Introduction to Trigonometry: questions with answers

From $\sin \theta = \frac{5}{13}$ the side opposite to $\theta$ is $5k$, the hypotenuse is $13k$, and the other leg is $12k$. $\sec \theta$ is a RECIPROCAL ratio: the hypotenuse on top, $\dfrac{ 13k }{ 12k } = \frac{13}{12}$. The value $\frac{12}{13}$ is $\cos \theta$, the ratio that $\sec \theta$ is the reciprocal of; it has the right two sides but is inverted once too few.

Using $\sin^2 \theta + \cos^2 \theta = 1$, we get $\cos^2 \theta = 1 - \frac{81}{1681} = \frac{1600}{1681}$. Since $\theta$ is acute, $\cos \theta$ is positive, so $\cos \theta = \frac{40}{41}$. Therefore, $\sin \theta + \cos \theta = \frac{9}{41} + \frac{40}{41} = \frac{49}{41}$.

From the table, $\cos 60^\circ = \frac{1}{2}$, so $A + B = 60^\circ$; and $\sin 30^\circ = \frac{1}{2}$, so $A - B = 30^\circ$. These are the SUM and the DIFFERENCE of the two angles, not the angles themselves. Adding the two equations, $2A = 90^\circ$, so $A = 45^\circ$; subtracting, $2B = 30^\circ$, so $B = 15^\circ$. Check: $45^\circ + 15^\circ = 60^\circ$ and $45^\circ - 15^\circ = 30^\circ$. Reporting $A = 60^\circ$, $B = 30^\circ$ stops at the two table readings.

From the table, $\tan 60^\circ = \sqrt{3}$, so $A + B = 60^\circ$; and $\cot 30^\circ = \sqrt{3}$, so $A - B = 30^\circ$. These are the SUM and the DIFFERENCE of the two angles, not the angles themselves. Adding the two equations, $2A = 90^\circ$, so $A = 45^\circ$; subtracting, $2B = 30^\circ$, so $B = 15^\circ$. Check: $45^\circ + 15^\circ = 60^\circ$ and $45^\circ - 15^\circ = 30^\circ$. Reporting $A = 60^\circ$, $B = 30^\circ$ stops at the two table readings.

The words opposite and adjacent are always relative to the angle being considered. The side opposite to $\angle R$ does not touch $R$ at all. The side adjacent to $\angle R$ is the leg that runs along one arm of the angle. The other arm is always the hypotenuse. At vertex $R$, the two sides that meet are $RQ$ and the hypotenuse $PR$. So the side adjacent to $\angle R$ is $QR$. The side $PQ$ is instead the side adjacent to the other acute angle, $\angle P$. The names swap when the angle changes.

First the hypotenuse: $AC = \sqrt{ 24^2 + 7^2 } = \sqrt{ 625 } = 25$ cm. Now read the sides FROM $C$: the side opposite to $C$ is $AB$ and the side adjacent to $C$ is $BC$. So $\sin C = \dfrac{ AB }{AC} = \frac{24}{25}$. The value $\frac{7}{25}$ is $\sin A$, using $BC$ as the opposite side. That is right for angle $A$, but the question asks about $C$, and the two legs swap roles between the angles.

The right angle is at $C$, so $A$ and $B$ are the two acute angles that share the same two legs, with the roles exchanged. The side opposite to $A$ ($BC$) is the side adjacent to $B$, and the side adjacent to $A$ ($AC$) is the side opposite to $B$. Hence $\sin A = \dfrac{BC}{AB}$ and $\cos B = \dfrac{BC}{AB}$ are the SAME quotient: $\cos B = \frac{8}{17}$. Writing $\frac{17}{8}$ applies the reciprocal rule ($\cos = 1 / \sin$) across two different angles; that rule only holds for one and the same angle.

Each term is a SQUARED ratio, so the table value must be squared before it is multiplied: $\sin^2 30^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$, $\operatorname{cosec}^2 30^\circ = 2^2 = 4$ and $\cos^2 60^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$. Then the expression is $1 \times \frac{1}{4} + 2 \times 4 - 1 \times \frac{1}{4} = 8$. Using the unsquared values $\frac{1}{2}$, $2$ and $\frac{1}{2}$ instead gives $4$ -- the square on a ratio applies to the whole value, $\sin^2 30^\circ$ means $(\sin 30^\circ)^2$.

At the ends of the range, $\sin 90^\circ = 1$ and $\cos 0^\circ = 1$ (while $\sin 0^\circ = 0$ and $\cos 90^\circ = 0$). So $\cos^2 0^\circ = 1$ and $\sin^2 90^\circ = 1$, and the expression is $3 \times 1 - 4 \times 1 = -1$. Reading $\cos^2 0^\circ$ as $0$ gives $-4$ and reading $\sin^2 90^\circ$ as $0$ gives $3$: those are the values at the OTHER end of the range.

Take out the common factor $5$: the expression is $5(\cot^2 A - \operatorname{cosec}^2 A)$. Now $\operatorname{cosec}^2 A - \cot^2 A = 1$, so the bracket with the terms the other way round is $-1$, and the value is $5 \times (-1) = -5$. The answer $5$ keeps the sign of the identity; the order of the two squares decides the sign, and here the tangent (or cotangent) term comes first.

Replace each part by the identity it matches: $\frac{\operatorname{cosec}^2 A - 1}{\cot A} = \frac{\cot^2 A}{\cot A} = \cot A$. Expressions like this look heavy but collapse as soon as $1 + \tan^2 A$, $\sec^2 A - 1$ and $1 - \cos^2 A$ are recognised as $\sec^2 A$, $\tan^2 A$ and $\sin^2 A$.

Split the fraction over its denominator, term by term. The first term gives a tangent or a cotangent; the constant gives $\dfrac{1}{ \sin \theta } = \operatorname{cosec} \theta$, NOT $1$. So $\frac{1 - \cos \theta}{\sin \theta} = \operatorname{cosec} \theta - \cot \theta$. In $1 - \cot \theta$ the $1$ has been carried over the line untouched, as though dividing $1$ by $\sin \theta$ left it as $1$.

Each reciprocal is the conjugate: $\dfrac{1}{a - b} = a + b$ and $\dfrac{1}{a + b} = a - b$ whenever $(a + b)(a - b) = 1$, which holds for these pairs. So the expression is $(\operatorname{cosec} \theta + \cot \theta) + (\operatorname{cosec} \theta - \cot \theta) = 2 \operatorname{cosec} \theta$. Inverting term by term gives $(\sin \theta - \tan \theta) + (\sin \theta + \tan \theta) = 2 \sin \theta$, which is wrong because $\dfrac{1}{a - b}$ is not $\dfrac{1}{a} - \dfrac{1}{b}$.

Using standard values, $\sin 30^\circ + \sin 30^\circ = \frac{1}{2} + \frac{1}{2} = 1$. Also, $\sin(A + B) = \sin 60^\circ = \frac{\sqrt{3}}{2}$. Subtracting gives $(1) - (\frac{\sqrt{3}}{2}) = \frac{2 - \sqrt{3}}{2}$. The nonzero difference shows that $\sin(A + B)$ and $\sin A + \sin B$ are unequal for these angles.

Work both sides out from the table. $\cos 2A = \cos 60^\circ = \frac{1}{2}$, while $2\,\cos A = 2 \cos 30^\circ = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$. So $\cos 2A < 2\,\cos A$: they are not equal. Doubling the angle does not double the ratio -- a trigonometric ratio is not proportional to its angle, and $\cos 2A = 2\,\cos A$ holds only when $A = 0^\circ$.