Compound, Multiple and Sum-to-Product Formulas, Worked Solutions

Ques­tion 1: Exact val­ues from sum and dif­fer­ence for­mu­las#

The prob­lem#

Find exactly: (a) $\sin 15^\circ$ (b) $\cos 105^\circ$ (c) $\tan 75^\circ$.

Under­stand­ing the prob­lem#

"Exactly" means in surds (square roots), not dec­i­mals from a cal­cu­la­tor. None of these angles is a stan­dard angle, but each is a sum or dif­fer­ence of stan­dard angles ($30^\circ$, $45^\circ$, $60^\circ$).

The idea#

Write $15^\circ = 45^\circ - 30^\circ$, $105^\circ = 60^\circ + 45^\circ$, $75^\circ = 45^\circ + 30^\circ$, and use the sum and dif­fer­ence for­mu­las with the known val­ues

$$\displaystyle \sin 30^\circ = \tfrac{1}{2},\ \cos 30^\circ = \tfrac{\sqrt{3}}{2},\ \sin 45^\circ = \cos 45^\circ = \tfrac{1}{\sqrt{2}},\ \sin 60^\circ = \tfrac{\sqrt{3}}{2},\ \cos 60^\circ = \tfrac{1}{2}.$$

Step-by-step solu­tion#

Part (a)

Step 1. Use $\sin(x - y) = \sin x \cos y - \cos x \sin y$ with $x = 45^\circ$, $y = 30^\circ$.

$$\displaystyle \begin{aligned} \sin 15^\circ &= \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ\\ &= \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\cdot\frac{1}{2}\\ &= \frac{\sqrt{3} - 1}{2\sqrt{2}} \end{aligned}$$

Part (b)

Step 1. Use $\cos(x + y) = \cos x \cos y - \sin x \sin y$ with $x = 60^\circ$, $y = 45^\circ$.

$$\displaystyle \begin{aligned} \cos 105^\circ &= \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ\\ &= \frac{1}{2}\cdot\frac{1}{\sqrt{2}} - \frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{2}}\\ &= \frac{1 - \sqrt{3}}{2\sqrt{2}} \end{aligned}$$

Part (c)

Step 1. Use $\displaystyle \tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$ with $\tan 45^\circ = 1$ and $\displaystyle \tan 30^\circ = \frac{1}{\sqrt{3}}$.

$$\displaystyle \tan 75^\circ = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$$

(We mul­ti­plied top and bot­tom by $\sqrt{3}$.)

Step 2. Ratio­nalise: mul­ti­ply top and bot­tom by $\sqrt{3} + 1$.

$$\displaystyle \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$$

Check­ing the answer#

(a) $\displaystyle \frac{1.732 - 1}{2.828} \approx 0.259$, and $\sin 15^\circ \approx 0.259$. (b) is neg­a­tive, as it must be since $105^\circ$ is in quad­rant II. (c) $2 + 1.732 = 3.732 \approx \tan 75^\circ$.

Answer#

(a) $\displaystyle \frac{\sqrt{3} - 1}{2\sqrt{2}}$ (b) $\displaystyle \frac{1 - \sqrt{3}}{2\sqrt{2}}$ (c) $2 + \sqrt{3}$

Ques­tion 2: Prov­ing a sum of two cosines#

The prob­lem#

Prove: $\displaystyle \cos\left(\frac{\pi}{4} + x\right) + \cos\left(\frac{\pi}{4} - x\right) = \sqrt{2}\cos x$.

Under­stand­ing the prob­lem#

We must show the left side (LHS) sim­pli­fies to the right side (RHS) for every $x$.

The idea#

Expand both cosines with the sum and dif­fer­ence for­mu­las. The $\sin$ terms have oppo­site signs and will can­cel.

