Conceptual problems: eliminating θ

This final les­son gath­ers the remain­ing ideas that make hard-look­ing prob­lems easy: a few rules from ratio and pro­por­tion, a dia­gram that cap­tures all the prod­uct rela­tions, some iden­ti­ties that are worth recog­nis­ing on sight, and the tech­nique of elim­i­nat­ing $\theta$, find­ing a rela­tion between quan­ti­ties that no longer involves the angle at all. It ends with cross-mul­ti­pli­ca­tion, which solves a pair of equa­tions in $\cos\theta$ and $\sin\theta$ at once.

Con­cept 5: ratio and pro­por­tion#

If $\displaystyle \frac{a}{b} = \frac{c}{d}$, then each ratio also equals

$$\displaystyle \frac{a + c}{b + d} = \frac{la + mc}{lb + md}$$

for any num­bers $l$, $m$ (as long as the denom­i­na­tors are not zero). And the com­po­nendo and div­i­dendo rule gives

$$\displaystyle \frac{a + b}{a - b} = \frac{c + d}{c - d}$$

Exam­ple 1#

If $\displaystyle \frac{\sin^4\theta}{a} + \frac{\cos^4\theta}{b} = \frac{1}{a + b}$, find $\displaystyle \frac{\sin^8\theta}{a^3} + \frac{\cos^8\theta}{b^3}$.

Write $s = \sin^2\theta$, so $\cos^2\theta = 1 - s$. Mul­ti­ply­ing the given equa­tion by $ab(a + b)$:

$$b(a + b)s^2 + a(a + b)(1 - s)^2 = ab$$

Expand­ing and col­lect­ing terms gives $(a + b)^2 s^2 - 2a(a + b)s + a^2 = 0$, that is,

$$\displaystyle \big((a + b)s - a\big)^2 = 0 \implies s = \frac{a}{a + b}, \quad 1 - s = \frac{b}{a + b}$$

So $\displaystyle \frac{\sin^2\theta}{a} = \frac{\cos^2\theta}{b} = \frac{1}{a + b}$ (the ratio rule above). Then

$$\displaystyle \frac{\sin^8\theta}{a^3} = a\left(\frac{\sin^2\theta}{a}\right)^4 = \frac{a}{(a + b)^4}, \qquad \frac{\cos^8\theta}{b^3} = \frac{b}{(a + b)^4}$$

and the sum is $\displaystyle \frac{a + b}{(a + b)^4} = \frac{1}{(a + b)^3}$.

Con­cept 6: the prod­uct hexa­gon#

Write the six ratios round a hexa­gon in this order: $\tan\theta$, $\sin\theta$, $\cos\theta$, $\cot\theta$, $\csc\theta$, $\sec\theta$. Then any ratio times the ratio two places fur­ther on gives the ratio between them:

Each line is checked by writ­ing every­thing in sines and cosines; for exam­ple $\displaystyle \csc\theta\tan\theta = \frac{1}{\sin\theta} \cdot \frac{\sin\theta}{\cos\theta} = \sec\theta$.

Con­cept 7: iden­ti­ties to recog­nise#

$$\tan^2\theta - \sin^2\theta = \tan^2\theta\sin^2\theta, \qquad \cot^2\theta - \cos^2\theta = \cot^2\theta\cos^2\theta$$

$$\displaystyle \sec^2\theta + \csc^2\theta = \sec^2\theta\csc^2\theta, \qquad \tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta} = \csc\theta\sec\theta$$

For the first: $\displaystyle \tan^2\theta - \sin^2\theta = \sin^2\theta\left(\frac{1}{\cos^2\theta} - 1\right) = \sin^2\theta\tan^2\theta$. The oth­ers are proved the same way.

Con­cept 8: elim­i­nat­ing θ#

To elim­i­nate $\theta$ from two equa­tions, com­bine them (usu­ally by adding, sub­tract­ing, squar­ing or mul­ti­ply­ing) so that an iden­tity such as $\sin^2\theta + \cos^2\theta = 1$ removes $\theta$.

Exam­ple 2#

If $m = \tan\theta + \sin\theta$ and $n = \tan\theta - \sin\theta$, show that $m^2 - n^2 = 4\sqrt{mn}$ (for acute $\theta$).

$m + n = 2\tan\theta$ and $m - n = 2\sin\theta$, so $m^2 - n^2 = (m + n)(m - n) = 4\tan\theta\sin\theta$.

Also $mn = \tan^2\theta - \sin^2\theta = \tan^2\theta\sin^2\theta$ (Con­cept 7), so $\sqrt{mn} = \tan\theta\sin\theta$. Hence $m^2 - n^2 = 4\sqrt{mn}$.

Exam­ple 3#

If $a\sec\alpha + b\tan\alpha = 1$ and $a^2\sec^2\alpha - b^2\tan^2\alpha = 5$, find $4a^2 - 9b^2$ in terms of $a$ and $b$.

The sec­ond equa­tion fac­torises: $(a\sec\alpha - b\tan\alpha)(a\sec\alpha + b\tan\alpha) = 5$, so $a\sec\alpha - b\tan\alpha = 5$. Adding and sub­tract­ing with the first equa­tion:

$$\displaystyle 2a\sec\alpha = 6 \implies \sec\alpha = \frac{3}{a}, \qquad 2b\tan\alpha = -4 \implies \tan\alpha = -\frac{2}{b}$$

Now $\sec^2\alpha - \tan^2\alpha = 1$ gives $\displaystyle \frac{9}{a^2} - \frac{4}{b^2} = 1$, so $9b^2 - 4a^2 = a^2b^2$, that is,

$$4a^2 - 9b^2 = -a^2b^2$$

Exam­ple 4#

If $m = \sin\theta + \cos\theta$ and $n = \sec\theta + \csc\theta$, elim­i­nate $\theta$.

