Chapter 8: Introduction To Trigonometry

If \(\sin A=\frac{1}{3}\), find the other five trigonometric ratios of acute angle \(A\).

1. Given \(\sin A=\frac{1}{3}\), so the opposite side and hypotenuse are in the ratio \(1:3\). Choose \(BC=k\) and \(AC=3k\), where \(k>0\).
   $$\frac{BC}{AC}=\frac{1}{3},\quad BC=k,\quad AC=3k$$
2. Apply the Pythagorean theorem to find the adjacent side.
   $$AB^2=AC^2-BC^2=(3k)^2-k^2=8k^2,\quad AB=2\sqrt{2}\,k$$
3. Use the side lengths to calculate the remaining ratios.
   $$\cos A=\frac{2\sqrt{2}}{3},\quad \tan A=\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4},\quad \cosec A=3,\quad \sec A=\frac{3}{2\sqrt{2}}=\frac{3\sqrt{2}}{4},\quad \cot A=2\sqrt{2}$$

Given \(\tan A=\frac{4}{3}\), find the other five trigonometric ratios.

1. Given \(\tan A=\frac{4}{3}\), take the opposite and adjacent sides as \(4k\) and \(3k\), where \(k>0\).
   $$\text{opposite}=4k,\quad \text{adjacent}=3k$$
2. Use Pythagoras' theorem to find the hypotenuse.
   $$h=\sqrt{(4k)^2+(3k)^2}=5k$$
3. Find sine and cosine.
   $$\sin A=\frac{4k}{5k}=\frac{4}{5},\quad \cos A=\frac{3k}{5k}=\frac{3}{5}$$
4. Take reciprocals to find the remaining ratios.
   $$\cot A=\frac{3}{4},\quad \cosec A=\frac{5}{4},\quad \sec A=\frac{5}{3}$$

Prove that if acute angles \(B\) and \(Q\) satisfy \(\sin B=\sin Q\), then \(B=Q\).

1. Suppose \(\sin B=\sin Q\) for acute angles \(B\) and \(Q\). Write sine as opposite side over hypotenuse in each right triangle.
   $$\frac{x}{h_1}=\frac{y}{h_2}$$
2. The equal ratios mean the opposite sides and hypotenuses have a common scale factor \(k\).
   $$\frac{h_1}{h_2}=\frac{x}{y}=k$$
3. By Pythagoras' theorem, the adjacent sides have the same scale factor.
   $$\frac{\sqrt{h_1^2-x^2}}{\sqrt{h_2^2-y^2}}=k$$
4. All corresponding sides are proportional, so the right triangles are similar and the corresponding angles are equal.
   $$B=Q$$

In a right triangle, the hypotenuse is \(29\) and the side adjacent to \(\theta\) is \(21\). Find \(\cos^2\theta+\sin^2\theta\) and \(\cos^2\theta-\sin^2\theta\).

1. The hypotenuse is \(29\) and the side adjacent to \(\theta\) is \(21\). Find the opposite side.
   $$p=\sqrt{29^2-21^2}=\sqrt{(29-21)(29+21)}=20$$
2. Write the sine and cosine values.
   $$\sin\theta=\frac{20}{29},\quad \cos\theta=\frac{21}{29}$$
3. Evaluate the sum of the squares.
   $$\cos^2\theta+\sin^2\theta=\frac{21^2+20^2}{29^2}=1$$
4. Evaluate the difference of the squares.
   $$\cos^2\theta-\sin^2\theta=\frac{21^2-20^2}{29^2}=\frac{41}{841}$$

In a right triangle, if \(\tan A=1\), verify that \(2\sin A\cos A=1\).

1. Given \(\tan A=1\), take both legs to have length \(k\).
   $$\text{opposite}=\text{adjacent}=k$$
2. Find the hypotenuse using Pythagoras' theorem.
   $$h=\sqrt{k^2+k^2}=k\sqrt{2}$$
3. Calculate sine and cosine.
   $$\sin A=\frac{k}{k\sqrt{2}}=\frac{1}{\sqrt{2}},\quad \cos A=\frac{k}{k\sqrt{2}}=\frac{1}{\sqrt{2}}$$
4. Substitute into the required expression.
   $$2\sin A\cos A=2\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)=1$$

In right triangle \(OPQ\), right-angled at \(P\), \(OP=7\) cm and \(OQ-PQ=1\) cm. Find \(\sin Q\) and \(\cos Q\).

1. Let \(PQ=x\) cm. Since \(OQ-PQ=1\) cm, \(OQ=x+1\) cm; also \(OP=7\) cm.
   $$OQ=x+1$$
2. Apply Pythagoras' theorem and solve for \(x\).
   $$(x+1)^2=x^2+7^2\quad\Longrightarrow\quad 2x+1=49\quad\Longrightarrow\quad x=24$$
3. Thus the hypotenuse is \(25\) cm. Relative to angle \(Q\), the opposite side is \(OP\) and the adjacent side is \(PQ\).
   $$OQ=25$$
4. Calculate the required ratios.
   $$\sin Q=\frac{OP}{OQ}=\frac{7}{25},\quad \cos Q=\frac{PQ}{OQ}=\frac{24}{25}$$

In a right triangle whose acute angles are both \(45^\circ\), express the hypotenuse and the sine of an acute angle in terms of a leg \(a\).