Step-by-step solu­tion#

Step 1. Expand each term.

$$\displaystyle \begin{aligned} \cos\left(\tfrac{\pi}{4} + x\right) &= \cos\tfrac{\pi}{4}\cos x - \sin\tfrac{\pi}{4}\sin x\\ \cos\left(\tfrac{\pi}{4} - x\right) &= \cos\tfrac{\pi}{4}\cos x + \sin\tfrac{\pi}{4}\sin x \end{aligned}$$

Step 2. Add. The $\displaystyle \sin\tfrac{\pi}{4}\sin x$ terms can­cel.

$$\displaystyle \text{LHS} = 2\cos\tfrac{\pi}{4}\cos x$$

Step 3. Put $\displaystyle \cos\tfrac{\pi}{4} = \tfrac{1}{\sqrt{2}}$.

$$\displaystyle \text{LHS} = 2 \cdot \frac{1}{\sqrt{2}}\cos x = \sqrt{2}\cos x = \text{RHS}$$

Check­ing the answer#

At $x = 0$: LHS $\displaystyle = 2\cos\tfrac{\pi}{4} = \sqrt{2}$ and RHS $= \sqrt{2}$.

Answer#

LHS $\displaystyle = 2\cos\tfrac{\pi}{4}\cos x = \sqrt{2}\cos x$, as required.

Ques­tion 3: A ratio of tan­gents#

The prob­lem#

Prove: $\displaystyle \frac{\tan\left(\frac{\pi}{4} + x\right)}{\tan\left(\frac{\pi}{4} - x\right)} = \left(\frac{1 + \tan x}{1 - \tan x}\right)^2$.

Under­stand­ing the prob­lem#

The RHS is writ­ten only in $\tan x$, so we should express both tan­gents on the LHS in terms of $\tan x$.

The idea#

Use $\displaystyle \tan(x \pm y) = \frac{\tan x \pm \tan y}{1 \mp \tan x \tan y}$ with $\displaystyle \tan\tfrac{\pi}{4} = 1$.

Step-by-step solu­tion#

Step 1. Expand the numer­a­tor.

$$\displaystyle \tan\left(\tfrac{\pi}{4} + x\right) = \frac{1 + \tan x}{1 - 1 \cdot \tan x} = \frac{1 + \tan x}{1 - \tan x}$$

Step 2. Expand the denom­i­na­tor.

$$\displaystyle \tan\left(\tfrac{\pi}{4} - x\right) = \frac{1 - \tan x}{1 + \tan x}$$

Step 3. Divide: divid­ing by a frac­tion is mul­ti­ply­ing by its rec­i­p­ro­cal.

$$\displaystyle \text{LHS} = \frac{1 + \tan x}{1 - \tan x} \times \frac{1 + \tan x}{1 - \tan x} = \left(\frac{1 + \tan x}{1 - \tan x}\right)^2 = \text{RHS}$$

Check­ing the answer#

At $x = 0$: LHS $\displaystyle = \frac{1}{1} = 1$, RHS $= 1^2 = 1$.

Answer#

Both tan­gents expand to $\displaystyle \frac{1 \pm \tan x}{1 \mp \tan x}$; their quo­tient is the RHS.

Ques­tion 4: Show­ing $\displaystyle A + B = \frac{\pi}{4}$#

The prob­lem#

$\displaystyle \tan A = \tfrac{1}{2}$, $\displaystyle \tan B = \tfrac{1}{3}$, with $A$ and $B$ both acute. Show $\displaystyle A + B = \tfrac{\pi}{4}$.

Under­stand­ing the prob­lem#

We are told the tan­gents of two angles. We need to find the angle $A + B$ itself.

The idea#

Find $\tan(A + B)$ with the sum for­mula. Then use the fact that $A + B$ lies in a range where only one angle has that tan­gent.

Step-by-step solu­tion#

Step 1. Apply the for­mula.

$$\displaystyle \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{6}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1$$

Step 2. Both $A$ and $B$ are acute, so $0 < A + B < \pi$. In this range, the only angle whose tan­gent is $1$ is $\displaystyle \tfrac{\pi}{4}$.

$$\displaystyle A + B = \frac{\pi}{4}$$

Check­ing the answer#

$\tan A = 0.5$ gives $A \approx 26.57^\circ$; $\tan B \approx 0.333$ gives $B \approx 18.43^\circ$; the sum is $45^\circ$.

Answer#

$\tan(A + B) = 1$ with $0 < A + B < \pi$, so $\displaystyle A + B = \frac{\pi}{4}$.

