Complimentary angles are pairs of angles who sum is \(\frac{\pi}{2}\text{.}\) Cofunctions are equal at their complimentary angles. This is stated more explicitly below.
The equivalence of the sine and cosine function at complementary angles can be established using the appropriate difference identities. Because the other four basic trigonometric functions can be expressed in terms of the sine and/or cosine functions, their cofunction equivalence at complimentary angles is an immediate consequence of that of the sine and cosine function. Let's establish the equivalences for the sine and cosine function.
The cosine and secant functions are even functions, while the other four trigonometric functions are all odd functions. The cause for this is illustrated in Figure 16.9.1. Two arcs of equal length are drawn in standard position, one traveling in the counterclockwise direction (which is measured as a positive value we will call \(t\)) and one traveling in the clockwise direction that is measured as \(-t\text{.}\)
Figure16.9.1. Arcs with Equal Length but Opposite Measure
It is apparent that the terminal points of the two arcs have equal \(x\)-coordinates (labeled as \(a\)) and opposite \(y\)-coordinates (labeled as \(b\) and \(-b\)). This gives us the following.
In the last section we established three double-angle identities for the cosine function. Two of these identities can be manipulated into half-angle identities for the sine and cosine functions.
We begin the derivation of the half-angle identity for the sine function by isolating the sine term in the identity \(\cos(2t)=1-2\sin^2(t)\text{.}\)
Let's first observe that \(2 \cdot 165^{\circ}=330^{\circ}\) which gives us \(165^{\circ}=\frac{330^{\circ}}{2}\text{.}\) Let's also note that when drawn in standard position, an angle with measurement \(165^{\circ}\) terminates in the second quadrant and, consequently, has a positive sine value. We are now ready to determine the sine value.
The derivation of the half-angle identity for the cosine function is similar to the derivation shown for the sine function and is left as an exercise. The result is stated below.
Before we apply the half-angle identity, let's also acknowledge that when drawn in standard position, an angle of measurement \(\frac{7\pi}{8}\) terminates in Quadrant II and, consequently, has a negative cosine value.
In order for the expression "\(1-\cos(u)\)"" to return a negative value, the cosine of \(u\) would have to have a value larger than one, which never happens. So the absolute value bars in the numerator of the last expression are unnecessary.
Now clearly the sine function has negative values as well as positive values, so at first glance the absolute value bars seem necessary. However, if we keep them then the expression \(\frac{1-\cos(u)}{\abs{\sin(u)}}\) will never have a negative value which is problematic in that the value of \(\tan\left(\frac{\alpha}{2}\right)\) will surely sometimes be negative.
It turns out that there is a remarkably opportune resolution to this seeming dilemma. If
\begin{equation*}
0^{\circ} \lt u \lt 180^{\circ}
\end{equation*}
so the value of \(\tan\left(\frac{\alpha}{2}\right)\) is also negative.
It is in fact the case that over the domain of the function \(y=\tan\left(\frac{\alpha}{2}\right)\text{,}\) the values of \(\sin(u)\) and \(\tan\left(\frac{\alpha}{2}\right)\) always have the same sign (or are both \(0\)). This leads to the following identity.
Apply the appropriate product-to-sum identity to the expression \(\sin\left(\frac{5\pi}{24}\right)\cos\left(\frac{\pi}{24}\right)\text{.}\) Completely simplify the result.
Adding the respective sides of our two substitution equations and then dividing both sides by \(2\) gives us the following.
\begin{equation*}
\frac{u+v}{2}=A
\end{equation*}
Subtracting the respective sides of the substitution equations (in the \(u-v\) direction) and then dividing both sides by \(2\) results in the following.
\begin{equation*}
\frac{u-v}{2}=B
\end{equation*}
Before making any actual substitutions, let's observe the following.
We begin by letting \(t=\frac{\alpha}{2}\text{.}\) Observe that this implies that \(2t=\alpha\text{.}\) Let's make the substitutions and then isolate the \(\cos\left(\frac{\alpha}{2}\right)\) term.
Apply the appropriate product-to-sum identity to the expression \(4\cos\left(97.5^{\circ}\right)\cos\left(-52.5^{\circ}\right)\) and completely simplify the result.
Apply the appropriate product-to-sum identity to the expression \(20\sin\left(195^{\circ}\right)\cos\left(105^{\circ}\right)\) and completely simplify the result.
Apply the appropriate sum-to-product identity to the expression \(\sin\left(\frac{5\pi}{12}\right)-\sin\left(\frac{\pi}{12}\right)\) and completely simplify the result.
Apply the appropriate sum-to-product identity to the expression \(\cos\left(255^{\circ}\right)+\cos\left(195^{\circ}\right)\) and completely simplify the result.
We begin by observing that \(112.5^{\circ}=\frac{225^{\circ}}{2}\text{.}\) Because the sign takes care of itself in the tangent half-angle identity,we can proceed directly to the determination of the exact value of \(\tan\left(122.5^{\circ}\right)\text{.}\)