1. Let the two equal legs of an isosceles right triangle be \(a\); find the hypotenuse and the sine of an acute angle.
   $$AB=BC=a$$
2. Apply Pythagoras to find the hypotenuse.
   $$AC=\sqrt{a^2+a^2}=a\sqrt{2}$$
3. Use opposite side divided by hypotenuse.
   $$\sin45^\circ=\frac{a}{a\sqrt{2}}=\frac{1}{\sqrt{2}}$$

Use an equilateral triangle of side \(2a\) to derive \(\sin30^\circ\).

1. Take an equilateral triangle of side \(2a\) and draw its altitude to form a right triangle.
   $$AB=2a$$
2. The altitude bisects the base.
   $$BD=a$$
3. Use Pythagoras to find the altitude.
   $$AD=\sqrt{(2a)^2-a^2}=a\sqrt{3}$$
4. For the \(30^\circ\) angle, the opposite side is \(a\) and the hypotenuse is \(2a\).
   $$\sin30^\circ=\frac{a}{2a}=\frac{1}{2}$$

Find \(\tan90^\circ\) and \(\operatorname{cosec}90^\circ\).

1. Use \(\tan A=\frac{\sin A}{\cos A}\) at \(A=90^\circ\).
   $$\tan90^\circ=\frac{1}{0}$$
2. Division by zero is undefined, so the tangent is not defined.
   $$\tan90^\circ\text{ is undefined}$$
3. The cosecant is the reciprocal of sine.
   $$\operatorname{cosec}90^\circ=\frac{1}{\sin90^\circ}=1$$

In \(\triangle ABC\), right-angled at \(B\), \(AB=5\) cm and \(\angle ACB=30^\circ\). Find \(BC\) and \(AC\).

1. In right triangle \(ABC\), \(AB=5\) cm and \(\angle ACB=30^\circ\). Side \(AB\) is opposite and \(BC\) is adjacent to angle \(C\).
   $$AB=5\text{ cm},\quad \angle ACB=30^\circ$$
2. Use tangent to find \(BC\).
   $$\tan30^\circ=\frac{AB}{BC}\quad\Rightarrow\quad\frac{1}{\sqrt{3}}=\frac{5}{BC}\quad\Rightarrow\quad BC=5\sqrt{3}\text{ cm}$$
3. Use sine to find the hypotenuse \(AC\).
   $$\sin30^\circ=\frac{AB}{AC}\quad\Rightarrow\quad\frac{1}{2}=\frac{5}{AC}\quad\Rightarrow\quad AC=10\text{ cm}$$

In \(\triangle PQR\), right-angled at \(Q\), \(PQ=3\) cm and \(PR=6\) cm. Determine \(\angle QPR\) and \(\angle PRQ\).

1. In right triangle \(PQR\), right-angled at \(Q\), use the given sides to find angle \(R\).
   $$PQ=3\text{ cm},\quad PR=6\text{ cm}$$
2. Since \(PQ\) is opposite angle \(R\) and \(PR\) is the hypotenuse, match the sine value.
   $$\sin R=\frac{PQ}{PR}=\frac{3}{6}=\frac{1}{2}$$
3. The standard angle with sine \(\frac{1}{2}\) is \(30^\circ\). The two acute angles of a right triangle sum to \(90^\circ\).
   $$\angle PRQ=30^\circ,\quad \angle QPR=90^\circ-30^\circ=60^\circ$$

If \(\sin(A-B)=\frac{1}{2}\), \(\cos(A+B)=\frac{1}{2}\), \(0^\circ<A+B\leq90^\circ\), and \(A>B\), find \(A\) and \(B\).

1. Since \(A>B\), the difference is positive; use the standard sine value in the permitted angle range.
   $$\sin(A-B)=\frac{1}{2}\quad\Rightarrow\quad A-B=30^\circ$$
2. The sum lies between \(0^\circ\) and \(90^\circ\), so use the corresponding cosine value.
   $$\cos(A+B)=\frac{1}{2}\quad\Rightarrow\quad A+B=60^\circ$$
3. Add and subtract the equations to solve for the angles.
   $$2A=90^\circ\Rightarrow A=45^\circ,\quad 2B=30^\circ\Rightarrow B=15^\circ$$

Prove that \(\sec A(1-\sin A)(\sec A+\tan A)=1\).

1. Rewrite secant and tangent using sine and cosine.
   $$\sec A(1-\sin A)(\sec A+\tan A)=\frac{1}{\cos A}(1-\sin A)\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)$$
2. Combine the factors and use the difference of squares.
   $$=\frac{(1-\sin A)(1+\sin A)}{\cos^2 A}=\frac{1-\sin^2 A}{\cos^2 A}$$
3. Apply \(\sin^2 A+\cos^2 A=1\).
   $$=\frac{\cos^2 A}{\cos^2 A}=1$$