Rewriting $sin(x+frac {pi} {6})cos(x)$ as $frac {1} {4}(2sin(2x+frac {pi} {6})+1)$ [closed]
$begingroup$
If I have a trigonometric expression like
$$sin(x+frac {pi} {6})cos(x)$$
what are the steps to simplify it to the following?
$$frac {1} {4}(2sin(2x+frac {pi} {6})+1)$$
trigonometry
edited Dec 3 '18 at 1:41
Blue
47.8k870152
asked Dec 3 '18 at 0:57
WereWere
33
$endgroup$
closed as off-topic by Saad, caverac, Jean-Claude Arbaut, Jyrki Lahtonen, Brahadeesh Dec 3 '18 at 4:21
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." – Saad, caverac, Jean-Claude Arbaut, Jyrki Lahtonen, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
If I have a trigonometric expression like
$$sin(x+frac {pi} {6})cos(x)$$
what are the steps to simplify it to the following?
$$frac {1} {4}(2sin(2x+frac {pi} {6})+1)$$
trigonometry
edited Dec 3 '18 at 1:41
Blue
47.8k870152
asked Dec 3 '18 at 0:57
WereWere
33
$endgroup$
closed as off-topic by Saad, caverac, Jean-Claude Arbaut, Jyrki Lahtonen, Brahadeesh Dec 3 '18 at 4:21
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." – Saad, caverac, Jean-Claude Arbaut, Jyrki Lahtonen, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Let $x=pi/6$. The first expression evaluates to $2sin(pi/3)=sqrt{3}$, the second to $(2sin(pi/2)+1)/4=3/4$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 1:06
$begingroup$
Sorry I made a mistake
$endgroup$
– Were
Dec 3 '18 at 1:11
$begingroup$
mathworld.wolfram.com/WernerFormulas.html and mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$endgroup$
– lab bhattacharjee
Dec 3 '18 at 2:28
add a comment |
$begingroup$
If I have a trigonometric expression like
$$sin(x+frac {pi} {6})cos(x)$$
what are the steps to simplify it to the following?
$$frac {1} {4}(2sin(2x+frac {pi} {6})+1)$$
trigonometry
edited Dec 3 '18 at 1:41
Blue
47.8k870152
asked Dec 3 '18 at 0:57
WereWere
33
$endgroup$
If I have a trigonometric expression like
$$sin(x+frac {pi} {6})cos(x)$$
what are the steps to simplify it to the following?
$$frac {1} {4}(2sin(2x+frac {pi} {6})+1)$$
trigonometry
trigonometry
edited Dec 3 '18 at 1:41
Blue
47.8k870152
asked Dec 3 '18 at 0:57
WereWere
33
edited Dec 3 '18 at 1:41
Blue
47.8k870152
asked Dec 3 '18 at 0:57
WereWere
33
edited Dec 3 '18 at 1:41
Blue
47.8k870152
edited Dec 3 '18 at 1:41
Blue
47.8k870152
edited Dec 3 '18 at 1:41
Blue
47.8k870152
47.8k870152
asked Dec 3 '18 at 0:57
WereWere
33
asked Dec 3 '18 at 0:57
WereWere
33
asked Dec 3 '18 at 0:57
WereWere
33
33
closed as off-topic by Saad, caverac, Jean-Claude Arbaut, Jyrki Lahtonen, Brahadeesh Dec 3 '18 at 4:21
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." – Saad, caverac, Jean-Claude Arbaut, Jyrki Lahtonen, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, caverac, Jean-Claude Arbaut, Jyrki Lahtonen, Brahadeesh Dec 3 '18 at 4:21
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." – Saad, caverac, Jean-Claude Arbaut, Jyrki Lahtonen, Brahadeesh
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Let $x=pi/6$. The first expression evaluates to $2sin(pi/3)=sqrt{3}$, the second to $(2sin(pi/2)+1)/4=3/4$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 1:06
$begingroup$
Sorry I made a mistake
$endgroup$
– Were
Dec 3 '18 at 1:11
$begingroup$
mathworld.wolfram.com/WernerFormulas.html and mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$endgroup$
– lab bhattacharjee
Dec 3 '18 at 2:28
add a comment |
$begingroup$
Let $x=pi/6$. The first expression evaluates to $2sin(pi/3)=sqrt{3}$, the second to $(2sin(pi/2)+1)/4=3/4$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 1:06
$begingroup$
Sorry I made a mistake
$endgroup$
– Were
Dec 3 '18 at 1:11
$begingroup$
mathworld.wolfram.com/WernerFormulas.html and mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$endgroup$
