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$tan(a) = 3/4$ and $tan (b) = 5/12$, what is $cos(a+b)$
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It is known that
$$tan(a) = frac34, ::: tan(b) = frac512 $$
with $a,b < fracpi2$.
What is $cos(a+b)$?
Attempt :
$$ cos(a+b) = cos(a) cos(b) - sin(a) sin(b) $$
And we can write $tan(a) = sin(a)/cos(a) = 0.3/0.4 $
and $ sin(b)/cos(b) = 0.05/0.12 $
so
$$ cos(a+b) = (0.4)(0.12) - (0.3) (0.05) = 33/1000$$
Is this correct? Thanks.
trigonometry triangle angle
asked Jul 17 at 1:00
Arief
1,3311522
add a comment |Â
up vote
5
down vote
favorite
It is known that
$$tan(a) = frac34, ::: tan(b) = frac512 $$
with $a,b < fracpi2$.
What is $cos(a+b)$?
Attempt :
$$ cos(a+b) = cos(a) cos(b) - sin(a) sin(b) $$
And we can write $tan(a) = sin(a)/cos(a) = 0.3/0.4 $
and $ sin(b)/cos(b) = 0.05/0.12 $
so
$$ cos(a+b) = (0.4)(0.12) - (0.3) (0.05) = 33/1000$$
Is this correct? Thanks.
trigonometry triangle angle
asked Jul 17 at 1:00
Arief
1,3311522
4
You cannot do that: $$fracsin(a)cos(a)=frac0.30.4Rightarrow sin(a)=0.3, cos(a)=0.4 $$. In fact, $sin^2(a)+cos^2(a)=0.09+0.16neq 1$
– Mateus Rocha
Jul 17 at 1:03
This is not since $sin^2(a)+cos^2(a)=1$
– abc...
Jul 17 at 1:05
1
You can solve for sin(a)(=x) and cos(a)(=y): $fracxy=frac34, x^2+y^2=1$
– abc...
Jul 17 at 1:05
You have $frac sin acos a = frac 34$ and $frac sin bcos b = frac 512$. Why would you think that means $cos bcos b = 4*12*.001$?
– fleablood
Jul 17 at 1:17
The 3 answers that have appeared so far have assumed that a,b are positive, but that is not in the Q. Did you mean to assume $a,b$ are positive? If not, the solution for $cos (a+b)$ is not unique.
– DanielWainfleet
Jul 17 at 1:53
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
It is known that
$$tan(a) = frac34, ::: tan(b) = frac512 $$
with $a,b < fracpi2$.
What is $cos(a+b)$?
Attempt :
$$ cos(a+b) = cos(a) cos(b) - sin(a) sin(b) $$
And we can write $tan(a) = sin(a)/cos(a) = 0.3/0.4 $
and $ sin(b)/cos(b) = 0.05/0.12 $
so
$$ cos(a+b) = (0.4)(0.12) - (0.3) (0.05) = 33/1000$$
Is this correct? Thanks.
trigonometry triangle angle
asked Jul 17 at 1:00
Arief
1,3311522
It is known that
$$tan(a) = frac34, ::: tan(b) = frac512 $$
with $a,b < fracpi2$.
What is $cos(a+b)$?
Attempt :
$$ cos(a+b) = cos(a) cos(b) - sin(a) sin(b) $$
And we can write $tan(a) = sin(a)/cos(a) = 0.3/0.4 $
and $ sin(b)/cos(b) = 0.05/0.12 $
so
$$ cos(a+b) = (0.4)(0.12) - (0.3) (0.05) = 33/1000$$
Is this correct? Thanks.
trigonometry triangle angle
asked Jul 17 at 1:00
Arief
1,3311522
asked Jul 17 at 1:00
Arief
1,3311522
asked Jul 17 at 1:00
Arief
1,3311522
asked Jul 17 at 1:00
Arief
1,3311522
1,3311522
4
You cannot do that: $$fracsin(a)cos(a)=frac0.30.4Rightarrow sin(a)=0.3, cos(a)=0.4 $$. In fact, $sin^2(a)+cos^2(a)=0.09+0.16neq 1$
– Mateus Rocha
Jul 17 at 1:03
This is not since $sin^2(a)+cos^2(a)=1$
– abc...
