Supplemental Videos.
The main topics of this section are also presented in the following videos:
Section 4.3 Half-Angle and Angle Sum and Difference Identities
In the last section we developed a few important trigonometric identities and some examples. In this section, we will continue working with a few special trigonometric identities and do some more examples. All of these examples can be derived from basic trigonometric principles. However, in this book we will focus not on deriving the identities but rather in using the identities to solve problems. For this reason we do not include derivations of the identities but instead refer you to other texts on trigonometry for such derivations.
Subsection 4.3.1 Half-Angle Identities
What happens to the cosine or sine of an angle if we cut the angle in half? You might expect the cosine or sine to be cut in half, but this is not the case. Instead, we have the Half-Angle Identities.
Half-Angle Identities.
\begin{align*}
\cos\left(\frac{\theta}{2}\right) \amp = \pm \sqrt{\frac{1+\cos(\theta)}{2}} \\
\\
\sin\left(\frac{\theta}{2}\right) \amp = \pm \sqrt{\frac{1-\cos(\theta)}{2}}
\end{align*}
There are examples that we can solve using the Half-Angle Identities.
Example 4.13.
Find the exact value for \(\cos(15^{\circ})\text{.}\)
Solution.
We first note that \(15^{\circ}\) is half of \(30^{\circ}\text{.}\) Therefore,
\begin{equation*}
\cos(15^{\circ})=\cos\left(\frac{30^{\circ}}{2}\right) = \pm \sqrt{\frac{1+\cos(30^{\circ})}{2}}.
\end{equation*}
We can evaluate the cosine. Since 15 degrees is in the first quadrant, we need the positive result.
\begin{equation*}
\cos(15^{\circ}) = \sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{1}{2}+\frac{\sqrt{3}}{4}}
\end{equation*}
Subsection 4.3.2 Angle Sum and Difference Identities
The sum and difference of angles identities are often used to rewrite expressions in other forms, or to rewrite an angle in terms of simpler angles.
Angle Sum and Difference Identities.
\begin{align*}
\sin(\theta + \phi) \amp = \sin(\theta)\cos(\phi) + \sin(\phi)\cos(\theta) \\
\\
\sin(\theta - \phi) \amp = \sin(\theta)\cos(\phi) - \sin(\phi)\cos(\theta) \\
\\
\cos(\theta + \phi) \amp = \cos(\theta)\cos(\phi) - \sin(\theta)\sin(\phi) \\
\\
\cos(\theta - \phi) \amp = \cos(\theta)\cos(\phi) + \sin(\theta)\sin(\phi)
\end{align*}
Example 4.14.
Find the exact value of \(\cos(75^{\circ})\)
Solution.
Since we know that \(75^{\circ}=30^{\circ}+45^{\circ}\) we can evaluate \(\cos(75^{\circ})\) using an angle sum identity.
\begin{align*}
\cos(30^{\circ} + 45^{\circ}) \amp = \cos(30^{\circ})\cos(45^{\circ}) - \sin(30^{\circ})\sin(45^{\circ}) \\
\amp = \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}-\frac{1}{2}\cdot\frac{\sqrt{2}}{2} \\
\amp = \frac{\sqrt{6}-\sqrt{2}}{4}
\end{align*}
The Angle Sum and Difference Identities can also be used to solve more complicated trigonometric equations.
Example 4.15.
Solve \(\sin(x)\sin(2x)+\cos(x)\cos(2x)=\frac{\sqrt{3}}{2}\)
Solution.
We first recognize that the left hand side of the equation is result of the difference of angles identity for cosine. Therefore,
\begin{align*}
\sin(x)\sin(2x)+\cos(x)\cos(2x)\amp=\frac{\sqrt{3}}{2}
\end{align*}
is equivalent to the equation
\begin{align*}
\cos(x-2x)\amp=\frac{\sqrt{3}}{2}
\end{align*}
Thus we have
\begin{align*}
\cos(x-2x)\amp=\frac{\sqrt{3}}{2} \\
\cos(-x)\amp=\frac{\sqrt{3}}{2} \\
\cos(x)\amp=\frac{\sqrt{3}}{2} \\
x\amp=\frac{\pi}{6}+2 \pi k, \frac{11 \pi}{6}+2\pi k
\end{align*}
The above examples demonstrate just a little bit of the power of the Half-Angle and Angle Sum and Difference Identities. If you go on to take a Calculus course you may also see these identities come up again as you calculate derivatives.