Ques­tion 5: Dou­ble angles from $\sin x$#

The prob­lem#

$\displaystyle \sin x = \tfrac{12}{13}$, $x$ in quad­rant II. Find $\sin 2x$, $\cos 2x$, $\tan 2x$.

Under­stand­ing the prob­lem#

We know $\sin x$ and the quad­rant of $x$. The dou­ble-angle for­mu­las also need $\cos x$, and the quad­rant tells us its sign: in quad­rant II cosine is neg­a­tive.

The idea#

Find $\cos x$ from $\sin^2 x + \cos^2 x = 1$, then use $\sin 2x = 2\sin x\cos x$, $\cos 2x = \cos^2 x - \sin^2 x$, $\displaystyle \tan 2x = \frac{\sin 2x}{\cos 2x}$.

Step-by-step solu­tion#

Step 1. Find $\cos x$.

$$\displaystyle \cos^2 x = 1 - \frac{144}{169} = \frac{25}{169} \;\Rightarrow\; \cos x = -\frac{5}{13} \quad(\text{negative in quadrant II})$$

Step 2. $\sin 2x$:

$$\displaystyle \sin 2x = 2 \cdot \frac{12}{13}\cdot\left(-\frac{5}{13}\right) = -\frac{120}{169}$$

Step 3. $\cos 2x$:

$$\displaystyle \cos 2x = \frac{25}{169} - \frac{144}{169} = -\frac{119}{169}$$

Step 4. $\tan 2x$:

$$\displaystyle \tan 2x = \frac{-120/169}{-119/169} = \frac{120}{119}$$

Check­ing the answer#

$\displaystyle \sin^2 2x + \cos^2 2x = \frac{14400 + 14161}{28561} = \frac{28561}{28561} = 1$.

Answer#

$\displaystyle \sin 2x = -\frac{120}{169}$, $\displaystyle \cos 2x = -\frac{119}{169}$, $\displaystyle \tan 2x = \frac{120}{119}$.

Com­mon mis­take to avoid#

Tak­ing $\displaystyle \cos x = +\tfrac{5}{13}$. Always use the quad­rant to fix the sign.

Ques­tion 6: Half angles from $\tan x$#

The prob­lem#

$\displaystyle \tan x = \tfrac{3}{4}$, $\displaystyle \pi < x < \tfrac{3\pi}{2}$. Find $\displaystyle \sin\tfrac{x}{2}$ and $\displaystyle \cos\tfrac{x}{2}$.

Under­stand­ing the prob­lem#

$x$ is in quad­rant III. The half-angle for­mu­las need $\cos x$. We also need the quad­rant of $\displaystyle \tfrac{x}{2}$ to choose signs.

The idea#

Find $\cos x$, then use $\displaystyle \sin^2\tfrac{x}{2} = \frac{1 - \cos x}{2}$ and $\displaystyle \cos^2\tfrac{x}{2} = \frac{1 + \cos x}{2}$.

Step-by-step solu­tion#

Step 1. Find $\cos x$. From $1 + \tan^2 x = \sec^2 x$:

$$\displaystyle \sec^2 x = 1 + \frac{9}{16} = \frac{25}{16} \;\Rightarrow\; \sec x = -\frac{5}{4} \;\Rightarrow\; \cos x = -\frac{4}{5}$$

(neg­a­tive, because cosine is neg­a­tive in quad­rant III).

Step 2. Locate $\displaystyle \tfrac{x}{2}$. Halv­ing $\displaystyle \pi < x < \tfrac{3\pi}{2}$ gives $\displaystyle \tfrac{\pi}{2} < \tfrac{x}{2} < \tfrac{3\pi}{4}$: quad­rant II. So $\displaystyle \sin\tfrac{x}{2} > 0$ and $\displaystyle \cos\tfrac{x}{2} < 0$.

