Exact trigonometric values Worked Solutions — GCSE Maths

Derive the $30$ and $60$ degree values from half an equilateral triangle.

Cut an equilateral triangle of side $2$ in half. The half has a hypotenuse of $2$, a base of $1$ and, by Pythagoras, a height of $\sqrt{2^2 - 1^2} = \sqrt{3}$. Its angles are $30^\circ$, $60^\circ$ and $90^\circ$, so every exact value for $30^\circ$ and $60^\circ$ is read straight off it.

Write the ratio for sin $30$ degrees using the sides of that triangle.

For the $30^\circ$ angle in that triangle, opposite over hypotenuse gives $\frac{1}{2}$, already in simplest form with a rational denominator.

State the exact value.

The exact value is $\frac{1}{2}$, in simplest form with a rational denominator.

Derive the $30$ and $60$ degree values from half an equilateral triangle.

Cut an equilateral triangle of side $2$ in half. The half has a hypotenuse of $2$, a base of $1$ and, by Pythagoras, a height of $\sqrt{2^2 - 1^2} = \sqrt{3}$. Its angles are $30^\circ$, $60^\circ$ and $90^\circ$, so every exact value for $30^\circ$ and $60^\circ$ is read straight off it.

Write the ratio for cos $30$ degrees using the sides of that triangle.

For the $30^\circ$ angle in that triangle, adjacent over hypotenuse gives $\frac{\sqrt{3}}{2}$, already in simplest form with a rational denominator.

State the exact value.

The exact value is $\frac{\sqrt{3}}{2}$, in simplest form with a rational denominator.

Derive the $30$ and $60$ degree values from half an equilateral triangle.

Cut an equilateral triangle of side $2$ in half. The half has a hypotenuse of $2$, a base of $1$ and, by Pythagoras, a height of $\sqrt{2^2 - 1^2} = \sqrt{3}$. Its angles are $30^\circ$, $60^\circ$ and $90^\circ$, so every exact value for $30^\circ$ and $60^\circ$ is read straight off it.

Write the ratio for tan $30$ degrees using the sides of that triangle.

For the $30^\circ$ angle in that triangle, opposite over adjacent gives $\frac{\sqrt{3}}{3}$, already in simplest form with a rational denominator.

State the exact value.

The exact value is $\frac{\sqrt{3}}{3}$, in simplest form with a rational denominator.

Derive the $45$ degree values from a right-angled isosceles triangle.

Take a right-angled triangle with two sides of $1$. Its two other angles are equal, so each is $45^\circ$, and its hypotenuse is $\sqrt{1^2 + 1^2} = \sqrt{2}$. So $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ and $\tan 45^\circ = \frac{1}{1} = 1$.

Write the ratio for sin $45$ degrees using the sides of that triangle.

For the $45^\circ$ angle in that triangle, opposite over hypotenuse gives $\frac{\sqrt{2}}{2}$, already in simplest form with a rational denominator.

State the exact value.

The exact value is $\frac{\sqrt{2}}{2}$, in simplest form with a rational denominator.

Derive the $45$ degree values from a right-angled isosceles triangle.

Take a right-angled triangle with two sides of $1$. Its two other angles are equal, so each is $45^\circ$, and its hypotenuse is $\sqrt{1^2 + 1^2} = \sqrt{2}$. So $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ and $\tan 45^\circ = \frac{1}{1} = 1$.

Write the ratio for cos $45$ degrees using the sides of that triangle.

For the $45^\circ$ angle in that triangle, adjacent over hypotenuse gives $\frac{\sqrt{2}}{2}$, already in simplest form with a rational denominator.

State the exact value.

The exact value is $\frac{\sqrt{2}}{2}$, in simplest form with a rational denominator.