Exercises 4.3.3 Exercises
1. Evaluating Half Angles.
Use the half-angle formulas to evaluate the following exactly.
(a) \(\sin(\frac{\pi}{12})\)=?
-
\(\displaystyle -\sqrt{\frac{1-\sqrt{3}/2}{2}}\)
-
\(\displaystyle -\sqrt{\frac{1+\sqrt{3}/2}{2}}\)
-
\(\displaystyle -\sqrt{\frac{1-1/2}{2}}\)
-
\(\displaystyle \sqrt{\frac{1+1/2}{2}}\)
-
\(\displaystyle \sqrt{\frac{1+\sqrt{3}/2}{2}}\)
-
\(\displaystyle \sqrt{\frac{1+\sqrt{2}/2}{2}}\)
-
\(\displaystyle -\sqrt{\frac{1+\sqrt{2}/2}{2}}\)
-
\(\displaystyle \sqrt{\frac{1-\sqrt{2}/2}{2}}\)
-
\(\displaystyle -\sqrt{\frac{1+1/2}{2}}\)
-
\(\displaystyle \sqrt{\frac{1-1/2}{2}}\)
-
\(\displaystyle \sqrt{\frac{1-\sqrt{3}/2}{2}}\)
-
\(\displaystyle -\sqrt{\frac{1-\sqrt{2}/2}{2}}\)
-
None of the above
(b) cos(\(\frac{9\pi}{8}\))=? Careful! Draw a sketch. In which quadrant is \(\frac{9\pi}{8}\text{?}\)
-
\(\displaystyle \sqrt{\frac{1-1/2}{2}}\)
-
\(\displaystyle \sqrt{\frac{1+1/2}{2}}\)
-
\(\displaystyle -\sqrt{\frac{1-1/2}{2}}\)
-
\(\displaystyle -\sqrt{\frac{1+\sqrt{2}/2}{2}}\)
-
\(\displaystyle -\sqrt{\frac{1-\sqrt{2}/2}{2}}\)
-
\(\displaystyle -\sqrt{\frac{1+1/2}{2}}\)
-
\(\displaystyle \sqrt{\frac{1+\sqrt{3}/2}{2}}\)
-
\(\displaystyle \sqrt{\frac{1-\sqrt{2}/2}{2}}\)
-
\(\displaystyle \sqrt{\frac{1-\sqrt{3}/2}{2}}\)
-
\(\displaystyle \sqrt{\frac{1+\sqrt{2}/2}{2}}\)
-
\(\displaystyle -\sqrt{\frac{1+\sqrt{3}/2}{2}}\)
-
\(\displaystyle -\sqrt{\frac{1-\sqrt{3}/2}{2}}\)
-
None of the above
2. Evaluating Sums of Angles Angles.
Use the angle sum and difference formulas to evaluate the following exactly. Note that there may be more than one way to evaluate each.
(a) \(\cos(\frac{7\pi}{12})\)=?
-
\(\displaystyle \frac{\sqrt{2}-\sqrt{3}}{4}\)
-
\(\displaystyle \frac{-\sqrt{2}-\sqrt{6}}{4}\)
-
\(\displaystyle \frac{\sqrt{3}-\sqrt{6}}{4}\)
-
\(\displaystyle \frac{\sqrt{6}-\sqrt{3}}{4}\)
-
\(\displaystyle \frac{\sqrt{2}+\sqrt{6}}{4}\)
-
\(\displaystyle \frac{\sqrt{6}-\sqrt{2}}{4}\)
-
\(\displaystyle \frac{\sqrt{3}+\sqrt{6}}{4}\)
-
\(\displaystyle \frac{\sqrt{3}-\sqrt{2}}{4}\)
-
\(\displaystyle \frac{\sqrt{2}+\sqrt{3}}{4}\)
-
\(\displaystyle \frac{\sqrt{2}-\sqrt{6}}{4}\)
-
\(\displaystyle \frac{-\sqrt{3}-\sqrt{6}}{4}\)
-
\(\displaystyle \frac{-\sqrt{2}-\sqrt{3}}{4}\)
-
None of the above
(b) \(\cos(\frac{11\pi}{12})\)=?
