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Can I input negative angles into the cosine half-angle formula?

Use the half-angle formula for cosine to compute $cos(theta/2)$ given $cos(theta)=63/68$ where $0ltthetaltpi/2$Calculating arbitrary sines/cosinesA few questions regarding the cosine difference identitySimplify the expression by using a Double-Angle Formula or a Half-Angle Formula.The correct half angle formula?Tangent half angle formulaDerive the formula for the cosine of the difference of two angles from the dot product formulaDoes atan2(mean sine, mean cosine) approximate the mean angle?Half-Angle Formula with DecimalsWhy is the cosine of a right angle, 90 degrees, equal to zero?

0

1

$begingroup$

So the cosine half-angle formula says:

Now, we know that co-terminal angles have equal cosines. Consider that $cos (7pi/4)$ = $cos(-pi/4)$. However, if you apply the half angle formula to $(7pi/4)$ you get a different answer than if you apply the half angle formula to $(-pi/4)$. Does the half angle formula require that your angle being inputted be positive, so that applying the half-angle formula to $(-pi/4)$ would mean plugging in the positive coterminal version, $(7pi/4)$, to the formula? (In which case the contradiction disappears)

algebra-precalculus trigonometry

share|cite|improve this question

edited Mar 20 at 18:39

Will

asked Mar 20 at 18:38

Will Will

515

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Look at the $pm$ .
  $endgroup$
  – Lord Shark the Unknown
  Mar 20 at 18:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $cos(pi-x)=-cos x$
  $endgroup$
  – lab bhattacharjee
  Mar 20 at 18:41
  

  
  
  
  

add a comment | 

0

1

$begingroup$

So the cosine half-angle formula says:

Now, we know that co-terminal angles have equal cosines. Consider that $cos (7pi/4)$ = $cos(-pi/4)$. However, if you apply the half angle formula to $(7pi/4)$ you get a different answer than if you apply the half angle formula to $(-pi/4)$. Does the half angle formula require that your angle being inputted be positive, so that applying the half-angle formula to $(-pi/4)$ would mean plugging in the positive coterminal version, $(7pi/4)$, to the formula? (In which case the contradiction disappears)

algebra-precalculus trigonometry

share|cite|improve this question

edited Mar 20 at 18:39

Will

asked Mar 20 at 18:38

Will Will

515

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Look at the $pm$ .
  $endgroup$
  – Lord Shark the Unknown
  Mar 20 at 18:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $cos(pi-x)=-cos x$
  $endgroup$
  – lab bhattacharjee
  Mar 20 at 18:41
  

  
  
  
  

add a comment | 

$begingroup$

So the cosine half-angle formula says:

Now, we know that co-terminal angles have equal cosines. Consider that $cos (7pi/4)$ = $cos(-pi/4)$. However, if you apply the half angle formula to $(7pi/4)$ you get a different answer than if you apply the half angle formula to $(-pi/4)$. Does the half angle formula require that your angle being inputted be positive, so that applying the half-angle formula to $(-pi/4)$ would mean plugging in the positive coterminal version, $(7pi/4)$, to the formula? (In which case the contradiction disappears)

algebra-precalculus trigonometry

share|cite|improve this question

edited Mar 20 at 18:39

Will

asked Mar 20 at 18:38

Will Will

515

$endgroup$

share|cite|improve this question

edited Mar 20 at 18:39

Will

asked Mar 20 at 18:38

Will Will

515

share|cite|improve this question

edited Mar 20 at 18:39

Will

asked Mar 20 at 18:38

Will Will

515

asked Mar 20 at 18:38

Will Will

515

asked Mar 20 at 18:38

Will Will

515

2 Answers
2

active

oldest

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2

$begingroup$

For any angle $theta$, we have$$frac1+costheta2=frac1+cosleft(2fractheta2right)2=frac1+cos^2left(fractheta2right)-sin^2left(fractheta2right)2=cos^2left(fractheta2right),$$since $1-sin^2left(fractheta2right)=cos^2left(fractheta2right).$ So, yes, that formula is always valid.

share|cite|improve this answer

answered Mar 20 at 18:46

José Carlos SantosJosé Carlos Santos

172k23132240

$endgroup$

add a comment | 

2

$begingroup$

Yes, but you need to be mindful of the resulting sign of the answer. If that negative angle ends up in the first or 4th quadrant use the positive sign, if it is in the 2nd or 3rd use negative. This formula is very delicate, if you use the following property of cosines: $cos(-x)=cos(x)$, then you will realize that the same analysis is required for the positive values.