Jul 17 at 1:05
1
You can solve for sin(a)(=x) and cos(a)(=y): $fracxy=frac34, x^2+y^2=1$
– abc...
Jul 17 at 1:05
You have $frac sin acos a = frac 34$ and $frac sin bcos b = frac 512$. Why would you think that means $cos bcos b = 4*12*.001$?
– fleablood
Jul 17 at 1:17
The 3 answers that have appeared so far have assumed that a,b are positive, but that is not in the Q. Did you mean to assume $a,b$ are positive? If not, the solution for $cos (a+b)$ is not unique.
– DanielWainfleet
Jul 17 at 1:53
add a comment |Â
4
You cannot do that: $$fracsin(a)cos(a)=frac0.30.4Rightarrow sin(a)=0.3, cos(a)=0.4 $$. In fact, $sin^2(a)+cos^2(a)=0.09+0.16neq 1$
– Mateus Rocha
Jul 17 at 1:03
This is not since $sin^2(a)+cos^2(a)=1$
– abc...
Jul 17 at 1:05
1
You can solve for sin(a)(=x) and cos(a)(=y): $fracxy=frac34, x^2+y^2=1$
– abc...
Jul 17 at 1:05
You have $frac sin acos a = frac 34$ and $frac sin bcos b = frac 512$. Why would you think that means $cos bcos b = 4*12*.001$?
– fleablood
Jul 17 at 1:17
The 3 answers that have appeared so far have assumed that a,b are positive, but that is not in the Q. Did you mean to assume $a,b$ are positive? If not, the solution for $cos (a+b)$ is not unique.
– DanielWainfleet
Jul 17 at 1:53
4
4
You cannot do that: $$fracsin(a)cos(a)=frac0.30.4Rightarrow sin(a)=0.3, cos(a)=0.4 $$. In fact, $sin^2(a)+cos^2(a)=0.09+0.16neq 1$
– Mateus Rocha
Jul 17 at 1:03
You cannot do that: $$fracsin(a)cos(a)=frac0.30.4Rightarrow sin(a)=0.3, cos(a)=0.4 $$. In fact, $sin^2(a)+cos^2(a)=0.09+0.16neq 1$
– Mateus Rocha
Jul 17 at 1:03
This is not since $sin^2(a)+cos^2(a)=1$
– abc...
Jul 17 at 1:05
This is not since $sin^2(a)+cos^2(a)=1$
– abc...
Jul 17 at 1:05
1
1
You can solve for sin(a)(=x) and cos(a)(=y): $fracxy=frac34, x^2+y^2=1$
– abc...
Jul 17 at 1:05
You can solve for sin(a)(=x) and cos(a)(=y): $fracxy=frac34, x^2+y^2=1$
– abc...
Jul 17 at 1:05
You have $frac sin acos a = frac 34$ and $frac sin bcos b = frac 512$. Why would you think that means $cos bcos b = 4*12*.001$?
– fleablood
Jul 17 at 1:17
You have $frac sin acos a = frac 34$ and $frac sin bcos b = frac 512$. Why would you think that means $cos bcos b = 4*12*.001$?
– fleablood
Jul 17 at 1:17
The 3 answers that have appeared so far have assumed that a,b are positive, but that is not in the Q. Did you mean to assume $a,b$ are positive? If not, the solution for $cos (a+b)$ is not unique.
– DanielWainfleet
Jul 17 at 1:53
The 3 answers that have appeared so far have assumed that a,b are positive, but that is not in the Q. Did you mean to assume $a,b$ are positive? If not, the solution for $cos (a+b)$ is not unique.
– DanielWainfleet
Jul 17 at 1:53
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
No, this can't be correct. Remember that $sin^2x+cos^2x=1$ for all $x$; your values for the sine and cosine of $a$ and $b$ do not satisfy this relation.
