Trigonometric Functions, Worked Examples (KBAT)

Expand the target form with the addition formula for sine:

$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$

So $R\sin(x+\alpha)=R\sin x\cos\alpha+R\cos x\sin\alpha$. Match the coefficients of $\sin x$ and $\cos x$ against $3\sin x+4\cos x$:

$$R\cos\alpha=3,\qquad R\sin\alpha=4$$

Square and add the two equations; the right side collapses because $\cos^{2}\alpha+\sin^{2}\alpha=1$:

$$R^{2}\cos^{2}\alpha+R^{2}\sin^{2}\alpha=3^{2}+4^{2}\ \Rightarrow\ R^{2}=25\ \Rightarrow\ R=5$$

Divide the two equations to remove $R$ and find $\alpha$:

$$\dfrac{R\sin\alpha}{R\cos\alpha}=\tan\alpha=\dfrac{4}{3}\ \Rightarrow\ \alpha=53.13^{\circ}$$

Both $R\cos\alpha$ and $R\sin\alpha$ are positive, so $\alpha$ is in the first quadrant, consistent with $0^{\circ}<\alpha<90^{\circ}$. The expression is therefore:

$$3\sin x+4\cos x=5\sin(x+53.13^{\circ})$$

Now use this to solve the equation. Replace the left side and isolate the sine:

$$5\sin(x+53.13^{\circ})=2.5\ \Rightarrow\ \sin(x+53.13^{\circ})=0.5$$

Let $u=x+53.13^{\circ}$. As $x$ runs from $0^{\circ}$ to $360^{\circ}$, $u$ runs from $53.13^{\circ}$ to $413.13^{\circ}$.

Solve $\sin u=0.5$ in that interval; the basic angle is $30^{\circ}$ and sine is positive in the first and second quadrants, giving candidates $30^{\circ},\ 150^{\circ},\ 390^{\circ}$. Keep only those inside $[53.13^{\circ},413.13^{\circ}]$:

$$u=150^{\circ}\quad\text{or}\quad u=390^{\circ}$$

The candidate $u=30^{\circ}$ is below $53.13^{\circ}$, so it is discarded. Subtract $53.13^{\circ}$ from each remaining value to recover $x$:

$$x=150^{\circ}-53.13^{\circ}=96.87^{\circ}\quad\text{or}\quad x=390^{\circ}-53.13^{\circ}=336.87^{\circ}$$

The two terms use different angles, $2x$ and $x$, so they cannot be combined yet. Rewrite $\cos 2x$ using the double-angle form written purely in sine, so the whole equation is in $\sin x$:

$$\cos 2A=1-2\sin^{2}A$$

Substitute $\cos 2x=1-2\sin^{2}x$ into the equation:

$$1-2\sin^{2}x+\sin x=0$$

Multiply through by $-1$ and arrange as a quadratic in $\sin x$:

$$2\sin^{2}x-\sin x-1=0$$

Factorise with $s=\sin x$:

$$(2s+1)(s-1)=0\ \Rightarrow\ \sin x=-\dfrac{1}{2}\ \text{or}\ \sin x=1$$

For $\sin x=-\dfrac{1}{2}$: the basic angle is $30^{\circ}$, and sine is negative in the third and fourth quadrants:

$$x=180^{\circ}+30^{\circ}=210^{\circ}\quad\text{or}\quad x=360^{\circ}-30^{\circ}=330^{\circ}$$

For $\sin x=1$: sine reaches its greatest value once in a full turn:

$$x=90^{\circ}$$

(a) Replace each double angle with a formula. For the numerator use $\sin 2\theta=2\sin\theta\cos\theta$; for the denominator choose the version of $\cos 2\theta$ that makes $1+\cos 2\theta$ collapse into a single square:

$$\sin 2A=2\sin A\cos A$$

$$\cos 2A=2\cos^{2}A-1$$

With that choice, $1+\cos 2\theta=1+(2\cos^{2}\theta-1)=2\cos^{2}\theta$. Substitute both results into the left-hand side:

$$\text{LHS}=\dfrac{2\sin\theta\cos\theta}{2\cos^{2}\theta}$$

Cancel the common factor $2\cos\theta$:

