Trigonometric Periodicity Identities

The period of a function is the smallest amount it can be shifted while remaining the same function. In more formal terms, it is the smallest $p$ such that $f(n+p)=f(n)$ for all $n$. Intuitively, the period is a measure of a function "repeating" itself.

Trigonometric functions are the simplest examples of periodic functions, as they repeat themselves due to their interpretation on the unit circle.

The periodicity identities of trigonometric functions tell us that shifting the graph of a trigonometric function by a certain amount results in the same function.

The $\sin x$ and $\cos x$ functions as well as their respective reciprocals $\csc x$ and $\sec x$ all have a period of $2\pi,$ while $\tan x$ and $\cot x$ have a period of $\pi$. So the periodicity identities of the functions are

$\begin{aligned} \sin(x)&=\sin(x+2\pi) &\quad \csc(x)&=\csc(x+2\pi)\\ \cos(x)&=\cos(x+2\pi) &\quad \sec(x)&=\sec(x+2\pi)\\ \tan(x)&=\tan(x+\pi) &\quad \cot(x)&=\cot(x+\pi). \end{aligned}$

Analogously,

$\begin{aligned} \sin(x)&=-\sin(x+\pi) &\quad \csc(x)&=-\csc(x+\pi)\\ \cos(x)&=-\cos(x+\pi) &\quad \sec(x)&=-\sec(x+\pi). \end{aligned}$

So basically, if we know the value of the function from $0$ to $2\pi$ for the first 3 functions, we can find the value of the function at any value. More clearly, we can think of the functions as the values of a unit circle.

The above figure shows that the sine and cosine functions repeat every time we go around the unit circle. In fact we can go around $4\pi, 6\pi, 8\pi, \ldots, 2k\pi$ for a positive integer $k$ and still get the same function. So a more general form of the periodicity identity will be:

$\begin{aligned} \sin(x)&=\sin(x+2k\pi) &\quad \csc(x)&=\csc(x+2k\pi)\\ \cos(x)&=\cos(x+2k\pi) &\quad \sec(x)&=\sec(x+2k\pi)\\ \tan(x)&=\tan(x+k\pi) &\quad \cot(x)&=\cot(x+k\pi). \end{aligned}$

Find the value of $2\sin(11\pi+x)$.

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We have

$\begin{aligned} 2\sin(11\pi +x) &=2\sin(2\cdot5\pi +\pi +x)\\ &=2\sin(\pi +x)\\ &=-2\sin x. \ _\square \end{aligned}$

Find the value of $2\cos \left(\frac{19}{3}\pi\right)+\sin \left(\frac{7}{2}\pi\right)$.

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We have

$\begin{aligned} 2 \cos\left( \frac { 19 }{ 3 } \pi \right) +\sin\left( \frac { 7 }{ 2 } \pi \right) &=2\cos\left( 6\pi +\frac { \pi }{ 3 } \right) +\sin\left( 3\pi +\frac { \pi }{ 2 } \right) \\ &=2\cos\left( \frac { \pi }{ 3 } \right) +\sin\left( 2\pi +\pi +\frac { \pi }{ 2 } \right) \\ &=2\cos\left( \frac { \pi }{ 3 } \right) +\sin\left( \pi +\frac { \pi }{ 2 } \right) \\ &=2\cos\left( \frac { \pi }{ 3 } \right) -\sin\left( \frac { \pi }{ 2 } \right) \\ &=1-1\\&=0. \ _\square \end{aligned}$

Find the value of $\displaystyle \frac{1-\cos^{2}(7\pi+x)}{\cos^{2}(-8\pi-x)}$.

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We have

$\begin{aligned} \frac { 1-\cos^{ 2 }(7\pi +x) }{ \cos^{ 2 }(-8\pi -x) } &=\frac { 1-{ \cos }^{ 2 }(6\pi +\pi +x) }{ { \cos }^{ 2 }(-x) } \\ &=\frac { 1-{ \cos }^{ 2 }(\pi +x) }{ { \cos }^{ 2 }x } \\ &=\frac { 1-{ (\cos x) }^{ 2 } }{ { \cos }^{ 2 }x } \\ &=\frac { 1-{ \cos }^{ 2 }x }{ { \cos }^{ 2 }x } \\ &=\frac { {\sin }^{ 2 }x }{ { \cos }^{ 2 }x } \\&={ \tan }^{ 2 }x. \ _\square \end{aligned}$

• Trigonometry
• Functional Equations - Periodic Functions