It turns out that
$$sin a=frac35qquadcos a=frac45$$
$$sin b=frac513qquadcos b=frac1213$$
and thus
$$cos(a+b)=cos acos b-sin asin b=frac45cdotfrac1213-frac35cdotfrac513=frac3365$$
answered Jul 17 at 1:04
Parcly Taxel
33.6k136588
add a comment |Â
up vote
3
down vote
$$ tan(a+b)= frac tan a +tan b1-tan a tan b = frac 3/4 +5/121-(3/4)(5/12)= frac 5633$$
$$ sec^2(a+b)=1+tan^2(a+b)=frac 42251089=(frac 6533)^2$$
$$cos(a+b)= 33/65$$
answered Jul 17 at 1:34
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
1
down vote
Solve for sin(a), cos(a), sin(b) and cos(b) (like I said in comments):
$sin(a)=frac35$
$cos(a)=frac45$
$sin(b)=frac513$
$cos(b)=frac1213$
Thus $$cos(a+b)=frac45timesfrac1213-frac35timesfrac513=frac3365$$
answered Jul 17 at 1:11
abc...
1,190524
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
No, this can't be correct. Remember that $sin^2x+cos^2x=1$ for all $x$; your values for the sine and cosine of $a$ and $b$ do not satisfy this relation.
It turns out that
$$sin a=frac35qquadcos a=frac45$$
$$sin b=frac513qquadcos b=frac1213$$
and thus
$$cos(a+b)=cos acos b-sin asin b=frac45cdotfrac1213-frac35cdotfrac513=frac3365$$
answered Jul 17 at 1:04
Parcly Taxel
33.6k136588
add a comment |Â
up vote
4
down vote
accepted
No, this can't be correct. Remember that $sin^2x+cos^2x=1$ for all $x$; your values for the sine and cosine of $a$ and $b$ do not satisfy this relation.
It turns out that
$$sin a=frac35qquadcos a=frac45$$
$$sin b=frac513qquadcos b=frac1213$$
and thus
$$cos(a+b)=cos acos b-sin asin b=frac45cdotfrac1213-frac35cdotfrac513=frac3365$$
answered Jul 17 at 1:04
Parcly Taxel
33.6k136588
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
No, this can't be correct. Remember that $sin^2x+cos^2x=1$ for all $x$; your values for the sine and cosine of $a$ and $b$ do not satisfy this relation.
It turns out that
$$sin a=frac35qquadcos a=frac45$$
$$sin b=frac513qquadcos b=frac1213$$
and thus
$$cos(a+b)=cos acos b-sin asin b=frac45cdotfrac1213-frac35cdotfrac513=frac3365$$
answered Jul 17 at 1:04
Parcly Taxel
33.6k136588
No, this can't be correct. Remember that $sin^2x+cos^2x=1$ for all $x$; your values for the sine and cosine of $a$ and $b$ do not satisfy this relation.