-
\(\displaystyle \frac{\sqrt{3}-\sqrt{6}}{4}\)
-
\(\displaystyle \frac{\sqrt{3}+\sqrt{6}}{4}\)
-
\(\displaystyle \frac{\sqrt{2}-\sqrt{6}}{4}\)
-
\(\displaystyle \frac{\sqrt{6}-\sqrt{3}}{4}\)
-
\(\displaystyle \frac{-\sqrt{2}-\sqrt{6}}{4}\)
-
\(\displaystyle \frac{\sqrt{2}+\sqrt{6}}{4}\)
-
\(\displaystyle \frac{\sqrt{2}+\sqrt{3}}{4}\)
-
\(\displaystyle \frac{-\sqrt{2}-\sqrt{3}}{4}\)
-
\(\displaystyle \frac{-\sqrt{3}-\sqrt{6}}{4}\)
-
\(\displaystyle \frac{\sqrt{2}-\sqrt{3}}{4}\)
-
\(\displaystyle \frac{\sqrt{3}-\sqrt{2}}{4}\)
-
\(\displaystyle \frac{\sqrt{6}-\sqrt{2}}{4}\)
-
None of the above
3. Verifying Identities.
(a) Let’s verify the identity \(\tan^2(\frac{\theta}{2})=\frac{1-\cos(\theta)}{1+\cos(\theta)}\text{.}\) What is a good first step?
-
\(\displaystyle \tan(2\theta)=\frac{1-\cos(\theta)}{1+\cos(\theta)}\)
-
\(\displaystyle \tan(\frac{\theta}{2})=\frac{1-\cos(\theta)}{1+\cos(\theta)}\)
-
\(\displaystyle \frac{\cos^2(\frac{\theta}{2})}{\sin^2(\frac{\theta}{2})}=\frac{1-\cos(\theta)}{1+\cos(\theta)}\)
-
\(\displaystyle \frac{\sin^2(\frac{\theta}{2})}{\cos^2(\frac{\theta}{2})}=\frac{1-\cos(\theta)}{1+\cos(\theta)}\)
-
None of the above
How can we use trig identities to rewrite our last equation?
-
\(\displaystyle \frac{2\sin(\theta)\cos(\theta)}{\cos^2(\theta)-\sin^2(\theta)}=\frac{1-\cos(\theta)}{1+\cos(\theta)}\)
-
\(\displaystyle \frac{\cos^2(\theta)-\sin^2(\theta)}{2\sin(\theta)\cos(\theta)}=\frac{1-\cos(\theta)}{1+\cos(\theta)}\)
-
\(\displaystyle \frac{2\sin(\theta)\cos(\theta)}{\sin^2(\theta)-\cos^2(\theta)}=\frac{1-\cos(\theta)}{1+\cos(\theta)}\)
-
\(\displaystyle \frac{\sin^2(\theta)-\cos^2(\theta)}{2\sin(\theta)\cos(\theta)}=\frac{1-\cos(\theta)}{1+\cos(\theta)}\)
-
\(\displaystyle \frac{\Big(\pm\sqrt{\frac{1}{2}(1+\cos(\theta))}\Big)^2}{\Big(\pm\sqrt{\frac{1}{2}(1-\cos(\theta))}\Big)^2}=\frac{1-\cos(\theta)}{1+\cos(\theta)}\)
-
\(\displaystyle \frac{\Big(\pm\sqrt{\frac{1}{2}(1-\cos(\theta))}\Big)^2}{\Big(\pm\sqrt{\frac{1}{2}(1+\cos(\theta))}\Big)^2}=\frac{1-\cos(\theta)}{1+\cos(\theta)}\)
-
None of the above
How can we simplify the previous equation?