Step 3. Com­pute $\displaystyle \sin\tfrac{x}{2}$.

$$\displaystyle \sin^2\frac{x}{2} = \frac{1 - (-\frac{4}{5})}{2} = \frac{9/5}{2} = \frac{9}{10} \;\Rightarrow\; \sin\frac{x}{2} = \frac{3}{\sqrt{10}}$$

Step 4. Com­pute $\displaystyle \cos\tfrac{x}{2}$.

$$\displaystyle \cos^2\frac{x}{2} = \frac{1 + (-\frac{4}{5})}{2} = \frac{1/5}{2} = \frac{1}{10} \;\Rightarrow\; \cos\frac{x}{2} = -\frac{1}{\sqrt{10}}$$

Check­ing the answer#

$\displaystyle \frac{9}{10} + \frac{1}{10} = 1$. Also $\displaystyle \cos x = 2\cos^2\tfrac{x}{2} - 1 = \tfrac{2}{10} - 1 = -\tfrac{4}{5}$. Cor­rect.

Answer#

$\displaystyle \sin\frac{x}{2} = \frac{3}{\sqrt{10}}$, $\displaystyle \cos\frac{x}{2} = -\frac{1}{\sqrt{10}}$.

Ques­tion 7: A sum-to-prod­uct iden­tity#

The prob­lem#

Prove: $\displaystyle \frac{\sin 5x - \sin 3x}{\cos 5x + \cos 3x} = \tan x$.

Under­stand­ing the prob­lem#

The numer­a­tor is a dif­fer­ence of sines and the denom­i­na­tor a sum of cosines. The answer $\tan x$ sug­gests a com­mon fac­tor will can­cel.

The idea#

Use $\displaystyle \sin C - \sin D = 2\cos\tfrac{C + D}{2}\sin\tfrac{C - D}{2}$ and $\displaystyle \cos C + \cos D = 2\cos\tfrac{C + D}{2}\cos\tfrac{C - D}{2}$ with $C = 5x$, $D = 3x$, so $\displaystyle \tfrac{C + D}{2} = 4x$ and $\displaystyle \tfrac{C - D}{2} = x$.

Step-by-step solu­tion#

Step 1. Numer­a­tor: $\sin 5x - \sin 3x = 2\cos 4x \sin x$.

Step 2. Denom­i­na­tor: $\cos 5x + \cos 3x = 2\cos 4x \cos x$.

Step 3. Can­cel $2\cos 4x$.

$$\displaystyle \text{LHS} = \frac{2\cos 4x \sin x}{2\cos 4x \cos x} = \frac{\sin x}{\cos x} = \tan x = \text{RHS}$$

Check­ing the answer#

At $\displaystyle x = \tfrac{\pi}{12}$ you can check numer­i­cally: both sides are about $0.268$.

Answer#

LHS $\displaystyle = \frac{2\cos 4x\sin x}{2\cos 4x\cos x} = \tan x$.

Ques­tion 8: $\cos 4x$ in terms of $\sin x$ and $\cos x$#

The prob­lem#

Prove: $\cos 4x = 1 - 8\sin^2 x \cos^2 x$.

Under­stand­ing the prob­lem#

$4x$ is dou­ble $2x$, and $2x$ is dou­ble $x$. So we can apply dou­ble-angle for­mu­las twice.

The idea#

Use $\cos 2\theta = 1 - 2\sin^2\theta$ with $\theta = 2x$, then $\sin 2x = 2\sin x\cos x$.

Step-by-step solu­tion#

Step 1. With $\theta = 2x$:

$$\cos 4x = 1 - 2\sin^2 2x$$

Step 2. Replace $\sin 2x$ by $2\sin x \cos x$ and square.

$$\sin^2 2x = 4\sin^2 x\cos^2 x$$

Step 3. Sub­sti­tute.

$$\cos 4x = 1 - 2 \cdot 4\sin^2 x\cos^2 x = 1 - 8\sin^2 x\cos^2 x$$

Check­ing the answer#

At $\displaystyle x = \tfrac{\pi}{4}$: $\cos\pi = -1$; RHS $\displaystyle = 1 - 8 \cdot \tfrac{1}{2}\cdot\tfrac{1}{2} = -1$.

Answer#

$\cos 4x = 1 - 2\sin^2 2x = 1 - 8\sin^2 x\cos^2 x$.