– lab bhattacharjee
Dec 3 '18 at 2:28
$begingroup$
Let $x=pi/6$. The first expression evaluates to $2sin(pi/3)=sqrt{3}$, the second to $(2sin(pi/2)+1)/4=3/4$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 1:06
$begingroup$
Let $x=pi/6$. The first expression evaluates to $2sin(pi/3)=sqrt{3}$, the second to $(2sin(pi/2)+1)/4=3/4$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 1:06
$begingroup$
Sorry I made a mistake
$endgroup$
– Were
Dec 3 '18 at 1:11
$begingroup$
Sorry I made a mistake
$endgroup$
– Were
Dec 3 '18 at 1:11
$begingroup$
mathworld.wolfram.com/WernerFormulas.html and mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$endgroup$
– lab bhattacharjee
Dec 3 '18 at 2:28
$begingroup$
mathworld.wolfram.com/WernerFormulas.html and mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$endgroup$
– lab bhattacharjee
Dec 3 '18 at 2:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
From $sin(a+b)=sin(a)cos(b)+sin(b)cos(a)$ and $sin(a-b)=sin(a)cos(b)-sin(b)cos(a)$ you get
$$sin(a)cos(b)=frac12(sin(a+b)+sin(a-b))$$
Hence
$$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+sin(pi/6))$$
And since $sin(pi/6)=1/2$,
$$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+frac12)=frac14(2sin(2x+pi/6)+1)$$
answered Dec 3 '18 at 1:18
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From $sin(a+b)=sin(a)cos(b)+sin(b)cos(a)$ and $sin(a-b)=sin(a)cos(b)-sin(b)cos(a)$ you get
$$sin(a)cos(b)=frac12(sin(a+b)+sin(a-b))$$
Hence
$$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+sin(pi/6))$$
And since $sin(pi/6)=1/2$,
$$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+frac12)=frac14(2sin(2x+pi/6)+1)$$
answered Dec 3 '18 at 1:18
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
$endgroup$
add a comment |
$begingroup$
From $sin(a+b)=sin(a)cos(b)+sin(b)cos(a)$ and $sin(a-b)=sin(a)cos(b)-sin(b)cos(a)$ you get
$$sin(a)cos(b)=frac12(sin(a+b)+sin(a-b))$$
Hence
$$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+sin(pi/6))$$
And since $sin(pi/6)=1/2$,
$$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+frac12)=frac14(2sin(2x+pi/6)+1)$$
answered Dec 3 '18 at 1:18
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
$endgroup$
add a comment |
$begingroup$
From $sin(a+b)=sin(a)cos(b)+sin(b)cos(a)$ and $sin(a-b)=sin(a)cos(b)-sin(b)cos(a)$ you get
$$sin(a)cos(b)=frac12(sin(a+b)+sin(a-b))$$
Hence
$$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+sin(pi/6))$$
And since $sin(pi/6)=1/2$,
$$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+frac12)=frac14(2sin(2x+pi/6)+1)$$
answered Dec 3 '18 at 1:18
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
$endgroup$
From $sin(a+b)=sin(a)cos(b)+sin(b)cos(a)$ and $sin(a-b)=sin(a)cos(b)-sin(b)cos(a)$ you get
$$sin(a)cos(b)=frac12(sin(a+b)+sin(a-b))$$
Hence
$$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+sin(pi/6))$$
And since $sin(pi/6)=1/2$,
$$sin(x+pi/6)cos(x)=frac12(sin(2x+pi/6)+frac12)=frac14(2sin(2x+pi/6)+1)$$
answered Dec 3 '18 at 1:18
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
answered Dec 3 '18 at 1:18
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
answered Dec 3 '18 at 1:18
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
answered Dec 3 '18 at 1:18
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
14.7k63464
add a comment |
add a comment |
$begingroup$
Let $x=pi/6$. The first expression evaluates to $2sin(pi/3)=sqrt{3}$, the second to $(2sin(pi/2)+1)/4=3/4$.
$endgroup$
– Jean-Claude Arbaut
Dec 3 '18 at 1:06
$begingroup$
Sorry I made a mistake
$endgroup$
– Were
Dec 3 '18 at 1:11
$begingroup$
mathworld.wolfram.com/WernerFormulas.html and mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$endgroup$
– lab bhattacharjee
Dec 3 '18 at 2:28