The tangent function

The line $\displaystyle {x}={1}$ touches the unit circle at the point $\displaystyle (1,0)$: it is a tangent to the circle. The ray from the origin through $\displaystyle P(t)$, extended, meets the tangent at a point $\displaystyle T$.

The tangent of $\displaystyle t$, $\displaystyle \tan t$, is the $\displaystyle y$-coordinate of $\displaystyle T$.

On the board: the unit circle, the tangent line $\displaystyle {x}={1}$, $\displaystyle {P}={P(0.9)}$, and the ray through it to $\displaystyle T$. When the point $\displaystyle P$ is on the $\displaystyle y$-axis, that is $\displaystyle {t}={\frac{\pi}{2}+\pi k}$, the ray is parallel to the tangent line and does not meet it: there $\displaystyle t$ has no tangent.

The triangles $\displaystyle OQP$ and $\displaystyle O(1,0)T$ on the board are similar: both have a right angle and share the angle at $\displaystyle O$. So their sides are in the same ratio:

• in $\displaystyle OQP$: the upright leg $\displaystyle \sin t$ and the level leg $\displaystyle \cos t$;
• in $\displaystyle O(1,0)T$: the upright leg $\displaystyle \tan t$ and the level leg $\displaystyle 1$.

The equality holds in every quadrant, for every $\displaystyle t$ with $\displaystyle \cos t\neq0$.

From the quotient and the table of values:

• The sign: the tangent is positive in the first and third quadrants (sine and cosine have the same sign), and negative in the second and fourth.
• $\displaystyle P(t+\pi)$ is the reflection of $\displaystyle P(t)$ in the origin: the same ray, the same point $\displaystyle T$. So the period of the tangent is $\displaystyle \pi$: $\displaystyle {\tan(t+\pi)}={\tan t}$.
• $\displaystyle {\tan(-t)}={\frac{-\sin t}{\cos t}}={-\tan t}$: the tangent is an odd function.

On the board: $\displaystyle {y}={\tan x}$, and the dashed lines $\displaystyle {x}={\pm\frac{\pi}{2}}$, $\displaystyle {x}={\pm\frac{3\pi}{2}}$.

• Domain: $\displaystyle x\neq\frac{\pi}{2}+\pi k$; range: every number.
• As $\displaystyle x$ comes close to $\displaystyle \frac{\pi}{2}$ from the left, $\displaystyle \cos x$ is close to $\displaystyle 0$ and positive, and $\displaystyle \tan x$ grows without limit; from the right – it falls without limit. The line $\displaystyle {x}={\frac{\pi}{2}}$ is a vertical asymptote, and after it – every $\displaystyle \pi$ – another one.
• Between two asymptotes the graph rises all the time.
• Zeros: $\displaystyle {x}={\pi k}$.

Divide both sides of $\displaystyle {\sin^2t+\cos^2t}={1}$ by $\displaystyle \cos^2t$ (where $\displaystyle \cos t\neq0$):

With it you go from the tangent to the cosine. For example, if $\displaystyle {\tan t}={\frac{3}{4}}$ and $\displaystyle P(t)$ is in the first quadrant: $\displaystyle {\frac{1}{\cos^2t}}={1+\frac{9}{16}}={\frac{25}{16}}$, so $\displaystyle {\cos t}={\frac{4}{5}}$ and $\displaystyle {\sin t}={\tan t\cdot\cos t}={\frac{3}{5}}$.

Find the exact value.

$\displaystyle {\tan t}={2}$, and the point $\displaystyle P(t)$ is in the third quadrant.

$\displaystyle x$ is $\displaystyle \cos t$. Solve the equation the identity $\displaystyle {1+\tan^2t}={\frac{1}{\cos^2t}}$ gives.

What is $\displaystyle \cos t$, and what is $\displaystyle \sin t$? Explain the signs.

Write the expression as one trigonometric function of $\displaystyle x$.

Mark on the board all the zeros of $\displaystyle {y}={\tan x}$ in the part shown.