$$=\dfrac{\sin\theta}{\cos\theta}=\tan\theta=\text{RHS}$$

(b) Put $\theta=15^{\circ}$, so $2\theta=30^{\circ}$. The right side becomes $\tan 15^{\circ}$, and the left side uses the exact values $\sin 30^{\circ}=\dfrac{1}{2}$ and $\cos 30^{\circ}=\dfrac{\sqrt{3}}{2}$:

$$\tan 15^{\circ}=\dfrac{\sin 30^{\circ}}{1+\cos 30^{\circ}}=\dfrac{\tfrac{1}{2}}{1+\tfrac{\sqrt{3}}{2}}$$

Write the denominator as a single fraction $\dfrac{2+\sqrt{3}}{2}$, then divide the fractions:

$$=\dfrac{\tfrac{1}{2}}{\tfrac{2+\sqrt{3}}{2}}=\dfrac{1}{2+\sqrt{3}}$$

Rationalise by multiplying top and bottom by the conjugate $2-\sqrt{3}$:

$$=\dfrac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}=\dfrac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$$

Sketch the quadrant of each angle first. $A$ is obtuse, so $\cos A$ is negative; $B$ is acute, so $\sin B$ is positive.

Use $\sin^{2}\theta+\cos^{2}\theta=1$ to find each missing ratio.

$$\cos A=-\sqrt{1-\left(\dfrac{3}{5}\right)^{2}}=-\dfrac{4}{5}\qquad \sin B=\sqrt{1-\left(\dfrac{12}{13}\right)^{2}}=\dfrac{5}{13}$$

(a) Expand $\cos(A+B)$ with the addition formula and substitute all four ratios:

$$\cos(A+B)=\cos A\cos B-\sin A\sin B$$

$$=\left(-\dfrac{4}{5}\right)\left(\dfrac{12}{13}\right)-\left(\dfrac{3}{5}\right)\left(\dfrac{5}{13}\right)=-\dfrac{48}{65}-\dfrac{15}{65}=-\dfrac{63}{65}$$

(b) For $\tan(A-B)$, first convert both ratios to tangents:

$$\tan A=\dfrac{3/5}{-4/5}=-\dfrac{3}{4}\qquad \tan B=\dfrac{5/13}{12/13}=\dfrac{5}{12}$$

Substitute into the tangent subtraction formula:

$$\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}=\dfrac{-\tfrac{3}{4}-\tfrac{5}{12}}{1+\left(-\tfrac{3}{4}\right)\left(\tfrac{5}{12}\right)}=\dfrac{-\tfrac{7}{6}}{\tfrac{11}{16}}=-\dfrac{56}{33}$$

Both terms should be written in $\tan x$ alone. Replace $\sec^{2}x$ using the identity that links it to $\tan x$:

$$1+\tan^{2}x=\sec^{2}x$$

Substitute $\sec^{2}x=1+\tan^{2}x$ into the equation:

$$2(1+\tan^{2}x)-\tan x=8$$

Expand and arrange as a quadratic in $\tan x$:

$$2\tan^{2}x-\tan x-6=0$$

Solve with the quadratic formula, letting $t=\tan x$:

$$t=\dfrac{1\pm\sqrt{1+48}}{4}=\dfrac{1\pm7}{4}\ \Rightarrow\ \tan x=2\ \text{or}\ \tan x=-1.5$$

For $\tan x=2$ (positive, first and third quadrants), the basic angle is $63.43^{\circ}$:

$$x=63.43^{\circ}\quad\text{or}\quad x=180^{\circ}+63.43^{\circ}=243.43^{\circ}$$

For $\tan x=-1.5$ (negative, second and fourth quadrants), the basic angle is $56.31^{\circ}$:

$$x=180^{\circ}-56.31^{\circ}=123.69^{\circ}\quad\text{or}\quad x=360^{\circ}-56.31^{\circ}=303.69^{\circ}$$

