knowing: $ tan x=2-sqrt3$ , obtain: $ cos 2x$
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Knowing: $$ tan x=2-sqrt3 $$
Obtain: $$cos2x$$
I tried converting $tan x$ into it's sinus and cosine form and trying to square both sides to try to get the form of:
$$cos^2x-sin^2x$$
But I can't really get to this form without having extra expressions of sine or cosine, any ideas how to start this properly?
Taken out of one of the entry tests to Maths in TAU.
Solution:
$$cos2x=cos^2x-sin^2x=fraccos^2x-sin^2xsin^2x+cos^2x:fraccos^2xcos^2x=frac1-tan^2x1+tan^2x$$
$$ frac1-tan^2x1+tan^2x=frac-6+4sqrt38-4sqrt3=frac-3+2sqrt34-2sqrt3 $$
calculus algebra-precalculus trigonometry
edited Sep 9 at 13:24
GoodDeeds
10.2k21335
asked Sep 9 at 12:47
Maxim
696314
add a comment |Â
up vote
2
down vote
favorite
Knowing: $$ tan x=2-sqrt3 $$
Obtain: $$cos2x$$
I tried converting $tan x$ into it's sinus and cosine form and trying to square both sides to try to get the form of:
$$cos^2x-sin^2x$$
But I can't really get to this form without having extra expressions of sine or cosine, any ideas how to start this properly?
Taken out of one of the entry tests to Maths in TAU.
Solution:
$$cos2x=cos^2x-sin^2x=fraccos^2x-sin^2xsin^2x+cos^2x:fraccos^2xcos^2x=frac1-tan^2x1+tan^2x$$
$$ frac1-tan^2x1+tan^2x=frac-6+4sqrt38-4sqrt3=frac-3+2sqrt34-2sqrt3 $$
calculus algebra-precalculus trigonometry
edited Sep 9 at 13:24
GoodDeeds
10.2k21335
asked Sep 9 at 12:47
Maxim
696314
Hint form a right triangle assuming the adjacent side to be 1
– Karl
Sep 9 at 12:50
You sound as if you have done all you need if you have both $sin x$ and $cos x$. Why not edit your question with that working and the chances are you have just made a silly error that someone will spot for you
– Bruce
Sep 9 at 12:51
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Knowing: $$ tan x=2-sqrt3 $$
Obtain: $$cos2x$$
I tried converting $tan x$ into it's sinus and cosine form and trying to square both sides to try to get the form of:
$$cos^2x-sin^2x$$
But I can't really get to this form without having extra expressions of sine or cosine, any ideas how to start this properly?
Taken out of one of the entry tests to Maths in TAU.
Solution:
$$cos2x=cos^2x-sin^2x=fraccos^2x-sin^2xsin^2x+cos^2x:fraccos^2xcos^2x=frac1-tan^2x1+tan^2x$$
$$ frac1-tan^2x1+tan^2x=frac-6+4sqrt38-4sqrt3=frac-3+2sqrt34-2sqrt3 $$
calculus algebra-precalculus trigonometry
edited Sep 9 at 13:24
GoodDeeds
10.2k21335
asked Sep 9 at 12:47
Maxim
696314
Knowing: $$ tan x=2-sqrt3 $$
Obtain: $$cos2x$$
I tried converting $tan x$ into it's sinus and cosine form and trying to square both sides to try to get the form of:
$$cos^2x-sin^2x$$
But I can't really get to this form without having extra expressions of sine or cosine, any ideas how to start this properly?
Taken out of one of the entry tests to Maths in TAU.
