Trigonometry: Solutions of Sure Questions (Exercise 8.2)

This lesson covers worked questions from Exercise 8.2 on trigonometric ratios of the standard angles. We use the known values of sine, cosine and tangent at $30^\circ$, $45^\circ$ and $60^\circ$ to find angles, evaluate expressions and prove a result.

Angles of triangle $PQR$

In right triangle $PQR$ with $\angle Q = 90^\circ$, we are given $PQ = 3$ cm and $PR = 6$ cm. Relative to angle $R$, the side $PQ$ is opposite and $PR$ is the hypotenuse, so

$\sin R = \frac{PQ}{PR} = \frac{3}{6} = \frac{1}{2}.$

Since $\sin 30^\circ = \tfrac{1}{2}$, we get $\angle R = 30^\circ$. Using the angle sum of a triangle,

$\angle P + \angle Q + \angle R = 180^\circ,$ $\angle P + 90^\circ + 30^\circ = 180^\circ,$

so $\angle P = 60^\circ$. Therefore $\angle QPR = 60^\circ$ and $\angle PRQ = 30^\circ$.

Solving for two angles from sine and cosine

We are given $\sin(A - B) = \tfrac{1}{2}$ and $\cos(A + B) = \tfrac{1}{2}$, with $0^\circ < A + B \le 90^\circ$ and $A > B$. Find $A$ and $B$.

Since $\sin(A - B) = \tfrac{1}{2}$ and $\sin 30^\circ = \tfrac{1}{2}$,

$A - B = 30^\circ. \qquad (1)$

Since $\cos(A + B) = \tfrac{1}{2}$ and $\cos 60^\circ = \tfrac{1}{2}$,

$A + B = 60^\circ. \qquad (2)$

Adding $(1)$ and $(2)$ gives $2A = 90^\circ$, so $A = 45^\circ$. Substituting into $(2)$, $45^\circ + B = 60^\circ$, so $B = 15^\circ$.

$A = 45^\circ, \qquad B = 15^\circ.$

Solving for two angles from tangent equations

We are given $\tan(A + B) = \sqrt{3}$ and $\tan(A - B) = \tfrac{1}{\sqrt{3}}$, with $0^\circ < A + B \le 90^\circ$ and $A > B$.

Since $\tan 60^\circ = \sqrt{3}$ and $\tan 30^\circ = \tfrac{1}{\sqrt{3}}$,

$A + B = 60^\circ, \qquad A - B = 30^\circ.$

Adding gives $2A = 90^\circ$, so $A = 45^\circ$, and then $B = 60^\circ - 45^\circ = 15^\circ$.

$A = 45^\circ, \qquad B = 15^\circ.$

Evaluating an expression with special angles

We simplify an expression that reduces to

$\frac{\sqrt{3}}{2\sqrt{2}\,(1 + \sqrt{3})}.$

Multiply the numerator and denominator by $(1 - \sqrt{3})$ to rationalise:

$\frac{\sqrt{3}\,(1 - \sqrt{3})}{2\sqrt{2}\,(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{\sqrt{3}\,(1 - \sqrt{3})}{2\sqrt{2}\,(1 - 3)} = \frac{\sqrt{3} - 3}{-4\sqrt{2}}.$

Now multiply the numerator and denominator by $\sqrt{2}$:

$\frac{(\sqrt{3} - 3)\sqrt{2}}{-4 \cdot 2} = \frac{\sqrt{6} - 3\sqrt{2}}{-8} = \frac{3\sqrt{2} - \sqrt{6}}{8}.$

A fraction of squared ratios

We evaluate

$\frac{5\sin^2 30^\circ + 4\sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}.$

Using $\sin 30^\circ = \tfrac{1}{2}$, $\sec 30^\circ = \tfrac{2}{\sqrt{3}}$, $\tan 45^\circ = 1$ and $\cos 30^\circ = \tfrac{\sqrt{3}}{2}$, the numerator is

$5\left(\tfrac{1}{2}\right)^2 + 4\left(\tfrac{2}{\sqrt{3}}\right)^2 - 1^2 = \frac{5}{4} + \frac{16}{3} - 1.$

The denominator is $\left(\tfrac{1}{2}\right)^2 + \left(\tfrac{\sqrt{3}}{2}\right)^2 = \tfrac{1}{4} + \tfrac{3}{4} = 1$. Taking the lowest common denominator $12$ in the numerator,

$\frac{15 + 64 - 12}{12} = \frac{67}{12}.$

Proving the double-angle result for $\sin 60^\circ$

We prove that

$\frac{2\tan 30^\circ}{1 + \tan^2 30^\circ} = \sin 60^\circ.$

Left side

With $\tan 30^\circ = \tfrac{1}{\sqrt{3}}$,

$\frac{2 \cdot \tfrac{1}{\sqrt{3}}}{1 + \left(\tfrac{1}{\sqrt{3}}\right)^2} = \frac{\tfrac{2}{\sqrt{3}}}{1 + \tfrac{1}{3}} = \frac{\tfrac{2}{\sqrt{3}}}{\tfrac{4}{3}} = \frac{2}{\sqrt{3}} \cdot \frac{3}{4} = \frac{\sqrt{3}}{2}.$

Right side

$\sin 60^\circ = \tfrac{\sqrt{3}}{2}$. Both sides equal $\tfrac{\sqrt{3}}{2}$, so the result is proved.

A quick check on $\sin 2A = 2\sin A$

The equation $\sin 2A = 2\sin A$ holds when $A = 0^\circ$: the left side is $\sin 0^\circ = 0$ and the right side is $2\sin 0^\circ = 0$, so both sides agree.

A final evaluation with standard angles

We evaluate

$\frac{\sin 30^\circ + \tan 45^\circ - \sec 30^\circ}{\sec 30^\circ + \cos 60^\circ + \tan 45^\circ}.$

Using the standard values, the numerator is $\tfrac{1}{2} + 1 - \tfrac{2}{\sqrt{3}} = \tfrac{3}{2} - \tfrac{2}{\sqrt{3}}$ and the denominator is $\tfrac{2}{\sqrt{3}} + \tfrac{1}{2} + 1 = \tfrac{2}{\sqrt{3}} + \tfrac{3}{2}$. Writing each over the common denominator $2\sqrt{3}$,

$\frac{3\sqrt{3} - 4}{2\sqrt{3}} \div \frac{4 + 3\sqrt{3}}{2\sqrt{3}} = \frac{3\sqrt{3} - 4}{4 + 3\sqrt{3}}.$

Multiply the numerator and denominator by $(4 - 3\sqrt{3})$:

$\frac{(3\sqrt{3} - 4)(4 - 3\sqrt{3})}{(4 + 3\sqrt{3})(4 - 3\sqrt{3})} = \frac{24\sqrt{3} - 27 - 16}{16 - 27} = \frac{24\sqrt{3} - 43}{-11} = \frac{43 - 24\sqrt{3}}{11}.$

Key takeaways

• A single side ratio in a right triangle fixes one angle, and the angle sum then gives the rest.
• Sum-and-difference systems for two angles solve quickly by adding and subtracting the equations.
• Expressions in special angles are evaluated by substituting the standard values and then rationalising any surds.