(a) Write $75^{\circ}$ as the sum of two angles whose exact ratios are known:

$$75^{\circ}=45^{\circ}+30^{\circ}$$

Expand with the addition formula for sine and substitute the exact values $\sin45^{\circ}=\cos45^{\circ}=\dfrac{\sqrt2}{2}$, $\sin30^{\circ}=\dfrac{1}{2}$, $\cos30^{\circ}=\dfrac{\sqrt3}{2}$:

$$\sin75^{\circ}=\sin45^{\circ}\cos30^{\circ}+\cos45^{\circ}\sin30^{\circ}=\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}+\dfrac{\sqrt2}{2}\cdot\dfrac{1}{2}$$

$$=\dfrac{\sqrt6}{4}+\dfrac{\sqrt2}{4}=\dfrac{\sqrt6+\sqrt2}{4}$$

(b) $165^{\circ}=180^{\circ}-15^{\circ}$, so use the second-quadrant symmetry $\cos(180^{\circ}-\theta)=-\cos\theta$:

$$\cos165^{\circ}=\cos(180^{\circ}-15^{\circ})=-\cos15^{\circ}$$

Now link $\cos15^{\circ}$ to the result from (a) using the complementary-angle rule $\cos(90^{\circ}-\theta)=\sin\theta$:

$$\cos15^{\circ}=\cos(90^{\circ}-75^{\circ})=\sin75^{\circ}=\dfrac{\sqrt6+\sqrt2}{4}$$

$$\therefore\ \cos165^{\circ}=-\dfrac{\sqrt6+\sqrt2}{4}$$

(a) Apply the double-angle formula for cosine to the half angle $\dfrac{\theta}{2}$, so that $2\times\dfrac{\theta}{2}=\theta$:

$$\cos\theta=1-2\sin^{2}\dfrac{\theta}{2}$$

Substitute $\cos\theta=\dfrac{7}{25}$ and rearrange for $\sin^{2}\dfrac{\theta}{2}$:

$$\dfrac{7}{25}=1-2\sin^{2}\dfrac{\theta}{2}\ \Rightarrow\ \sin^{2}\dfrac{\theta}{2}=\dfrac{1}{2}\left(1-\dfrac{7}{25}\right)=\dfrac{9}{25}$$

Since $0^{\circ}<\theta<90^{\circ}$, the half angle $\dfrac{\theta}{2}$ lies between $0^{\circ}$ and $45^{\circ}$, a first-quadrant angle, so its sine is positive:

$$\sin\dfrac{\theta}{2}=\dfrac{3}{5}$$

(b) Find $\cos\dfrac{\theta}{2}$ with the Pythagorean identity, again taking the positive root:

$$\cos\dfrac{\theta}{2}=\sqrt{1-\dfrac{9}{25}}=\dfrac{4}{5}$$

Divide to get the tangent:

$$\tan\dfrac{\theta}{2}=\dfrac{\sin\tfrac{\theta}{2}}{\cos\tfrac{\theta}{2}}=\dfrac{3/5}{4/5}=\dfrac{3}{4}$$

(a) Combine the two fractions on the left over a common denominator:

$$\dfrac{1}{1-\sin\theta}+\dfrac{1}{1+\sin\theta}=\dfrac{(1+\sin\theta)+(1-\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}$$

The numerator collapses to $2$, and the denominator is a difference of two squares:

$$=\dfrac{2}{1-\sin^{2}\theta}$$

Replace $1-\sin^{2}\theta$ with $\cos^{2}\theta$ using the Pythagorean identity, then rewrite as a secant:

$$=\dfrac{2}{\cos^{2}\theta}=2\sec^{2}\theta=\text{RHS}$$

(b) The left side of the new equation is exactly the expression just proven, so replace it with $2\sec^{2}\theta$ and solve for $\cos\theta$:

$$2\sec^{2}\theta=8\ \Rightarrow\ \sec^{2}\theta=4\ \Rightarrow\ \cos^{2}\theta=\dfrac{1}{4}\ \Rightarrow\ \cos\theta=\pm\dfrac{1}{2}$$

For $\cos\theta=\dfrac{1}{2}$ (first and fourth quadrants) and $\cos\theta=-\dfrac{1}{2}$ (second and third quadrants), the basic angle is $60^{\circ}$ in every case:

$$\theta=60^{\circ},\ 120^{\circ},\ 240^{\circ},\ 300^{\circ}$$