Simply, be careful of the sign of the cos of your angle and ignore the negative sign in front of the angle.

share|cite|improve this answer

answered Mar 20 at 18:45

Al-Fahad Al-QadhiAl-Fahad Al-Qadhi

213

$endgroup$

add a comment | 

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2

$begingroup$

For any angle $theta$, we have$$frac1+costheta2=frac1+cosleft(2fractheta2right)2=frac1+cos^2left(fractheta2right)-sin^2left(fractheta2right)2=cos^2left(fractheta2right),$$since $1-sin^2left(fractheta2right)=cos^2left(fractheta2right).$ So, yes, that formula is always valid.

share|cite|improve this answer

answered Mar 20 at 18:46

José Carlos SantosJosé Carlos Santos

172k23132240

$endgroup$

add a comment | 

2

$begingroup$

For any angle $theta$, we have$$frac1+costheta2=frac1+cosleft(2fractheta2right)2=frac1+cos^2left(fractheta2right)-sin^2left(fractheta2right)2=cos^2left(fractheta2right),$$since $1-sin^2left(fractheta2right)=cos^2left(fractheta2right).$ So, yes, that formula is always valid.

share|cite|improve this answer

answered Mar 20 at 18:46

José Carlos SantosJosé Carlos Santos

172k23132240

$endgroup$

add a comment | 

$begingroup$

For any angle $theta$, we have$$frac1+costheta2=frac1+cosleft(2fractheta2right)2=frac1+cos^2left(fractheta2right)-sin^2left(fractheta2right)2=cos^2left(fractheta2right),$$since $1-sin^2left(fractheta2right)=cos^2left(fractheta2right).$ So, yes, that formula is always valid.

share|cite|improve this answer

answered Mar 20 at 18:46

José Carlos SantosJosé Carlos Santos

172k23132240

$endgroup$

share|cite|improve this answer

answered Mar 20 at 18:46

José Carlos SantosJosé Carlos Santos

172k23132240

answered Mar 20 at 18:46

José Carlos SantosJosé Carlos Santos

172k23132240

answered Mar 20 at 18:46

José Carlos SantosJosé Carlos Santos

172k23132240

2

$begingroup$

Yes, but you need to be mindful of the resulting sign of the answer. If that negative angle ends up in the first or 4th quadrant use the positive sign, if it is in the 2nd or 3rd use negative. This formula is very delicate, if you use the following property of cosines: $cos(-x)=cos(x)$, then you will realize that the same analysis is required for the positive values.

Simply, be careful of the sign of the cos of your angle and ignore the negative sign in front of the angle.

share|cite|improve this answer

answered Mar 20 at 18:45

Al-Fahad Al-QadhiAl-Fahad Al-Qadhi

213

$endgroup$

add a comment | 

2

$begingroup$

Yes, but you need to be mindful of the resulting sign of the answer. If that negative angle ends up in the first or 4th quadrant use the positive sign, if it is in the 2nd or 3rd use negative. This formula is very delicate, if you use the following property of cosines: $cos(-x)=cos(x)$, then you will realize that the same analysis is required for the positive values.

Simply, be careful of the sign of the cos of your angle and ignore the negative sign in front of the angle.

share|cite|improve this answer

answered Mar 20 at 18:45

Al-Fahad Al-QadhiAl-Fahad Al-Qadhi

213

$endgroup$

add a comment | 

$begingroup$

Yes, but you need to be mindful of the resulting sign of the answer. If that negative angle ends up in the first or 4th quadrant use the positive sign, if it is in the 2nd or 3rd use negative. This formula is very delicate, if you use the following property of cosines: $cos(-x)=cos(x)$, then you will realize that the same analysis is required for the positive values.

Simply, be careful of the sign of the cos of your angle and ignore the negative sign in front of the angle.

share|cite|improve this answer

answered Mar 20 at 18:45

Al-Fahad Al-QadhiAl-Fahad Al-Qadhi

213

$endgroup$

share|cite|improve this answer

answered Mar 20 at 18:45

Al-Fahad Al-QadhiAl-Fahad Al-Qadhi

213

answered Mar 20 at 18:45

Al-Fahad Al-QadhiAl-Fahad Al-Qadhi

213

answered Mar 20 at 18:45

Al-Fahad Al-QadhiAl-Fahad Al-Qadhi

213