It turns out that
$$sin a=frac35qquadcos a=frac45$$
$$sin b=frac513qquadcos b=frac1213$$
and thus
$$cos(a+b)=cos acos b-sin asin b=frac45cdotfrac1213-frac35cdotfrac513=frac3365$$
answered Jul 17 at 1:04
Parcly Taxel
33.6k136588
answered Jul 17 at 1:04
Parcly Taxel
33.6k136588
answered Jul 17 at 1:04
Parcly Taxel
33.6k136588
answered Jul 17 at 1:04
Parcly Taxel
33.6k136588
33.6k136588
add a comment |Â
add a comment |Â
up vote
3
down vote
$$ tan(a+b)= frac tan a +tan b1-tan a tan b = frac 3/4 +5/121-(3/4)(5/12)= frac 5633$$
$$ sec^2(a+b)=1+tan^2(a+b)=frac 42251089=(frac 6533)^2$$
$$cos(a+b)= 33/65$$
answered Jul 17 at 1:34
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
3
down vote
$$ tan(a+b)= frac tan a +tan b1-tan a tan b = frac 3/4 +5/121-(3/4)(5/12)= frac 5633$$
$$ sec^2(a+b)=1+tan^2(a+b)=frac 42251089=(frac 6533)^2$$
$$cos(a+b)= 33/65$$
answered Jul 17 at 1:34
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$$ tan(a+b)= frac tan a +tan b1-tan a tan b = frac 3/4 +5/121-(3/4)(5/12)= frac 5633$$
$$ sec^2(a+b)=1+tan^2(a+b)=frac 42251089=(frac 6533)^2$$
$$cos(a+b)= 33/65$$
answered Jul 17 at 1:34
Mohammad Riazi-Kermani
27.5k41852
$$ tan(a+b)= frac tan a +tan b1-tan a tan b = frac 3/4 +5/121-(3/4)(5/12)= frac 5633$$
$$ sec^2(a+b)=1+tan^2(a+b)=frac 42251089=(frac 6533)^2$$
$$cos(a+b)= 33/65$$
answered Jul 17 at 1:34
Mohammad Riazi-Kermani
27.5k41852
edited Jul 17 at 3:42
answered Jul 17 at 1:34
Mohammad Riazi-Kermani
27.5k41852
answered Jul 17 at 1:34
Mohammad Riazi-Kermani
27.5k41852
answered Jul 17 at 1:34
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
add a comment |Â
add a comment |Â
up vote
1
down vote
Solve for sin(a), cos(a), sin(b) and cos(b) (like I said in comments):
$sin(a)=frac35$
$cos(a)=frac45$
$sin(b)=frac513$
$cos(b)=frac1213$
Thus $$cos(a+b)=frac45timesfrac1213-frac35timesfrac513=frac3365$$
answered Jul 17 at 1:11
abc...
1,190524
add a comment |Â
up vote
1
down vote
Solve for sin(a), cos(a), sin(b) and cos(b) (like I said in comments):
$sin(a)=frac35$
$cos(a)=frac45$
$sin(b)=frac513$
$cos(b)=frac1213$
Thus $$cos(a+b)=frac45timesfrac1213-frac35timesfrac513=frac3365$$
answered Jul 17 at 1:11
abc...
1,190524
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Solve for sin(a), cos(a), sin(b) and cos(b) (like I said in comments):
$sin(a)=frac35$
$cos(a)=frac45$
$sin(b)=frac513$
$cos(b)=frac1213$
Thus $$cos(a+b)=frac45timesfrac1213-frac35timesfrac513=frac3365$$
answered Jul 17 at 1:11
abc...
1,190524
Solve for sin(a), cos(a), sin(b) and cos(b) (like I said in comments):
$sin(a)=frac35$
$cos(a)=frac45$
$sin(b)=frac513$
$cos(b)=frac1213$
Thus $$cos(a+b)=frac45timesfrac1213-frac35timesfrac513=frac3365$$
answered Jul 17 at 1:11
abc...
1,190524
answered Jul 17 at 1:11
abc...
1,190524
answered Jul 17 at 1:11
abc...
1,190524
answered Jul 17 at 1:11
abc...
1,190524
1,190524
add a comment |Â
add a comment |Â
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4
You cannot do that: $$fracsin(a)cos(a)=frac0.30.4Rightarrow sin(a)=0.3, cos(a)=0.4 $$. In fact, $sin^2(a)+cos^2(a)=0.09+0.16neq 1$
– Mateus Rocha
Jul 17 at 1:03
This is not since $sin^2(a)+cos^2(a)=1$
– abc...
Jul 17 at 1:05
1
You can solve for sin(a)(=x) and cos(a)(=y): $fracxy=frac34, x^2+y^2=1$
– abc...
Jul 17 at 1:05
You have $frac sin acos a = frac 34$ and $frac sin bcos b = frac 512$. Why would you think that means $cos bcos b = 4*12*.001$?
– fleablood
Jul 17 at 1:17
The 3 answers that have appeared so far have assumed that a,b are positive, but that is not in the Q. Did you mean to assume $a,b$ are positive? If not, the solution for $cos (a+b)$ is not unique.
– DanielWainfleet
Jul 17 at 1:53