Solution:
$$cos2x=cos^2x-sin^2x=fraccos^2x-sin^2xsin^2x+cos^2x:fraccos^2xcos^2x=frac1-tan^2x1+tan^2x$$
$$ frac1-tan^2x1+tan^2x=frac-6+4sqrt38-4sqrt3=frac-3+2sqrt34-2sqrt3 $$
calculus algebra-precalculus trigonometry
calculus algebra-precalculus trigonometry
edited Sep 9 at 13:24
GoodDeeds
10.2k21335
asked Sep 9 at 12:47
Maxim
696314
edited Sep 9 at 13:24
GoodDeeds
10.2k21335
asked Sep 9 at 12:47
Maxim
696314
edited Sep 9 at 13:24
GoodDeeds
10.2k21335
edited Sep 9 at 13:24
GoodDeeds
10.2k21335
edited Sep 9 at 13:24
GoodDeeds
10.2k21335
10.2k21335
asked Sep 9 at 12:47
Maxim
696314
asked Sep 9 at 12:47
Maxim
696314
asked Sep 9 at 12:47
Maxim
696314
696314
Hint form a right triangle assuming the adjacent side to be 1
– Karl
Sep 9 at 12:50
You sound as if you have done all you need if you have both $sin x$ and $cos x$. Why not edit your question with that working and the chances are you have just made a silly error that someone will spot for you
– Bruce
Sep 9 at 12:51
add a comment |Â
Hint form a right triangle assuming the adjacent side to be 1
– Karl
Sep 9 at 12:50
You sound as if you have done all you need if you have both $sin x$ and $cos x$. Why not edit your question with that working and the chances are you have just made a silly error that someone will spot for you
– Bruce
Sep 9 at 12:51
Hint form a right triangle assuming the adjacent side to be 1
– Karl
Sep 9 at 12:50
Hint form a right triangle assuming the adjacent side to be 1
– Karl
Sep 9 at 12:50
You sound as if you have done all you need if you have both $sin x$ and $cos x$. Why not edit your question with that working and the chances are you have just made a silly error that someone will spot for you
– Bruce
Sep 9 at 12:51
You sound as if you have done all you need if you have both $sin x$ and $cos x$. Why not edit your question with that working and the chances are you have just made a silly error that someone will spot for you
– Bruce
Sep 9 at 12:51
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
HINT
Recall that by half-angle identities
$$cos 2x = frac1-tan^2 x1+tan^2 x$$
answered Sep 9 at 12:50
gimusi
74.3k73889
Your ship travels directly, but mine has double routes. :-))
– mrs
Sep 9 at 12:53
@mrs Yes you gave the way to obtain that! It's nice :)
– gimusi
Sep 9 at 12:53
add a comment |Â
up vote
5
down vote
One may use the following identities:
$cos^2(x)=frac11+tan^2(x)$
$cos(2x)=2cos^2(x)-1$
answered Sep 9 at 12:49
mrs
58.5k750143
add a comment |Â
up vote
3
down vote
HINT:
$$tan^2x=fracsin^2xcos^2x=(2-sqrt3)^2=13-4sqrt3$$ and $$cos2x=cos^2x-sin^2x=cos^2x(1-tan^2x)$$ and use this with the identity $1+tan^2x=sec^2x$
answered Sep 9 at 12:50
TheSimpliFire
11k62255
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
HINT
Recall that by half-angle identities
$$cos 2x = frac1-tan^2 x1+tan^2 x$$
answered Sep 9 at 12:50
gimusi
74.3k73889
Your ship travels directly, but mine has double routes. :-))
– mrs
Sep 9 at 12:53
@mrs Yes you gave the way to obtain that! It's nice :)
– gimusi
Sep 9 at 12:53
add a comment |Â
up vote
4
down vote
accepted
HINT
Recall that by half-angle identities
$$cos 2x = frac1-tan^2 x1+tan^2 x$$
answered Sep 9 at 12:50
gimusi
74.3k73889
Your ship travels directly, but mine has double routes. :-))
– mrs
Sep 9 at 12:53
@mrs Yes you gave the way to obtain that! It's nice :)
– gimusi
Sep 9 at 12:53
