every integer $k$, you can only determine $cos(a+b)$ and $cos(a-b)$ up to a sign.
answered Nov 27 at 5:51
William McGonagall
1337
add a comment |
up vote
0
down vote
Strictly speaking, no. As $tan(x)=tan(x+kpi)$ for every integer $k$, you can only determine $cos(a+b)$ and $cos(a-b)$ up to a sign.
answered Nov 27 at 5:51
William McGonagall
1337
add a comment |
up vote
0
down vote
up vote
0
down vote
Strictly speaking, no. As $tan(x)=tan(x+kpi)$ for every integer $k$, you can only determine $cos(a+b)$ and $cos(a-b)$ up to a sign.
answered Nov 27 at 5:51
William McGonagall
1337
Strictly speaking, no. As $tan(x)=tan(x+kpi)$ for every integer $k$, you can only determine $cos(a+b)$ and $cos(a-b)$ up to a sign.
answered Nov 27 at 5:51
William McGonagall
1337
answered Nov 27 at 5:51
William McGonagall
1337
answered Nov 27 at 5:51
William McGonagall
1337
answered Nov 27 at 5:51
William McGonagall
1337
1337
add a comment |
add a comment |
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