Ques­tion 9: Three sines into a prod­uct#

The prob­lem#

Prove: $\displaystyle \sin 3x + \sin 2x - \sin x = 4\sin x \cos\frac{x}{2}\cos\frac{3x}{2}$.

Under­stand­ing the prob­lem#

The RHS has a fac­tor $\sin x$, so we should group terms on the LHS that pro­duce $\sin x$ as a fac­tor.

The idea#

Pair $\sin 3x - \sin x$ (sum-to-prod­uct), write $\sin 2x = 2\sin x\cos x$, take out $2\sin x$, and then use $\cos C + \cos D$ on what is left.

Step-by-step solu­tion#

Step 1. Use $\sin C - \sin D$ with $C = 3x$, $D = x$.

$$\displaystyle \sin 3x - \sin x = 2\cos\frac{3x + x}{2}\sin\frac{3x - x}{2} = 2\cos 2x \sin x$$

Step 2. Add $\sin 2x = 2\sin x \cos x$ and take out the com­mon fac­tor $2\sin x$.

$$\text{LHS} = 2\sin x\cos 2x + 2\sin x\cos x = 2\sin x\,(\cos 2x + \cos x)$$

Step 3. Use $\cos C + \cos D$ with $C = 2x$, $D = x$.

$$\displaystyle \cos 2x + \cos x = 2\cos\frac{3x}{2}\cos\frac{x}{2}$$

Step 4. Sub­sti­tute.

$$\displaystyle \text{LHS} = 2\sin x \cdot 2\cos\frac{3x}{2}\cos\frac{x}{2} = 4\sin x\cos\frac{x}{2}\cos\frac{3x}{2} = \text{RHS}$$

Check­ing the answer#

At $\displaystyle x = \tfrac{\pi}{2}$: LHS $= -1 + 0 - 1 = -2$; RHS $\displaystyle = 4 \cdot 1 \cdot \tfrac{1}{\sqrt{2}}\cdot\left(-\tfrac{1}{\sqrt{2}}\right) = -2$.

Answer#

LHS $\displaystyle = 2\sin x(\cos 2x + \cos x) = 4\sin x\cos\frac{x}{2}\cos\frac{3x}{2}$.

Ques­tion 10: The exact value of $\sin 18^\circ$#

The prob­lem#

Find $\sin 18^\circ$ using $\sin 2\theta = \cos 3\theta$ for $\theta = 18^\circ$.

Under­stand­ing the prob­lem#

First see why the hint is true: with $\theta = 18^\circ$, $2\theta = 36^\circ$ and $3\theta = 54^\circ$, and $36^\circ + 54^\circ = 90^\circ$. So $\sin 36^\circ = \cos(90^\circ - 36^\circ) = \cos 54^\circ$.

The idea#

Expand both sides with the dou­ble- and triple-angle for­mu­las, turn every­thing into $s = \sin\theta$, and solve the result­ing qua­dratic.

Step-by-step solu­tion#

Step 1. Expand.

$$2\sin\theta\cos\theta = 4\cos^3\theta - 3\cos\theta$$

Step 2. $\cos 18^\circ \ne 0$, so divide both sides by $\cos\theta$.

$$2\sin\theta = 4\cos^2\theta - 3$$

Step 3. Replace $\cos^2\theta$ by $1 - \sin^2\theta$ and write $s = \sin\theta$.

$$2s = 4(1 - s^2) - 3 = 1 - 4s^2$$

Step 4. Rearrange into a qua­dratic and solve with the for­mula.

$$\displaystyle \begin{aligned} 4s^2 + 2s - 1 &= 0\\ s &= \frac{-2 \pm \sqrt{4 + 16}}{8} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4} \end{aligned}$$

Step 5. $18^\circ$ is acute, so $\sin 18^\circ > 0$. Choose the pos­i­tive root.

$$\displaystyle \sin 18^\circ = \frac{\sqrt{5} - 1}{4}$$

Check­ing the answer#

$\displaystyle \frac{2.236 - 1}{4} = 0.309$, and $\sin 18^\circ \approx 0.309$.

Answer#

$\displaystyle \sin 18^\circ = \frac{\sqrt{5} - 1}{4}$.