Values from a Condition, Maximum and Minimum

Given $\sin\theta + \cos\theta$, Find the Product

When the sum or the difference of $\sin\theta$ and $\cos\theta$ is given as a number, square it. The squares add up to 1, and what is left is twice the product:

$$(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta \qquad (\sin\theta - \cos\theta)^2 = 1 - 2\sin\theta\cos\theta$$

So

$$\sin\theta + \cos\theta = p \Rightarrow \sin\theta\cos\theta = \frac{p^2 - 1}{2} \qquad \sin\theta - \cos\theta = q \Rightarrow \sin\theta\cos\theta = \frac{1 - q^2}{2}$$

Adding the two squares gives $(\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2 = 2$, which links the sum and the difference directly: $p^2 + q^2 = 2$.

From the product, everything else. Take $\sin\theta + \cos\theta = \frac{47}{37}$. Then

$$\sin\theta\cos\theta = \frac{1}{2}\left(\frac{2209}{1369} - 1\right) = \frac{420}{1369}$$

• $\tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta} = \frac{1369}{420}$
• $(\sin\theta - \cos\theta)^2 = 2 - \frac{2209}{1369} = \frac{529}{1369}$, so $\sin\theta - \cos\theta = \pm\frac{23}{37}$. Both signs are possible, since the angle may have $\sin\theta = \frac{35}{37}$, $\cos\theta = \frac{12}{37}$ or the other way round.
• $\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)$, from $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$.
• $\sin^4\theta + \cos^4\theta = 1 - 2(\sin\theta\cos\theta)^2$.

A quick look at the data. Since $\sin\theta\cos\theta$ is never more than $\frac{1}{2}$ (shown in the part on the product, further down), $p^2 = 1 + 2\sin\theta\cos\theta$ is at most 2. So a question with $\sin\theta + \cos\theta = \frac{8}{5}$ has no answer: $\frac{8}{5} = 1.6$ is more than $\sqrt{2} \approx 1.414$.

Given $\sec\theta + \tan\theta$ or $\operatorname{cosec}\theta + \cot\theta$

From Identities and Simplification, $(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1$. So

$$\sec\theta + \tan\theta = p \Rightarrow \sec\theta - \tan\theta = \frac{1}{p}$$

Add the two equations and halve, then subtract them and halve:

$$\sec\theta = \frac{1}{2}\left(p + \frac{1}{p}\right) = \frac{p^2 + 1}{2p} \qquad \tan\theta = \frac{1}{2}\left(p - \frac{1}{p}\right) = \frac{p^2 - 1}{2p}$$

Dividing one by the other, $\sin\theta = \frac{\tan\theta}{\sec\theta} = \frac{p^2 - 1}{p^2 + 1}$.

The triangle explains all of these. Sides 2p, $p^2 - 1$ and $p^2 + 1$ always make a right triangle, since $(2p)^2 + (p^2 - 1)^2 = (p^2 + 1)^2$. Every ratio can be read off it, including $\cos\theta = \frac{2p}{p^2 + 1}$.

Take $\sec\theta + \tan\theta = 3$. Then $\sec\theta - \tan\theta = \frac{1}{3}$, $\sec\theta = \frac{5}{3}$, $\tan\theta = \frac{4}{3}$ and $\sin\theta = \frac{4}{5}$: the 3, 4, 5 triangle.

The cosec and cot version. If $\operatorname{cosec}\theta + \cot\theta = p$, then $\operatorname{cosec}\theta - \cot\theta = \frac{1}{p}$, $\operatorname{cosec}\theta = \frac{p^2 + 1}{2p}$, $\cot\theta = \frac{p^2 - 1}{2p}$ and $\cos\theta = \frac{p^2 - 1}{p^2 + 1}$. With p = 5, $\cos\theta = \frac{24}{26} = \frac{12}{13}$.

The difference given. Turn it over first. $\sec\theta - \tan\theta = \frac{2}{7}$ means $\sec\theta + \tan\theta = \frac{7}{2}$, so $p = \frac{7}{2}$ and

$$\sin\theta = \frac{p^2 - 1}{p^2 + 1} = \frac{49 - 4}{49 + 4} = \frac{45}{53}$$

after multiplying the top and bottom by 4.

