Tangent and the Exact Ratios

Write out $\dfrac{\text{opposite}}{\text{hypotenuse}}$ divided by $\dfrac{\text{adjacent}}{\text{hypotenuse}}$ as a single fraction, and simplify it. What have you ended up with, and what does that tell you about how the three ratios are related?

Tangent is not independent: $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$. And two triangles, from a square and an equilateral triangle, generate every exact ratio for $30°$, $45°$ and $60°$.

In a right-angled triangle the three ratios are

$$\sin\theta = \frac{o}{h}, \qquad \cos\theta = \frac{a}{h}, \qquad \tan\theta = \frac{o}{a}$$

writing $o$, $a$ and $h$ for the opposite, adjacent and hypotenuse. Divide the first by the second:

$$\frac{\sin\theta}{\cos\theta} = \frac{o/h}{a/h} = \frac{o}{h} \times \frac{h}{a} = \frac{o}{a} = \tan\theta$$

The hypotenuse cancels, which is the whole trick. So tangent was never a third independent ratio.

The same holds on the unit circle, where the point is $(\cos\theta, \sin\theta)$. The tangent is then

$$\tan\theta = \frac{\text{second coordinate}}{\text{first coordinate}}$$

which is the gradient of the radius joining the origin to that point, since gradient is rise over run. Tangent is the steepness of the radius, and that reading is what Lesson 8 uses to connect trigonometry to the gradient of a line.

Take a square of side $1$ and cut it along a diagonal. The result is a right-angled triangle with two legs of length $1$ and two equal angles.

The two acute angles are equal and sum to $90°$, so each is $45°$. Pythagoras gives the hypotenuse:

$$h^2 = 1^2 + 1^2 = 2 \quad \Longrightarrow \quad h = \sqrt{2}$$

Now read the ratios for $45°$ straight off the triangle. The side opposite is $1$, the side adjacent is $1$, and the hypotenuse is $\sqrt{2}$:

$$\sin 45° = \frac{1}{\sqrt{2}}, \qquad \cos 45° = \frac{1}{\sqrt{2}}, \qquad \tan 45° = \frac{1}{1} = 1$$

Sine and cosine are equal, which makes sense: the triangle is symmetric, so swapping which leg is opposite changes nothing.

The form $\dfrac{1}{\sqrt{2}}$ is often rewritten with no surd in the denominator, by multiplying top and bottom by $\sqrt{2}$:

$$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

Both forms are correct and equal to about $0.707$. Use whichever the question uses, and be able to move between them.

Take an equilateral triangle of side $2$ and cut it down the middle, from one vertex to the midpoint of the opposite side.

Each angle of an equilateral triangle is $60°$, and the cut bisects one of them into two angles of $30°$. The cut is perpendicular to the base, so the half-triangle is right-angled, with hypotenuse $2$ and short side $1$, being half of the original side. Pythagoras gives the third side:

$$x^2 = 2^2 - 1^2 = 3 \quad \Longrightarrow \quad x = \sqrt{3}$$

So the triangle has sides $1$, $\sqrt{3}$ and $2$, with the $30°$ angle opposite the side of length $1$ and the $60°$ angle opposite the side of length $\sqrt{3}$.

Reading off both angles:

$$\sin 30° = \tfrac{1}{2}, \qquad \cos 30° = \tfrac{\sqrt{3}}{2}, \qquad \tan 30° = \tfrac{1}{\sqrt{3}}$$

$$\sin 60° = \tfrac{\sqrt{3}}{2}, \qquad \cos 60° = \tfrac{1}{2}, \qquad \tan 60° = \sqrt{3}$$

Notice the pattern: the sine of one angle equals the cosine of the other, because the side opposite one is the side adjacent to the other. That relationship is proved in general in Lesson 5.

Collected together, with the sines written over a common denominator to show the pattern:

$$\sin 30° = \frac{1}{2}, \qquad \sin 45° = \frac{\sqrt{2}}{2}, \qquad \sin 60° = \frac{\sqrt{3}}{2}$$

$$\cos 30° = \frac{\sqrt{3}}{2}, \qquad \cos 45° = \frac{\sqrt{2}}{2}, \qquad \cos 60° = \frac{1}{2}$$

$$\tan 30° = \frac{1}{\sqrt{3}}, \qquad \tan 45° = 1, \qquad \tan 60° = \sqrt{3}$$

Written like this the sines read $\dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}$ as the angle increases, and the cosines read the same three values backwards. That pattern is a useful check, though not a proof.

The tangents can be recovered from the other two without any triangle at all, using the quotient relationship. For instance

$$\tan 60° = \frac{\sin 60°}{\cos 60°} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$$

The recommended method remains: draw the two triangles, label them, and read what you need. It takes seconds, it cannot be misremembered, and it works under exam pressure when a memorised table often does not.

Five items on tangent and the special angles. Work each one, then reveal the answer.

1. 1 State $\tan 45°$.

   Show answer
   Both legs of the half-square are equal.$1$

2. 2 State $\cos 60°$ exactly.

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   Adjacent over hypotenuse in the $30$-$60$-$90$ triangle.$\tfrac{1}{2}$

3. 3 State $\sin 60°$ exactly.

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   Opposite over hypotenuse: $\sqrt{3}$ over $2$.$\tfrac{\sqrt{3}}{2}$

4. 4 Why is $\tan 90°$ undefined?

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   It is a quotient with cosine underneath.$\cos 90° = 0$

5. 5 Rationalise $\dfrac{1}{\sqrt{3}}$.

   Show answer
   Multiply top and bottom by $\sqrt{3}$.$\tfrac{\sqrt{3}}{3}$