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
HINT
Recall that by half-angle identities
$$cos 2x = frac1-tan^2 x1+tan^2 x$$
answered Sep 9 at 12:50
gimusi
74.3k73889
HINT
Recall that by half-angle identities
$$cos 2x = frac1-tan^2 x1+tan^2 x$$
answered Sep 9 at 12:50
gimusi
74.3k73889
answered Sep 9 at 12:50
gimusi
74.3k73889
answered Sep 9 at 12:50
gimusi
74.3k73889
answered Sep 9 at 12:50
gimusi
74.3k73889
74.3k73889
Your ship travels directly, but mine has double routes. :-))
– mrs
Sep 9 at 12:53
@mrs Yes you gave the way to obtain that! It's nice :)
– gimusi
Sep 9 at 12:53
add a comment |Â
Your ship travels directly, but mine has double routes. :-))
– mrs
Sep 9 at 12:53
@mrs Yes you gave the way to obtain that! It's nice :)
– gimusi
Sep 9 at 12:53
Your ship travels directly, but mine has double routes. :-))
– mrs
Sep 9 at 12:53
Your ship travels directly, but mine has double routes. :-))
– mrs
Sep 9 at 12:53
@mrs Yes you gave the way to obtain that! It's nice :)
– gimusi
Sep 9 at 12:53
@mrs Yes you gave the way to obtain that! It's nice :)
– gimusi
Sep 9 at 12:53
add a comment |Â
up vote
5
down vote
One may use the following identities:
$cos^2(x)=frac11+tan^2(x)$
$cos(2x)=2cos^2(x)-1$
answered Sep 9 at 12:49
mrs
58.5k750143
add a comment |Â
up vote
5
down vote
One may use the following identities:
$cos^2(x)=frac11+tan^2(x)$
$cos(2x)=2cos^2(x)-1$
answered Sep 9 at 12:49
mrs
58.5k750143
add a comment |Â
up vote
5
down vote
up vote
5
down vote
One may use the following identities:
$cos^2(x)=frac11+tan^2(x)$
$cos(2x)=2cos^2(x)-1$
answered Sep 9 at 12:49
mrs
58.5k750143
One may use the following identities:
$cos^2(x)=frac11+tan^2(x)$
$cos(2x)=2cos^2(x)-1$
answered Sep 9 at 12:49
mrs
58.5k750143
answered Sep 9 at 12:49
mrs
58.5k750143
answered Sep 9 at 12:49
mrs
58.5k750143
answered Sep 9 at 12:49
mrs
58.5k750143
58.5k750143
add a comment |Â
add a comment |Â
up vote
3
down vote
HINT:
$$tan^2x=fracsin^2xcos^2x=(2-sqrt3)^2=13-4sqrt3$$ and $$cos2x=cos^2x-sin^2x=cos^2x(1-tan^2x)$$ and use this with the identity $1+tan^2x=sec^2x$
answered Sep 9 at 12:50
TheSimpliFire
11k62255
add a comment |Â
up vote
3
down vote
HINT:
$$tan^2x=fracsin^2xcos^2x=(2-sqrt3)^2=13-4sqrt3$$ and $$cos2x=cos^2x-sin^2x=cos^2x(1-tan^2x)$$ and use this with the identity $1+tan^2x=sec^2x$
answered Sep 9 at 12:50
TheSimpliFire
11k62255
add a comment |Â
up vote
3
down vote
up vote
3
down vote
HINT:
$$tan^2x=fracsin^2xcos^2x=(2-sqrt3)^2=13-4sqrt3$$ and $$cos2x=cos^2x-sin^2x=cos^2x(1-tan^2x)$$ and use this with the identity $1+tan^2x=sec^2x$
answered Sep 9 at 12:50
TheSimpliFire
11k62255
HINT:
$$tan^2x=fracsin^2xcos^2x=(2-sqrt3)^2=13-4sqrt3$$ and $$cos2x=cos^2x-sin^2x=cos^2x(1-tan^2x)$$ and use this with the identity $1+tan^2x=sec^2x$
answered Sep 9 at 12:50
TheSimpliFire
11k62255
answered Sep 9 at 12:50
TheSimpliFire
11k62255
answered Sep 9 at 12:50
TheSimpliFire
11k62255
answered Sep 9 at 12:50
TheSimpliFire
11k62255
11k62255
add a comment |Â
add a comment |Â
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Hint form a right triangle assuming the adjacent side to be 1
– Karl
Sep 9 at 12:50
You sound as if you have done all you need if you have both $sin x$ and $cos x$. Why not edit your question with that working and the chances are you have just made a silly error that someone will spot for you
– Bruce
Sep 9 at 12:51