[SSC Tip] $\frac{2p}{p^2 + 1}$ is $\cos\theta$. When $\sin\theta$ is asked, that value is the wrong option to watch for; look at which side of the triangle is opposite $\theta$.

A Ratio Plus Its Reciprocal Equal to 2

For a positive number x, $x + \frac{1}{x} = 2$ means $x^2 - 2x + 1 = 0$, that is $(x - 1)^2 = 0$, so x = 1 (see Chapter 11). Applied to a ratio and its reciprocal:

Given	Forces	So $\theta$ is
$\tan\theta + \cot\theta = 2$	$\tan\theta = 1$	45°
$\sin\theta + \operatorname{cosec}\theta = 2$	$\sin\theta = 1$	90°
$\cos\theta + \sec\theta = 2$	$\cos\theta = 1$	0°

After that every power of the ratio is 1, so the questions built on these are quick:

• $\tan\theta + \cot\theta = 2 \Rightarrow \tan^7\theta + \cot^7\theta = 1 + 1 = 2$, and also $\sin\theta + \cos\theta = \sqrt{2}$.
• $\sin\theta + \operatorname{cosec}\theta = 2 \Rightarrow \sin^5\theta + \operatorname{cosec}^5\theta = 2$, and also $\cos\theta = 0$.

The graph shows the other side of the same fact: for an acute angle, $\tan\theta + \cot\theta$ is never below 2. So a condition such as $\tan\theta + \cot\theta = \frac{3}{2}$ has no acute angle at all.

Other values of $\tan\theta + \cot\theta$. Since $\tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta}$, a given value hands you the product at once. $\tan\theta + \cot\theta = 6$ gives $\sin\theta\cos\theta = \frac{1}{6}$, and then $(\sin\theta + \cos\theta)^2 = 1 + \frac{1}{3} = \frac{4}{3}$. Squaring the condition gives $\tan^2\theta + \cot^2\theta = 6^2 - 2 = 34$, because the middle term is $2\tan\theta\cot\theta = 2$.

A Condition Like $\sin\theta + \sin^2\theta = 1$

Move one term across and look for an identity:

$$\sin\theta = 1 - \sin^2\theta = \cos^2\theta$$

So the condition really says $\sin\theta = \cos^2\theta$, and anything in $\cos^2\theta$ can now be turned into sines. For instance

$$\cos^2\theta + \cos^4\theta = \sin\theta + (\cos^2\theta)^2 = \sin\theta + \sin^2\theta = 1$$

Higher powers. Cube the result: $(\cos^2\theta + \cos^4\theta)^3 = 1$. Expanded with $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, that is

$$\cos^6\theta + 3\cos^8\theta + 3\cos^{10}\theta + \cos^{12}\theta = 1$$

A question that shows a long sum like this is really asking you to spot the cube. With coefficients 1, 2, 1 it is a square instead.

Other ratios. With $\cos^2\theta = \sin\theta$, $\cot^2\theta = \frac{\cos^2\theta}{\sin^2\theta} = \frac{\sin\theta}{\sin^2\theta} = \operatorname{cosec}\theta$. Squaring, $\cot^4\theta = \operatorname{cosec}^2\theta = 1 + \cot^2\theta$, so $\cot^4\theta - \cot^2\theta = 1$.

The cos version. $\cos\theta + \cos^2\theta = 1$ gives $\cos\theta = \sin^2\theta$, and everything above repeats with sin and cos swapped; for example $\sin^2\theta + \sin^4\theta = 1$.

Such a condition does belong to a real angle. $\sin\theta + \sin^2\theta = 1$ is a quadratic in $\sin\theta$ whose positive root is $\sin\theta = \frac{\sqrt{5} - 1}{2} \approx 0.618$, which is about 38°. You never need the angle itself; the identity does all the work.