-
\(\displaystyle (2\sin(\theta)\cos(\theta))=\frac{\cos^2(\theta)-\sin^2(\theta)}{1+\cos(\theta)}\)
-
\(\displaystyle \frac{\sin^2(\theta)}{2}=\frac{1-\cos(\theta)}{1+\cos(\theta)}\)
-
\(\displaystyle \frac{\frac{1}{2}(1+\cos(\theta))}{\frac{1}{2}(1-\cos(\theta))}=\frac{1-\cos(\theta)}{1+\cos(\theta)}\)
-
\(\displaystyle \frac{\cos^2(\theta)}{2}=\frac{1-\cos(\theta)}{1+\cos(\theta)}\)
-
\(\displaystyle \frac{\frac{1}{2}(1-\cos(\theta))}{\frac{1}{2}(1+\cos(\theta))}=\frac{1-\cos(\theta)}{1+\cos(\theta)}\)
-
\(\displaystyle \frac{(1+\cos(\theta))^2}{\cos(\theta)}=\frac{1-\cos(\theta)}{1+\cos(\theta)}\)
-
None of the above
How can we finish the proof?
-
After simplifying further, we obtain the half angle formulas for sine and cosine.
-
After simplifying further, both sides become equal.
-
After simplifying further, we obtain the double angle formulas for cosine.
-
After simplifying further, we obtain the half angle formulas for sine.
-
After simplifying further, we obtain the double angle formulas for sine.
-
After simplifying further, we obtain the half angle formulas for cosine.
-
After simplifying further, we obtain the Pythagorean Identity.
-
After simplifying further, we obtain the double angle formulas for sine and cosine.
-
None of the above
(b) Evaluate tan(\(\frac{\pi}{12}\)) exactly.
-
\(\displaystyle -\sqrt{\frac{1-1/2}{1+1/2}}\)
-
\(\displaystyle -\sqrt{\frac{1-\sqrt{3}/2}{1+\sqrt{3}/2}}\)
-
\(\displaystyle \sqrt{\frac{1-\sqrt{3}/2}{1+\sqrt{2}/2}}\)
-
\(\displaystyle -\sqrt{\frac{1-\sqrt{3}/2}{1+\sqrt{2}/2}}\)
-
\(\displaystyle \sqrt{\frac{1-\sqrt{2}/2}{1+\sqrt{2}/2}}\)
-
\(\displaystyle \sqrt{\frac{1-1/2}{1+1/2}}\)
-
\(\displaystyle \sqrt{\frac{1-\sqrt{3}/2}{1+\sqrt{3}/2}}\)
-
\(\displaystyle -\sqrt{\frac{1-\sqrt{2}/2}{1+\sqrt{2}/2}}\)
-
None of the above
4. Application: Ferris Wheel.
Suppose Asha is on a Ferris wheel of diameter 20 meters. She boarded the Ferris wheel at ground level, at the 6:00 position, and the wheel has turned counter-clockwise \(\frac{7\pi}{8}\) radians. How high off the ground is she? Give both an exact solution and a decimal approximation. [Hint: Try sketching a picture of the situation first.]
The exact height (given in meters) is...
-
\(\displaystyle 10+10\sqrt{\frac{1-\sqrt{2}/2}{2}}\)
-
\(\displaystyle 10\sqrt{\frac{1-\sqrt{2}/2}{2}}\)
-
\(\displaystyle 10+\sqrt{\frac{1-\sqrt{2}/2}{2}}\)
-
\(\displaystyle 10-10\sqrt{\frac{1+\sqrt{2}/2}{2}}\)
-
\(\displaystyle 10-10\sqrt{\frac{1-\sqrt{2}/2}{2}}\)
-
\(\displaystyle 10+\sqrt{\frac{1+\sqrt{2}/2}{2}}\)
-
\(\displaystyle 10+10\sqrt{\frac{1+\sqrt{2}/2}{2}}\)
-
\(\displaystyle 10-\sqrt{\frac{1-\sqrt{2}/2}{2}}\)
-
\(\displaystyle -10\sqrt{\frac{1+\sqrt{2}/2}{2}}\)
-
\(\displaystyle 10-\sqrt{\frac{1+\sqrt{2}/2}{2}}\)
-
None of the above
The decimal approximation of Asha’s height is meters.
5. Calculating Angle Sums and Differenecs.
Suppose angles \(A\) and \(B\) are in the first quadrant and \(\sin(A)=\frac{1}{4}\) and \(\sin(B)=\)\(\frac{3}{5}\text{.}\)
(a) Find \(\cos(A)\) and \(\cos(B)\) exactly.
