Given $sin(x)$, find $sin(frac{x}{2}) cos(frac{5x}{2})$ [closed]
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Knowing that
$pi < 2x < 2pi$
and
$$sin(x) = frac{4}{5},$$
find
$$sinleft(frac{x}{2}right) cosleft(frac{5x}{2}right) = ?$$
trigonometry transformation
closed as off-topic by Nosrati, Davide Giraudo, amWhy, Leucippus, Cesareo Nov 23 at 1:10
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up vote
-1
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Knowing that
$pi < 2x < 2pi$
and
$$sin(x) = frac{4}{5},$$
find
$$sinleft(frac{x}{2}right) cosleft(frac{5x}{2}right) = ?$$
trigonometry transformation
closed as off-topic by Nosrati, Davide Giraudo, amWhy, Leucippus, Cesareo Nov 23 at 1:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Davide Giraudo, amWhy, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
mathworld.wolfram.com/WernerFormulas.html
– lab bhattacharjee
Nov 22 at 14:17
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up vote
-1
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up vote
-1
down vote
favorite
Knowing that
$pi < 2x < 2pi$
and
$$sin(x) = frac{4}{5},$$
find
$$sinleft(frac{x}{2}right) cosleft(frac{5x}{2}right) = ?$$
trigonometry transformation
Knowing that
$pi < 2x < 2pi$
and
$$sin(x) = frac{4}{5},$$
find
$$sinleft(frac{x}{2}right) cosleft(frac{5x}{2}right) = ?$$
trigonometry transformation
trigonometry transformation
edited Nov 22 at 14:13
Tianlalu
3,0051938
3,0051938
asked Nov 22 at 14:07
critical_mass
81
81
closed as off-topic by Nosrati, Davide Giraudo, amWhy, Leucippus, Cesareo Nov 23 at 1:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Davide Giraudo, amWhy, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Nosrati, Davide Giraudo, amWhy, Leucippus, Cesareo Nov 23 at 1:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Nosrati, Davide Giraudo, amWhy, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
mathworld.wolfram.com/WernerFormulas.html
– lab bhattacharjee
Nov 22 at 14:17
add a comment |
1
mathworld.wolfram.com/WernerFormulas.html
– lab bhattacharjee
Nov 22 at 14:17
1
1
mathworld.wolfram.com/WernerFormulas.html
– lab bhattacharjee
Nov 22 at 14:17
mathworld.wolfram.com/WernerFormulas.html
– lab bhattacharjee
Nov 22 at 14:17
add a comment |
2 Answers
2
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HINT
Use that
$$sin theta cos varphi = frac12{{sin(theta + varphi) + frac12 sin(theta - varphi)} }$$
and
$sin (2theta) =2sintheta costheta$
$sin (3theta) =3sintheta - 4sin^3theta$
moreover from the given
- $cos x=-sqrt{1-sin^2 x}$
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$sin(x/2)cos(5x/2)\
=sin(x/2)cos(2x+x/2)\=sin(x/2){cos(2x)cos(x/2)-sin(2x)sin(x/2)}\=cos(2x)cos(x/2)sin(x/2)-sin(2x)sin^2(x/2)\=1/2(1-2sin^2(x))sin(x)-sin(x)cos(x)(1-cos(x))$
As $pi/2<x<pi$, $sin(x)=4/5$ implies $cos(x)=-3/5$. So the final answer is $sin(x/2)cos(5x/2)=82/125.$
Welcome to to MathSE. Type$sin x$
,$cos x$
,$tan x$
,$csc x$
,$sec x$
, and$cot x$
to produce $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, and $cot x$, respectively.