A Value of $\tan\theta$ Given: Divide Through by $\cos\theta$

Suppose $5\tan\theta = 4$ and you want $\frac{5\sin\theta - 3\cos\theta}{5\sin\theta + 2\cos\theta}$. Every term on the top and the bottom has exactly one sin or cos in it. Divide each term by $\cos\theta$:

$$\frac{5\tan\theta - 3}{5\tan\theta + 2} = \frac{4 - 3}{4 + 2} = \frac{1}{6}$$

There is no need to find $\sin\theta$ and $\cos\theta$ separately.

The rule. When every term has the same total power of sin and cos, either all first powers or all second powers, divide by that power of $\cos\theta$ to bring in $\tan\theta$, or of $\sin\theta$ to bring in $\cot\theta$.

• $\cot\theta$ given: with $3\cot\theta = 2$, divide by $\sin\theta$: $\frac{3\cos\theta + \sin\theta}{3\cos\theta - \sin\theta} = \frac{3\cot\theta + 1}{3\cot\theta - 1} = \frac{2 + 1}{2 - 1} = 3$.
• Second powers: with $\tan\theta = 3$, divide by $\cos^2\theta$: $\frac{2\sin^2\theta + \sin\theta\cos\theta}{\cos^2\theta + 3\sin\theta\cos\theta} = \frac{2\tan^2\theta + \tan\theta}{1 + 3\tan\theta} = \frac{18 + 3}{1 + 9} = \frac{21}{10}$.
• A lone 1: write it as $\sin^2\theta + \cos^2\theta$ so that every term is a second power. With $\tan\theta = 3$,

$$\sin^2\theta - \cos^2\theta = \frac{\sin^2\theta - \cos^2\theta}{\sin^2\theta + \cos^2\theta} = \frac{\tan^2\theta - 1}{\tan^2\theta + 1} = \frac{8}{10} = \frac{4}{5}$$

and in the same way $\sin\theta\cos\theta = \frac{\tan\theta}{1 + \tan^2\theta} = \frac{3}{10}$.

Working backwards. If $\frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = 7$, divide the top and bottom by $\cos\theta$: $\frac{\tan\theta + 1}{\tan\theta - 1} = 7$, so $\tan\theta + 1 = 7\tan\theta - 7$ and $\tan\theta = \frac{4}{3}$. A statement like $4\sin\theta = 3\cos\theta$ is simply $\tan\theta = \frac{3}{4}$ in another form.

Divide every term, top and bottom. Dividing only the top changes the value of the fraction. When the terms are not all of one power, as in $\frac{1 + \sin\theta}{\cos\theta}$, draw the right triangle instead (see Angles, Radians and Trigonometric Ratios).

Squares That Add Up: $a\cos\theta - b\sin\theta$ and $a\sin\theta + b\cos\theta$

Expand both squares:

$$(a\cos\theta - b\sin\theta)^2 = a^2\cos^2\theta - 2ab\sin\theta\cos\theta + b^2\sin^2\theta$$

$$(a\sin\theta + b\cos\theta)^2 = a^2\sin^2\theta + 2ab\sin\theta\cos\theta + b^2\cos^2\theta$$

Add them. The middle terms cancel, and $a^2(\cos^2\theta + \sin^2\theta) + b^2(\sin^2\theta + \cos^2\theta)$ is left:

$$(a\cos\theta - b\sin\theta)^2 + (a\sin\theta + b\cos\theta)^2 = a^2 + b^2$$

So when one of the pair is given as c, the other is $\pm\sqrt{a^2 + b^2 - c^2}$. For an acute angle with a and b positive, $a\sin\theta + b\cos\theta$ is positive, so it takes the + sign.

Take $\sqrt{3}\cos\theta - \sin\theta = 1$. Here $a^2 + b^2 = 3 + 1 = 4$, so $(\sqrt{3}\sin\theta + \cos\theta)^2 = 4 - 1 = 3$ and $\sqrt{3}\sin\theta + \cos\theta = \sqrt{3}$. The angle 30° fits the condition, and it checks: $\sqrt{3} \times \frac{\sqrt{3}}{2} - \frac{1}{2} = 1$, while $\sqrt{3} \times \frac{1}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}$.

