4.1.1 Basic Knowledge
We will use the definition of sine and cosine via the unit circle. We will also use the values of all six trigonometric functions at special angles (whenever defined), e.g., \(\displaystyle \frac{\pi}{6}\), \(\displaystyle \frac{\pi}{4}\), \(\displaystyle \frac{\pi}{3}\), \(\displaystyle \frac{\pi}{2}\), \(\pi\), and their multiples. Make sure to draw the two essential triangles to obtain these values.
The Pythagorean identities are used often:
\[\sin^2 x + \cos^2 x=1 \;\;\;\;\; \tan^2 x+1 = \sec^2 x \;\;\;\;\; 1 + \cot^2x= \csc^2x.\]
On Your Own 1. (a) Prove \(\sin^2 x + \cos^2 x=1\);
(b) use part (a) to show the validity of \(\tan^2 x+1 = \sec^2 x\) and \(1 + \cot^2x= \csc^2x.\).
4.1.2 Addition Difference Formulas
\[\sin (\alpha \pm \beta )=\sin \alpha \cos \beta \pm \sin \beta \cos \alpha \;\;\; \text{ and } \;\;\; \cos (\alpha \pm \beta )=\cos
\alpha \cos \beta \mp \sin \alpha \sin \beta.\]
Example 1. Derive the addition formula for cotangent from the addition formulas for sine and cosine.
Solution.
\begin{align*}
\cot (x+h)&=\frac{\cos (x+h)}{\sin (x+h)}
=\frac{\cos x \cos h-\sin x \sin h}{\sin x \cos h + \sin h \cos x}
\,\underset{(*)}{=}\,\frac{\displaystyle \frac{\cos x \cos h-\sin x \sin h}{\sin x \sin h}}{\displaystyle \frac{\sin x \cos h+ \sin h \cos x}{\sin x \sin h}}=\\
& =\frac{\displaystyle \frac{\cos x \cos h}{\sin x \sin h}-1}{\displaystyle \frac{\cos h}{\sin h}+\frac{\cos x}{\sin x}}
=\frac{\cot x \cot h-1}{\cot h + \cot x}
\end{align*}
\((*)\) It is natural to divide numerator and denominator by \(\sin x \sin h\) because the aim is to get a new expression for \(\cot (x + h)\) in terms of \(\cot x\) and \(\cot h\).
On Your Own 2. Show that \(\displaystyle \tan(x+h)= \frac{\tan x + \tan h}{1-\tan x \tan h}\).
Double Angle Identities
\[\sin 2x = 2\sin x \cos x\;\; \text{and} \;\; \cos 2x = \cos^2x-\sin^2 x= 2\cos^2x-1=1-2\sin^2x.\]
On Your Own 3. Prove the above double angle identities. Hint: \(\cos 2 x = \cos (x+x)\).
4.1.3 Products-to-Sum Identities
Given \(x,y \in \mathbb{R}\),
\begin{align*}
\sin x\cos y& = \frac{{\sin (x + y) + \sin (x – y)}}{2},\\
\cos x\cos y &= \frac{{\cos (x + y) + \cos (x – y)}}{2}, \\
\sin x\sin y &= \frac{{\cos (x – y) – \cos (x + y)}}{2}.
\end{align*}
For example, the proof of the third identity is as follows:
\begin{align*}
\frac{\cos (x – y) – \cos (x + y)}{2} &= \frac{\cos x\cos y + \sin x\sin y – (\cos x\cos y – \sin x\sin y)}{2} =\\
&= \frac{2\sin x\sin y}{2} = \sin x\sin y,
\end{align*}
On Your Own 4. Prove the remaining two identities.
4.1.4 Sum-to-Product Identities
Given \(x,y \in \mathbb{R}\), we have
\[\renewcommand{\arraystretch}{2}
\begin{array}{cll}
\displaystyle \sin x + \sin y = 2\sin \frac{x+y}{2}\cos \frac{x – y}{2}, &\hspace{.25in} &
\displaystyle \sin x – \sin y = 2\sin \frac{x – y}{2}\cos \frac{x+y}{2}, \\
\displaystyle \cos x + \cos y = 2\cos \frac{x+y}{2}\cos \frac{x – y}{2}, & &
\displaystyle \cos x – \cos y = – 2\sin \frac{x+y}{2}\sin \frac{x – y}{2}.
\end{array}
\]
Observation 1. \(\displaystyle x = \frac{x + y}{2} + \frac{x – y}{2},\;\;y = \frac{x + y}{2} – \frac{x – y}{2} \; \text{for all} \; x,y \in \mathbb{R}\).
Using the observation, we have
\[\sin x – \sin y
= \sin \left(\frac{x+y}{2}+\frac{x-y}{2}\right) – \sin \left(\frac{x+y}{2}-\frac{x-y}{2}\right) = \]
\[ =\left(\sin \frac{x+y}{2} \cos \frac{x-y}{2} + \sin \frac{x-y}{2} \cos \frac{x+y}{2} \right)
-\left( \sin \frac{x+y}{2} \cos \frac{x-y}{2} – \sin \frac{x-y}{2} \cos \frac{x+y}{2} \right) =\]
\[= 2 \sin \frac{x-y}{2} \cos \frac{x+y}{2}.\]
On Your Own 5. Prove the remaining three identities.