– N. F. Taussig
Nov 22 at 14:34
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
HINT
Use that
$$sin theta cos varphi = frac12{{sin(theta + varphi) + frac12 sin(theta - varphi)} }$$
and
$sin (2theta) =2sintheta costheta$
$sin (3theta) =3sintheta - 4sin^3theta$
moreover from the given
- $cos x=-sqrt{1-sin^2 x}$
add a comment |
up vote
4
down vote
accepted
HINT
Use that
$$sin theta cos varphi = frac12{{sin(theta + varphi) + frac12 sin(theta - varphi)} }$$
and
$sin (2theta) =2sintheta costheta$
$sin (3theta) =3sintheta - 4sin^3theta$
moreover from the given
- $cos x=-sqrt{1-sin^2 x}$
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
HINT
Use that
$$sin theta cos varphi = frac12{{sin(theta + varphi) + frac12 sin(theta - varphi)} }$$
and
$sin (2theta) =2sintheta costheta$
$sin (3theta) =3sintheta - 4sin^3theta$
moreover from the given
- $cos x=-sqrt{1-sin^2 x}$
HINT
Use that
$$sin theta cos varphi = frac12{{sin(theta + varphi) + frac12 sin(theta - varphi)} }$$
and
$sin (2theta) =2sintheta costheta$
$sin (3theta) =3sintheta - 4sin^3theta$
moreover from the given
- $cos x=-sqrt{1-sin^2 x}$
edited Nov 22 at 14:13
answered Nov 22 at 14:10
gimusi
92.3k84495
92.3k84495
add a comment |
add a comment |
up vote
2
down vote
$sin(x/2)cos(5x/2)\
=sin(x/2)cos(2x+x/2)\=sin(x/2){cos(2x)cos(x/2)-sin(2x)sin(x/2)}\=cos(2x)cos(x/2)sin(x/2)-sin(2x)sin^2(x/2)\=1/2(1-2sin^2(x))sin(x)-sin(x)cos(x)(1-cos(x))$
As $pi/2<x<pi$, $sin(x)=4/5$ implies $cos(x)=-3/5$. So the final answer is $sin(x/2)cos(5x/2)=82/125.$
Welcome to to MathSE. Type$sin x$
,$cos x$
,$tan x$
,$csc x$
,$sec x$
, and$cot x$
to produce $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, and $cot x$, respectively.
– N. F. Taussig
Nov 22 at 14:34
add a comment |
up vote
2
down vote
$sin(x/2)cos(5x/2)\
=sin(x/2)cos(2x+x/2)\=sin(x/2){cos(2x)cos(x/2)-sin(2x)sin(x/2)}\=cos(2x)cos(x/2)sin(x/2)-sin(2x)sin^2(x/2)\=1/2(1-2sin^2(x))sin(x)-sin(x)cos(x)(1-cos(x))$
As $pi/2<x<pi$, $sin(x)=4/5$ implies $cos(x)=-3/5$. So the final answer is $sin(x/2)cos(5x/2)=82/125.$
Welcome to to MathSE. Type$sin x$
,$cos x$
,$tan x$
,$csc x$
,$sec x$
, and$cot x$
to produce $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, and $cot x$, respectively.
– N. F. Taussig
Nov 22 at 14:34
add a comment |
up vote
2
down vote
up vote
2
down vote
$sin(x/2)cos(5x/2)\
=sin(x/2)cos(2x+x/2)\=sin(x/2){cos(2x)cos(x/2)-sin(2x)sin(x/2)}\=cos(2x)cos(x/2)sin(x/2)-sin(2x)sin^2(x/2)\=1/2(1-2sin^2(x))sin(x)-sin(x)cos(x)(1-cos(x))$
As $pi/2<x<pi$, $sin(x)=4/5$ implies $cos(x)=-3/5$. So the final answer is $sin(x/2)cos(5x/2)=82/125.$
$sin(x/2)cos(5x/2)\
=sin(x/2)cos(2x+x/2)\=sin(x/2){cos(2x)cos(x/2)-sin(2x)sin(x/2)}\=cos(2x)cos(x/2)sin(x/2)-sin(2x)sin^2(x/2)\=1/2(1-2sin^2(x))sin(x)-sin(x)cos(x)(1-cos(x))$
As $pi/2<x<pi$, $sin(x)=4/5$ implies $cos(x)=-3/5$. So the final answer is $sin(x/2)cos(5x/2)=82/125.$
answered Nov 22 at 14:23
John_Wick
1,119111
1,119111
Welcome to to MathSE. Type$sin x$
,$cos x$
,$tan x$
,$csc x$
,$sec x$
, and$cot x$
to produce $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, and $cot x$, respectively.
– N. F. Taussig
Nov 22 at 14:34
add a comment |
Welcome to to MathSE. Type$sin x$
,$cos x$
,$tan x$
,$csc x$
,$sec x$
, and$cot x$
to produce $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, and $cot x$, respectively.
– N. F. Taussig
Nov 22 at 14:34
Welcome to to MathSE. Type
$sin x$
, $cos x$
, $tan x$
, $csc x$
, $sec x$
, and $cot x$
to produce $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, and $cot x$, respectively.– N. F. Taussig
Nov 22 at 14:34
Welcome to to MathSE. Type
$sin x$
, $cos x$
, $tan x$
, $csc x$
, $sec x$
, and $cot x$
to produce $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, and $cot x$, respectively.– N. F. Taussig
Nov 22 at 14:34
add a comment |
1
mathworld.wolfram.com/WernerFormulas.html
– lab bhattacharjee
Nov 22 at 14:17