Matching the pair. The two expressions have the same a and b, with sin and cos swapped and the minus sign moved. $3\cos\theta - 4\sin\theta$ pairs with $3\sin\theta + 4\cos\theta$, not with $4\sin\theta + 3\cos\theta$. Check the match before using the rule.

As an elimination. If $x = a\sin\theta + b\cos\theta$ and $y = a\cos\theta - b\sin\theta$, then $x^2 + y^2 = a^2 + b^2$ for every $\theta$.

When the given value is the largest possible. If $3\sin\theta + 4\cos\theta = 5$, the partner is $\sqrt{25 - 25} = 0$. So $3\cos\theta - 4\sin\theta = 0$, which means $\tan\theta = \frac{3}{4}$. This is exactly the angle at which $3\sin\theta + 4\cos\theta$ is greatest, as the part on greatest and least values shows.

Conditions in $\sin^2\theta$ and $\cos^2\theta$, and Rearranging a Condition

$3\sin^2\theta + 7\cos^2\theta = 4$ has two squares in it, but they are tied by $\sin^2\theta + \cos^2\theta = 1$. Replace one of them:

$$3(1 - \cos^2\theta) + 7\cos^2\theta = 4 \Rightarrow 3 + 4\cos^2\theta = 4 \Rightarrow \cos^2\theta = \frac{1}{4}$$

So $\cos\theta = \frac{1}{2}$, $\theta = 60^\circ$ and $\tan\theta = \sqrt{3}$. A quick way to see the second step: three of the seven $\cos^2\theta$ join the three $\sin^2\theta$ to make 3.

Sec and tan are tied by $\sec^2\theta = 1 + \tan^2\theta$, and the same idea works: $\sec^2\theta + \tan^2\theta = 7$ gives $1 + 2\tan^2\theta = 7$, so $\tan^2\theta = 3$ and $\theta = 60^\circ$.

Rearranging a condition. If $\cos\theta + \sin\theta = \sqrt{2}\cos\theta$, then $\sin\theta = (\sqrt{2} - 1)\cos\theta$. Multiply both sides by $\sqrt{2} + 1$; since $(\sqrt{2} + 1)(\sqrt{2} - 1) = 1$,

$$(\sqrt{2} + 1)\sin\theta = \cos\theta \Rightarrow \cos\theta - \sin\theta = \sqrt{2}\sin\theta$$

The condition with cos on the right turns into the same shape with sin. On the way it also gives $\tan\theta = \sqrt{2} - 1$.

[Tier 2] Powers of sec from $\tan^2\theta$. Given $\tan^2\theta = 1 - e^2$, find $\sec\theta + \tan^3\theta\operatorname{cosec}\theta$. Write

$$\tan^3\theta\operatorname{cosec}\theta = \tan^2\theta \times \frac{\tan\theta}{\sin\theta} = \tan^2\theta\sec\theta$$

Then the whole expression is $\sec\theta(1 + \tan^2\theta) = \sec^3\theta$. Since $\sec^2\theta = 1 + \tan^2\theta = 2 - e^2$, the answer is $(2 - e^2)^{\frac{3}{2}}$.

Greatest and Least Value of $a\sin\theta + b\cos\theta$

The rule from the squares that add up says, for every $\theta$,

$$(a\sin\theta + b\cos\theta)^2 = a^2 + b^2 - (a\cos\theta - b\sin\theta)^2 \le a^2 + b^2$$

since a square cannot be negative. Taking square roots,

$$-\sqrt{a^2 + b^2} \le a\sin\theta + b\cos\theta \le \sqrt{a^2 + b^2}$$

Both ends are actually reached. The bracket $a\cos\theta - b\sin\theta$ is 0 when $\tan\theta = \frac{a}{b}$, and that happens at two angles $180^\circ$ apart. At one of them the expression equals $\sqrt{a^2 + b^2}$; at the other, where sine and cosine have both changed sign, it equals $-\sqrt{a^2 + b^2}$.