$\displaystyle n = \frac{1}{\cos\theta} + \frac{1}{\sin\theta} = \frac{\sin\theta + \cos\theta}{\sin\theta\cos\theta} = \frac{m}{\sin\theta\cos\theta}$. Squar­ing $m$: $m^2 = 1 + 2\sin\theta\cos\theta$, so $\displaystyle \sin\theta\cos\theta = \frac{m^2 - 1}{2}$. There­fore

$$\displaystyle n = \frac{2m}{m^2 - 1}, \quad \text{that is,} \quad n(m^2 - 1) = 2m$$

Exam­ple 5#

If $m = \csc\theta - \sin\theta$ and $n = \sec\theta - \cos\theta$, find $(m^2n)^{2/3} + (mn^2)^{2/3}$.

$\displaystyle m = \frac{1 - \sin^2\theta}{\sin\theta} = \frac{\cos^2\theta}{\sin\theta}$ and $\displaystyle n = \frac{1 - \cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta}$.

$$\displaystyle m^2n = \frac{\cos^4\theta}{\sin^2\theta} \cdot \frac{\sin^2\theta}{\cos\theta} = \cos^3\theta, \qquad mn^2 = \frac{\cos^2\theta}{\sin\theta} \cdot \frac{\sin^4\theta}{\cos^2\theta} = \sin^3\theta$$

So the expres­sion is $\cos^2\theta + \sin^2\theta = 1$.

Exam­ple 6#

If $\displaystyle a\sin x = b\cos x = \frac{2c\tan x}{1 - \tan^2 x}$, find $\displaystyle \frac{(a^2 - b^2)^2}{a^2 + b^2}$.

From $a\sin x = b\cos x$, $\displaystyle \tan x = \frac{b}{a}$, so $\displaystyle \frac{2\tan x}{1 - \tan^2 x} = \frac{2b/a}{1 - b^2/a^2} = \frac{2ab}{a^2 - b^2}$. Then $\displaystyle b\cos x = \frac{2abc}{a^2 - b^2}$, so $\displaystyle \cos x = \frac{2ac}{a^2 - b^2}$. But $\displaystyle \tan x = \frac{b}{a}$ also gives $\displaystyle \cos x = \frac{a}{\sqrt{a^2 + b^2}}$ (tak­ing $x$ acute). Equat­ing,

$$\displaystyle \frac{a}{\sqrt{a^2 + b^2}} = \frac{2ac}{a^2 - b^2} \implies a^2 - b^2 = 2c\sqrt{a^2 + b^2} \implies \frac{(a^2 - b^2)^2}{a^2 + b^2} = 4c^2$$

Exam­ple 7#

If $\sin\alpha$, $\sin\beta$, $\sin\gamma$ are in A.P. and $\cos\alpha$, $\cos\beta$, $\cos\gamma$ are in G.P., find $\displaystyle \frac{\cos^2\alpha + \cos^2\gamma - 4\cos\alpha\cos\gamma}{1 - \sin\alpha\sin\gamma}$.

A.P. gives $2\sin\beta = \sin\alpha + \sin\gamma$; G.P. gives $\cos^2\beta = \cos\alpha\cos\gamma$. So

$$4\cos\alpha\cos\gamma = 4\cos^2\beta = 4 - 4\sin^2\beta = 4 - (\sin\alpha + \sin\gamma)^2$$

The numer­a­tor becomes $(1 - \sin^2\alpha) + (1 - \sin^2\gamma) - 4 + \sin^2\alpha + 2\sin\alpha\sin\gamma + \sin^2\gamma = -2 + 2\sin\alpha\sin\gamma = -2(1 - \sin\alpha\sin\gamma)$. The value is $-2$.

Con­cept 9: cross-mul­ti­pli­ca­tion#

For two equa­tions $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ (with $a_1b_2 - a_2b_1 \neq 0$),

$$\displaystyle \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$$

so $\displaystyle x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}$ and $\displaystyle y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}$.

Exam­ple 8#

Elim­i­nate $\theta$ from $a_1\cos\theta + b_1\sin\theta + c_1 = 0$ and $a_2\cos\theta + b_2\sin\theta + c_2 = 0$.

Treat $\cos\theta$ as $x$ and $\sin\theta$ as $y$:

$$\displaystyle \cos\theta = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}, \qquad \sin\theta = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}$$

Since $\cos^2\theta + \sin^2\theta = 1$,

$$(b_1c_2 - b_2c_1)^2 + (c_1a_2 - c_2a_1)^2 = (a_1b_2 - a_2b_1)^2$$

Prac­tice#

1. Prove that $\tan^2\theta - \sin^2\theta = \tan^2\theta\sin^2\theta$.
2. Prove that $\sec^2\theta + \csc^2\theta = \sec^2\theta\csc^2\theta$.
3. Prove that $\cot^2\theta - \cos^2\theta = \cot^2\theta\cos^2\theta$.
4. Elim­i­nate $\theta$ from $x = a\cos\theta$, $y = b\sin\theta$.
5. Elim­i­nate $\theta$ from $x = a\sec\theta$, $y = b\tan\theta$.
6. Elim­i­nate $\theta$ from $x = \csc\theta + \cot\theta$, $y = \csc\theta - \cot\theta$.
7. If $\cos\theta + \sin\theta = \sqrt{2}\cos\theta$, show that $\cos\theta - \sin\theta = \sqrt{2}\sin\theta$.
8. Use the hexa­gon to sim­plify $\sec\theta \cdot \sin\theta \cdot \cot\theta$.

Answers#