The problem is as follows:

Find the value of this function $$A=\left(\cos\frac{\omega}{2} +\cos\frac{\phi}{2}\right )^{2} +\left(\sin\frac{\omega}{2} -\sin\frac{\phi}{2}\right )^{2}$$ when $\omega=33^{\circ}{20}'$ and $\phi=56^{\circ}{40}'$.


$$A=\left(\cos\frac{\omega}{2} +\cos\frac{\phi}{2}\right)^{2} +\left(\sin\frac{\omega}{2} -\sin\frac{\phi}{2}\right)^{2}$$

By solving the power I obtained the following:

$$A= \cos^{2}\frac{\omega}{2} +\cos^{2}\frac{\phi}{2} +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} +\sin^{2}\frac{\omega}{2} +\sin^{2}\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2}$$

I noticed some familiar terms and using pitagoric identities then I rearranged the equation as follows:

\begin{align} A &= \cos^{2}\frac{\omega}{2} +\sin^{2}\frac{\omega}{2} +\cos^{2}\frac{\phi}{2} +\sin^{2}\frac{\phi}{2} +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2} \\ &= 1+1+2\cos\frac{\omega}{2}cos\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2} \end{align}

Since the latter terms are another way to write prosthapharesis formulas I did the following:

\begin{align} \cos\alpha +\cos\beta &= 2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} \\ \cos\alpha -\cos\beta &= -2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} \\ \\ \alpha+\beta &= \omega \\ \alpha-\beta &= \phi \end{align}

By solving the system I found: $\alpha=\frac{\omega+\phi}{2}$ and $\beta=\frac{\omega-\phi}{2}$.

Therefore by inserting these into the problem:

\begin{align} A &= 1+1 +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} +2\sin\frac{\omega}{2}\cos\frac{\phi}{2} \\ &= 2 +2\cos\frac{\omega}{2}\cos\frac{\phi}{2} -\left(-2\sin\frac{\omega}{2}\cos\frac{\phi}{2}\right) \\ &= 2 +\cos\frac{\omega+\phi}{2} +\cos\frac{\omega-\phi}{2} -\left(\cos\frac{\omega+\phi}{2} -\cos\frac{\omega-\phi}{2}\right) \end{align}

By cancelling elements,

$$A=2 +2\cos\frac{\omega-\phi}{2}$$

However I'm stuck at trying to evaluate these values:

\begin{align} \omega &= 33^{\circ}{20}'\; \phi=56^{\circ}{40}' \\ \omega-\phi &= \left(33+\frac{20}{60}\right) -\left(56+\frac{40}{60}\right) = -23-\frac{20}{60} \end{align}



However the latter answer does not appear in the alternative neither seems to be right. Is there something wrong on what I did?

  • 1
    $\begingroup$ Just a remark: Use \cos and \sin instead of cos and sin. $\endgroup$ – MrYouMath Oct 17 '17 at 8:52
  • $\begingroup$ @MrYouMath Thanks for the advise. I'll take note on that next time I post a similar question. I'm new on LaTex. $\endgroup$ – Chris Steinbeck Bell Oct 18 '17 at 2:31

There is a mistake in your expansion of the second bracket.

We should have

$A=\cos^2\frac{\omega}{2}+2\cos\frac{\omega}{2}\cos\frac{\phi}{2}+\cos^2 \frac{\phi}{2}+\sin^2\frac{\omega}{2}-2\sin\frac{\omega}{2}\sin\frac{\phi}{2}+\sin^2\frac{\phi}{2}$

$=2+2(\cos\frac{\omega}{2} \cos\frac{\phi}{2} -\sin\frac{\omega}{2}\sin\frac{\phi}{2})$ .

In the bracket we have the expansion for $\cos(\frac{\omega}{2}+\frac{\phi}{2})$ and fortunately $\omega+\phi=90^\circ$.

Thus $A=2+2\cos(45^\circ)$



Hint: You did a miscalculation:

$$A=\left (\cos\frac{\omega}{2}+\cos\frac{\phi}{2} \right )^{2}+\left (\sin\frac{\omega}{2}-\sin\frac{\phi}{2} \right )^{2}$$ $$=\cos ^2\frac{\omega}{2}+2\cos\frac{\omega}{2}\cos\frac{\phi}{2}+\cos^2 \frac{\phi}{2}+\sin ^2\frac{\omega}{2}-2\sin\frac{\omega}{2}\sin\frac{\phi}{2}+\sin^2 \frac{\phi}{2}$$ $$=2+2\cos\frac{\omega}{2}\cos\frac{\phi}{2}-2\sin\frac{\omega}{2}\sin\frac{\phi}{2}$$ $$=2+2\left(\cos\frac{\omega}{2}\cos\frac{\phi}{2}-\sin\frac{\omega}{2}\sin\frac{\phi}{2}\right)$$ $$=2+2\cos\left(\frac{\omega}{2}+\frac{\phi}{2}\right)$$

In the last step I used $\cos(x+y)=\cos x\cos y-\sin x \sin y$.

  • $\begingroup$ Mind including the "last step" in your answer?. I mean the one involving inserting the angle values in the last formula. Usually this seems the most logical and obvious thing to do. But what to do when the angles are not "integers". Was the procedure I used correct?. I mean to decompose them into fractions. The latter part I'm not sure. $\endgroup$ – Chris Steinbeck Bell Oct 18 '17 at 2:34

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