For $3\sin\theta + 4\cos\theta$ the greatest value is $\sqrt{9 + 16} = 5$, reached when $\tan\theta = \frac{3}{4}$, and the least is −5. The sum of the coefficients, 7, is never reached, because $\sin\theta$ and $\cos\theta$ cannot both be 1 at the same angle.

Signs and constants.

• $5\sin\theta - 12\cos\theta$ also lies between −13 and 13. A minus sign inside does not change $a^2 + b^2$.
• $3\sin\theta + 4\cos\theta + 7$ lies between 7 − 5 = 2 and 7 + 5 = 12. An added constant shifts the whole range.
• $\sin\theta + \cos\theta$ lies between $-\sqrt{2}$ and $\sqrt{2}$, with the top at $\theta = 45^\circ$.
• $\sqrt{3}\sin\theta + \cos\theta$ has greatest value $\sqrt{3 + 1} = 2$, at $\theta = 60^\circ$.

These limits are over all angles, which is what such questions ask for unless they restrict $\theta$.

[SSC Tip] Do not add the coefficients. The greatest value of $a\sin\theta + b\cos\theta$ is $\sqrt{a^2 + b^2}$, which is less than a + b whenever a and b are both positive; a + b is the wrong option to watch for.

Between a and b: $a\sin^2\theta + b\cos^2\theta$, and the Product $\sin\theta\cos\theta$

Write $a\sin^2\theta + b\cos^2\theta = b + (a - b)\sin^2\theta$. As $\sin^2\theta$ runs from 0 to 1, the value runs from b to a. So the expression always lies between a and b: the smaller of the two is its least value and the larger is its greatest.

$4\sin^2\theta + 9\cos^2\theta$ is 9 at $\theta = 0^\circ$ and 4 at $\theta = 90^\circ$, so its least value is 4 and its greatest is 9. Compare $4\sin\theta + 9\cos\theta$, with no squares, whose greatest value is $\sqrt{97}$ by the earlier rule. The squares change the method completely.

An added constant works as before: $5 + 3\sin^2\theta$ lies between 5 and 8.

The product. $(\sin\theta - \cos\theta)^2 \ge 0$ gives $1 - 2\sin\theta\cos\theta \ge 0$, so

$$\sin\theta\cos\theta \le \frac{1}{2}$$

with equality at $\theta = 45^\circ$. Squaring, $\sin^2\theta\cos^2\theta \le \frac{1}{4}$. This fixes the ranges of the power sums from Identities and Simplification:

• $\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta$ lies between $\frac{1}{2}$ and 1
• $\sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta$ lies between $\frac{1}{4}$ and 1

The least values come at 45° and the greatest at 0° or 90°.

Mixed powers. $\sin^2\theta + \cos^4\theta = 1 - \cos^2\theta + \cos^4\theta$. Put $x = \cos^2\theta$, which lies between 0 and 1:

$$1 - x + x^2 = \frac{3}{4} + \left(x - \frac{1}{2}\right)^2$$

So the value lies between $\frac{3}{4}$, when $\cos^2\theta = \frac{1}{2}$, and 1, when x is 0 or 1.

A Number Plus Its Reciprocal Is at Least 2

For any positive x, $x + \frac{1}{x} \ge 2$, because $x + \frac{1}{x} - 2 = \frac{(x - 1)^2}{x}$, which is never negative. Equality holds only at x = 1. With the ratios:

• $\sin^2\theta + \operatorname{cosec}^2\theta \ge 2$, equal only at $\theta = 90^\circ$
• $\cos^2\theta + \sec^2\theta \ge 2$, equal only at $\theta = 0^\circ$
• $\tan^2\theta + \cot^2\theta \ge 2$, equal only at $\theta = 45^\circ$

$a\tan^2\theta + b\cot^2\theta$. The two terms multiply to ab, a fixed number. When two positive numbers have a fixed product, their sum is least when they are equal, and then each is $\sqrt{ab}$ (Chapter 11). So

$$a\tan^2\theta + b\cot^2\theta \ge 2\sqrt{ab}$$

For instance $2\tan^2\theta + 8\cot^2\theta \ge 2\sqrt{16} = 8$. The least value 8 comes when $2\tan^2\theta = 8\cot^2\theta$, that is when $\tan^4\theta = 4$ and $\tan^2\theta = 2$.

