Is there a way to find $cos(a+b)$ and $cos(a-b)$ in terms of $tan a$ and $tan b$?
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
47.6k870151
asked Nov 27 at 4:50
Asghar Razzaqi
31
add a comment |
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
47.6k870151
asked Nov 27 at 4:50
Asghar Razzaqi
31
add a comment |
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
47.6k870151
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
47.6k870151
asked Nov 27 at 4:50
Asghar Razzaqi
31
edited Nov 27 at 4:53
Blue
47.6k870151
asked Nov 27 at 4:50
Asghar Razzaqi
31
edited Nov 27 at 4:53
Blue
47.6k870151
edited Nov 27 at 4:53
Blue
47.6k870151
edited Nov 27 at 4:53
Blue
47.6k870151
47.6k870151
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
2
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oldest
votes
$cos(a+b)=cos acos b- sin a sin b$
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
44211
add a comment |
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
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2 Answers
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$cos(a+b)=cos acos b- sin a sin b$
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
44211
add a comment |
$cos(a+b)=cos acos b- sin a sin b$
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
44211
add a comment |
$cos(a+b)=cos acos b- sin a sin b$
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
44211
$cos(a+b)=cos acos b- sin a sin b$
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
44211
edited Nov 27 at 5:11
answered Nov 27 at 5:06
Fareed AF
44211
answered Nov 27 at 5:06
Fareed AF
44211
answered Nov 27 at 5:06
Fareed AF
44211
44211
add a comment |
add a comment |
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 |
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 |
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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