\(\cos(A)=\)
\(\cos(B)=\)
(b) Find \(\sin(A+B)\) and \(\sin(A-B)\) exactly. The exact value of \(\sin(A+B)\) is...
-
\(\displaystyle \frac{1}{4}\cdot\frac{3}{5}-\frac{4}{5}\cdot\sqrt{1-\frac{1}{4^2}}\)
-
\(\displaystyle \frac{1}{4}\cdot\frac{5}{13}+\frac{12}{13}\cdot\sqrt{1-\frac{1}{4^2}}\)
-
\(\displaystyle \frac{1}{4}\cdot\frac{5}{13}-\frac{12}{13}\cdot\sqrt{1-\frac{1}{4^2}}\)
-
\(\displaystyle \frac{1}{4}\cdot\frac{4}{5}-\frac{3}{5}\cdot\sqrt{1-\frac{1}{4^2}}\)
-
\(\displaystyle \frac{1}{4}\cdot\frac{12}{13}-\frac{5}{13}\cdot\sqrt{1-\frac{1}{4^2}}\)
-
\(\displaystyle \frac{1}{4}\cdot\frac{12}{13}+\frac{5}{13}\cdot\sqrt{1-\frac{1}{4^2}}\)
-
\(\displaystyle \frac{1}{4}\cdot\frac{3}{5}+\frac{4}{5}\cdot\sqrt{1-\frac{1}{4^2}}\)
-
\(\displaystyle \frac{1}{4}\cdot\frac{4}{5}+\frac{3}{5}\cdot\sqrt{1-\frac{1}{4^2}}\)
-
None of the above
The exact value of \(\sin(A-B)\) is...
-
\(\displaystyle \frac{1}{4}\cdot\frac{4}{5}+\frac{3}{5}\cdot\sqrt{1-\frac{1}{4^2}}\)
-
\(\displaystyle \frac{1}{4}\cdot\frac{12}{13}+\frac{5}{13}\cdot\sqrt{1-\frac{1}{4^2}}\)
-
\(\displaystyle \frac{1}{4}\cdot\frac{3}{5}-\frac{4}{5}\cdot\sqrt{1-\frac{1}{4^2}}\)
-
\(\displaystyle \frac{1}{4}\cdot\frac{3}{5}+\frac{4}{5}\cdot\sqrt{1-\frac{1}{4^2}}\)
-
\(\displaystyle \frac{1}{4}\cdot\frac{5}{13}+\frac{12}{13}\cdot\sqrt{1-\frac{1}{4^2}}\)
-
\(\displaystyle \frac{1}{4}\cdot\frac{4}{5}-\frac{3}{5}\cdot\sqrt{1-\frac{1}{4^2}}\)
-
\(\displaystyle \frac{1}{4}\cdot\frac{5}{13}-\frac{12}{13}\cdot\sqrt{1-\frac{1}{4^2}}\)
-
\(\displaystyle \frac{1}{4}\cdot\frac{12}{13}-\frac{5}{13}\cdot\sqrt{1-\frac{1}{4^2}}\)
-
None of the above
6. Calculating Angle Sums and Differences.
(a) Using an angle-sum or angle-difference identity, rewrite the quantity \(4\cos(\frac{x}{2}+\frac{\pi}{3})\) as a sum or difference of two trigonometric functions.