$a\sec^2\theta + b\operatorname{cosec}^2\theta$. Change to tan and cot first:

$$a(1 + \tan^2\theta) + b(1 + \cot^2\theta) = a + b + a\tan^2\theta + b\cot^2\theta \ge a + b + 2\sqrt{ab} = (\sqrt{a} + \sqrt{b})^2$$

So $\sec^2\theta + 4\operatorname{cosec}^2\theta \ge (1 + 2)^2 = 9$.

All six squares. $\sin^2\theta + \cos^2\theta + \tan^2\theta + \cot^2\theta + \sec^2\theta + \operatorname{cosec}^2\theta$ is $1 + \tan^2\theta + \cot^2\theta + (1 + \tan^2\theta) + (1 + \cot^2\theta) = 3 + 2(\tan^2\theta + \cot^2\theta)$, which is at least 3 + 4 = 7.

Check that the equal point can happen. The bound $2\sqrt{ab}$ is the true least value only if the two terms can be equal. For $\sin^2\theta + 4\operatorname{cosec}^2\theta$, equality would need $\sin^2\theta = 2$, which is impossible. Writing $x = \sin^2\theta$, the expression $x + \frac{4}{x}$ keeps falling as x rises to 1, so its least value comes at $\sin^2\theta = 1$ and is 1 + 4 = 5, not 4. With tan and cot the problem never arises, because $\tan^2\theta$ can be any positive number.

Values a Ratio Can Never Take

In a right triangle the hypotenuse is the longest side, and that fixes where each ratio can lie.

• $\sin\theta$ and $\cos\theta$ lie from −1 to 1 (from 0 to 1 for acute angles). So $\sin\theta = \frac{7}{6}$ and $\cos\theta = \sqrt{2}$ are impossible.
• $\sec\theta$ and $\operatorname{cosec}\theta$ are never strictly between −1 and 1. $\sec\theta = \frac{3}{5}$ and $\operatorname{cosec}\theta = \frac{2}{3}$ are impossible, while $\sec\theta = 2$ is fine.
• $\tan\theta$ and $\cot\theta$ can be any real number. $\tan\theta = 100$ is possible; $\theta$ is then just under 90°.

Rejecting a root. $2\sin^2\theta + 3\sin\theta - 2 = 0$ factorises as $(2\sin\theta - 1)(\sin\theta + 2) = 0$. $\sin\theta = -2$ is impossible, so $\sin\theta = \frac{1}{2}$ and $\theta = 30^\circ$ for an acute angle.

Conditions with letters. $\sec^2\theta = \frac{4xy}{(x + y)^2}$, with x and y positive, can hold only when x = y. The reason: $(x + y)^2 - 4xy = (x - y)^2 \ge 0$, so the right side is at most 1, while $\sec^2\theta$ is at least 1. Both can be true only when the right side is exactly 1. In the same way $\sin\theta = \frac{x^2 + y^2}{2xy}$ needs x = y, since $x^2 + y^2 \ge 2xy$. And $\cos\theta = x + \frac{1}{x}$ is impossible for every real x, because $x + \frac{1}{x}$ is either at least 2 or at most −2.

[SSC Tip] Look at the numbers before solving. A sine of 1.2, or $\sin\theta + \cos\theta = 1.42$ (just more than $\sqrt{2} \approx 1.414$), means the data are impossible, and an option of that kind can be struck out at once.

