Prove the identity $csc2theta=frac{secthetacsctheta}{2}$
$begingroup$
Prove the identity $$csc2theta=frac{secthetacsctheta}{2}$$
I've started by using a double angle identity, but I'm not sure how to continue from here or if this is right approach.
$$csc2theta=frac{1}{sin2theta}=frac{1}{2sinthetacostheta}$$
trigonometry
asked Dec 4 '18 at 19:17
H.LinkhornH.Linkhorn
458113
$endgroup$
add a comment |
$begingroup$
Prove the identity $$csc2theta=frac{secthetacsctheta}{2}$$
I've started by using a double angle identity, but I'm not sure how to continue from here or if this is right approach.
$$csc2theta=frac{1}{sin2theta}=frac{1}{2sinthetacostheta}$$
trigonometry
asked Dec 4 '18 at 19:17
H.LinkhornH.Linkhorn
458113
$endgroup$
$begingroup$
Your left-hand side should be $csc (2theta)$
$endgroup$
– paw88789
Dec 4 '18 at 19:20
add a comment |
$begingroup$
Prove the identity $$csc2theta=frac{secthetacsctheta}{2}$$
I've started by using a double angle identity, but I'm not sure how to continue from here or if this is right approach.
$$csc2theta=frac{1}{sin2theta}=frac{1}{2sinthetacostheta}$$
trigonometry
asked Dec 4 '18 at 19:17
H.LinkhornH.Linkhorn
458113
$endgroup$
Prove the identity $$csc2theta=frac{secthetacsctheta}{2}$$
I've started by using a double angle identity, but I'm not sure how to continue from here or if this is right approach.
$$csc2theta=frac{1}{sin2theta}=frac{1}{2sinthetacostheta}$$
trigonometry
trigonometry
asked Dec 4 '18 at 19:17
H.LinkhornH.Linkhorn
458113
asked Dec 4 '18 at 19:17
H.LinkhornH.Linkhorn
458113
edited Dec 4 '18 at 19:20
H.Linkhorn
asked Dec 4 '18 at 19:17
H.LinkhornH.Linkhorn
458113
asked Dec 4 '18 at 19:17
H.LinkhornH.Linkhorn
458113
asked Dec 4 '18 at 19:17
H.LinkhornH.Linkhorn
458113
458113
$begingroup$
Your left-hand side should be $csc (2theta)$
$endgroup$
– paw88789
Dec 4 '18 at 19:20
add a comment |
$begingroup$
Your left-hand side should be $csc (2theta)$
$endgroup$
– paw88789
Dec 4 '18 at 19:20
$begingroup$
Your left-hand side should be $csc (2theta)$
$endgroup$
– paw88789
Dec 4 '18 at 19:20
$begingroup$
Your left-hand side should be $csc (2theta)$
$endgroup$
– paw88789
Dec 4 '18 at 19:20
add a comment |
3 Answers
3
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oldest
votes
$begingroup$
What you did is correct. Now, you do:$$frac1{2sinthetacostheta}=frac{frac1{costheta}timesfrac1{sintheta}}2=frac{secthetacsctheta}2.$$
answered Dec 4 '18 at 19:20
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
add a comment |
$begingroup$
Your approach is right, you just need to do the last step to get the desired result: What are $frac{1}{sin(theta)}$ and $frac{1}{cos(theta)}$?
answered Dec 4 '18 at 19:19
BotondBotond
5,9002832
$endgroup$
add a comment |
$begingroup$
The thing you did is OK. The question which occurs is do you know how to prove $$sin (2x) = 2sin x cos x$$
(Hint: use adition theorem for function $sin $)
answered Dec 4 '18 at 19:21
greedoidgreedoid
44.8k1156111
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What you did is correct. Now, you do:$$frac1{2sinthetacostheta}=frac{frac1{costheta}timesfrac1{sintheta}}2=frac{secthetacsctheta}2.$$
answered Dec 4 '18 at 19:20
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
add a comment |
$begingroup$
What you did is correct. Now, you do:$$frac1{2sinthetacostheta}=frac{frac1{costheta}timesfrac1{sintheta}}2=frac{secthetacsctheta}2.$$
answered Dec 4 '18 at 19:20
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
add a comment |
$begingroup$
What you did is correct. Now, you do:$$frac1{2sinthetacostheta}=frac{frac1{costheta}timesfrac1{sintheta}}2=frac{secthetacsctheta}2.$$
answered Dec 4 '18 at 19:20
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
What you did is correct. Now, you do:$$frac1{2sinthetacostheta}=frac{frac1{costheta}timesfrac1{sintheta}}2=frac{secthetacsctheta}2.$$
answered Dec 4 '18 at 19:20
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 4 '18 at 19:20
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 4 '18 at 19:20
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 4 '18 at 19:20
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
add a comment |
add a comment |
$begingroup$
Your approach is right, you just need to do the last step to get the desired result: What are $frac{1}{sin(theta)}$ and $frac{1}{cos(theta)}$?
answered Dec 4 '18 at 19:19
BotondBotond
5,9002832
$endgroup$
add a comment |
$begingroup$
Your approach is right, you just need to do the last step to get the desired result: What are $frac{1}{sin(theta)}$ and $frac{1}{cos(theta)}$?
answered Dec 4 '18 at 19:19
BotondBotond
5,9002832
$endgroup$
add a comment |
$begingroup$
Your approach is right, you just need to do the last step to get the desired result: What are $frac{1}{sin(theta)}$ and $frac{1}{cos(theta)}$?
answered Dec 4 '18 at 19:19
BotondBotond
5,9002832
$endgroup$
Your approach is right, you just need to do the last step to get the desired result: What are $frac{1}{sin(theta)}$ and $frac{1}{cos(theta)}$?
answered Dec 4 '18 at 19:19
BotondBotond
5,9002832
answered Dec 4 '18 at 19:19
BotondBotond
5,9002832
answered Dec 4 '18 at 19:19
BotondBotond
5,9002832
answered Dec 4 '18 at 19:19
BotondBotond
5,9002832
5,9002832
add a comment |
add a comment |
$begingroup$
The thing you did is OK. The question which occurs is do you know how to prove $$sin (2x) = 2sin x cos x$$
(Hint: use adition theorem for function $sin $)
answered Dec 4 '18 at 19:21
greedoidgreedoid
44.8k1156111
$endgroup$
add a comment |
$begingroup$
The thing you did is OK. The question which occurs is do you know how to prove $$sin (2x) = 2sin x cos x$$
(Hint: use adition theorem for function $sin $)
answered Dec 4 '18 at 19:21
greedoidgreedoid
44.8k1156111
$endgroup$
add a comment |
$begingroup$
The thing you did is OK. The question which occurs is do you know how to prove $$sin (2x) = 2sin x cos x$$
(Hint: use adition theorem for function $sin $)
answered Dec 4 '18 at 19:21
greedoidgreedoid
44.8k1156111
$endgroup$
The thing you did is OK. The question which occurs is do you know how to prove $$sin (2x) = 2sin x cos x$$
(Hint: use adition theorem for function $sin $)
answered Dec 4 '18 at 19:21
greedoidgreedoid
44.8k1156111
answered Dec 4 '18 at 19:21
greedoidgreedoid
44.8k1156111
answered Dec 4 '18 at 19:21
greedoidgreedoid
44.8k1156111
answered Dec 4 '18 at 19:21
greedoidgreedoid
44.8k1156111
44.8k1156111
add a comment |
add a comment |
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$begingroup$
Your left-hand side should be $csc (2theta)$
$endgroup$
– paw88789
Dec 4 '18 at 19:20