-
\(\displaystyle 4\cdot\frac{\sqrt{3}}{2}\cos(\frac{x}{2})-4\cdot\frac{1}{2}\sin(\frac{x}{2})\)
-
\(\displaystyle 4\cdot\frac{\sqrt{2}}{2}\cos(\frac{x}{2})-4\cdot\frac{\sqrt{2}}{2}\sin(\frac{x}{2})\)
-
\(\displaystyle 4\cdot\frac{\sqrt{2}}{2}\cos(\frac{x}{2})+4\cdot\frac{\sqrt{2}}{2}\sin(\frac{x}{2})\)
-
\(\displaystyle 4\cdot\frac{\sqrt{3}}{2}\cos(\frac{x}{2})+4\cdot\frac{\sqrt{3}}{2}\sin(\frac{x}{2})\)
-
\(\displaystyle -4\cdot\frac{\sqrt{2}}{2}\cos(\frac{x}{2})-4\cdot\frac{\sqrt{2}}{2}\sin(\frac{x}{2})\)
-
\(\displaystyle -4\cdot\frac{1}{2}\cos(\frac{x}{2})-4\cdot\frac{\sqrt{3}}{2}\sin(\frac{x}{2})\)
-
\(\displaystyle 4\cdot\frac{\sqrt{3}}{2}\cos(\frac{x}{2})+4\cdot\frac{1}{2}\sin(\frac{x}{2})\)
-
\(\displaystyle -4\cdot\frac{\sqrt{3}}{2}\cos(\frac{x}{2})+4\cdot\frac{\sqrt{3}}{2}\sin(\frac{x}{2})\)
-
\(\displaystyle -4\cdot\frac{\sqrt{3}}{2}\cos(\frac{x}{2})+4\cdot\frac{1}{2}\sin(\frac{x}{2})\)
-
\(\displaystyle -4\cdot\frac{\sqrt{2}}{2}\cos(\frac{x}{2})+4\cdot\frac{\sqrt{2}}{2}\sin(\frac{x}{2})\)
-
\(\displaystyle 4\cdot\frac{1}{2}\cos(\frac{x}{2})+4\cdot\frac{\sqrt{3}}{2}\sin(\frac{x}{2})\)
-
\(\displaystyle -4\cdot\frac{1}{2}\cos(\frac{x}{2})+4\cdot\frac{\sqrt{3}}{2}\sin(\frac{x}{2})\)
-
\(\displaystyle 4\cdot\frac{1}{2}\cos(\frac{x}{2})-4\cdot\frac{\sqrt{3}}{2}\sin(\frac{x}{2})\)
-
\(\displaystyle 4\cdot\frac{\sqrt{3}}{2}\cos(\frac{x}{2})-4\cdot\frac{\sqrt{3}}{2}\sin(\frac{x}{2})\)
-
\(\displaystyle -4\cdot\frac{\sqrt{3}}{2}\cos(\frac{x}{2})-4\cdot\frac{\sqrt{3}}{2}\sin(\frac{x}{2})\)
-
\(\displaystyle -4\cdot\frac{\sqrt{3}}{2}\cos(\frac{x}{2})-4\cdot\frac{1}{2}\sin(\frac{x}{2})\)
-
None of the above
(b) Suppose \(\sin(x)=\)\(\frac{12}{13}\text{,}\) and \(x\) is between \(\frac{\pi}{2}\) and \(\pi\text{.}\) Find \(\cos(x)\text{.}\)
\(\cos(x)=\)
(c) With \(x\) as in part (b), find \(4\cos(\frac{x}{2}+\frac{\pi}{3})\) exactly.
-
-\(4\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}+4\cdot\frac{1}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}\)
-
-\(4\cdot\frac{1}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}-4\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}\)
-
\(\displaystyle 4\cdot\frac{1}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}+4\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}\)
-
\(\displaystyle 4\cdot\frac{1}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}+4\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}\)
-
-\(4\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}+4\cdot\frac{1}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}\)
-
-\(4\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}-4\cdot\frac{1}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}\)
-
-\(4\cdot\frac{1}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}-4\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}\)
-
-\(4\cdot\frac{1}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}+4\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}\)
-
\(\displaystyle 4\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}-4\cdot\frac{1}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}\)
-
\(\displaystyle 4\cdot\frac{1}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}-4\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}\)
-
\(\displaystyle 4\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}-4\cdot\frac{1}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}\)
-
\(\displaystyle 4\cdot\frac{1}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}-4\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}\)
-
-\(4\cdot\frac{1}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}+4\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}\)
-
-\(4\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}-4\cdot\frac{1}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}\)
-
\(\displaystyle 4\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}+4\cdot\frac{1}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}\)
-
\(\displaystyle 4\cdot\frac{\sqrt{3}}{2}\cdot\sqrt{\frac{1+\frac{5}{13}}{2}}+4\cdot\frac{1}{2}\cdot\sqrt{\frac{1-\frac{5}{13}}{2}}\)
-
None of the above