Solved Examples

Question 1

If $\sin\theta + \cos\theta = \frac{13}{10}$, what is the value of $\sin\theta\cos\theta$? (a) $\frac{69}{100}$ (b) $\frac{3}{10}$ (c) $\frac{69}{200}$ (d) $\frac{269}{200}$

Answer. I square both sides: $1 + 2\sin\theta\cos\theta = \frac{169}{100}$. So $2\sin\theta\cos\theta = \frac{69}{100}$ and $\sin\theta\cos\theta$ = (c) $\frac{69}{200}$. The value $\frac{69}{100}$ is twice the product, so I make sure to halve it.

Question 2

If $\sin\theta - \cos\theta = \frac{1}{4}$, what is the value of $\sin\theta\cos\theta$? (a) $\frac{15}{32}$ (b) $\frac{15}{16}$ (c) $\frac{1}{4}$ (d) $\frac{17}{32}$

Answer. Squaring, $1 - 2\sin\theta\cos\theta = \frac{1}{16}$. The minus sign in the middle matters here: $2\sin\theta\cos\theta = 1 - \frac{1}{16} = \frac{15}{16}$, so the product is (a) $\frac{15}{32}$.

Question 3

If $\sin\theta + \cos\theta = \frac{9}{7}$, what is the value of $\tan\theta + \cot\theta$? (a) $\frac{16}{49}$ (b) $\frac{9}{7}$ (c) $\frac{32}{49}$ (d) $\frac{49}{16}$

Answer. First the product: $1 + 2\sin\theta\cos\theta = \frac{81}{49}$, so $\sin\theta\cos\theta = \frac{16}{49}$. Then $\tan\theta + \cot\theta = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}$ = (d) $\frac{49}{16}$. Stopping at $\frac{16}{49}$ gives the product, which is only the halfway point.

Question 4

If $\sec\theta + \tan\theta = \frac{5}{2}$, what is the value of $\sin\theta$? (a) $\frac{21}{29}$ (b) $\frac{20}{29}$ (c) $\frac{29}{21}$ (d) $\frac{21}{20}$

Answer. The partner is $\sec\theta - \tan\theta = \frac{2}{5}$. Adding the two equations: $2\sec\theta = \frac{5}{2} + \frac{2}{5} = \frac{29}{10}$, so $\sec\theta = \frac{29}{20}$. Subtracting: $2\tan\theta = \frac{21}{10}$, so $\tan\theta = \frac{21}{20}$. Then $\sin\theta = \frac{\tan\theta}{\sec\theta}$ = (a) $\frac{21}{29}$. The triangle is 20, 21, 29, and $\frac{20}{29}$ would be $\cos\theta$.

Question 5

If $\operatorname{cosec}\theta + \cot\theta = 9$, what is the value of $\cos\theta$? (a) $\frac{9}{41}$ (b) $\frac{40}{41}$ (c) $\frac{41}{40}$ (d) $\frac{9}{40}$

Answer. $\operatorname{cosec}\theta - \cot\theta = \frac{1}{9}$. Adding, $2\operatorname{cosec}\theta = 9 + \frac{1}{9} = \frac{82}{9}$, so $\operatorname{cosec}\theta = \frac{41}{9}$; subtracting, $2\cot\theta = \frac{80}{9}$, so $\cot\theta = \frac{40}{9}$. Now $\cos\theta = \frac{\cot\theta}{\operatorname{cosec}\theta}$ = (b) $\frac{40}{41}$. The formula $\frac{p^2 - 1}{p^2 + 1} = \frac{80}{82}$ gives the same in one line.

Question 6

If $\sec\theta - \tan\theta = \frac{1}{7}$, what is the value of $\cos\theta$? (a) $\frac{24}{25}$ (b) $\frac{25}{7}$ (c) $\frac{7}{25}$ (d) $\frac{24}{7}$

Answer. I turn the difference over: $\sec\theta + \tan\theta = 7$. Adding the two, $2\sec\theta = 7 + \frac{1}{7} = \frac{50}{7}$, so $\sec\theta = \frac{25}{7}$. Then $\cos\theta$ is its reciprocal, (c) $\frac{7}{25}$.

Question 7

If $\tan\theta + \cot\theta = 2$, what is the value of $\tan^{100}\theta + \cot^{100}\theta$? (a) $2^{100}$ (b) 2 (c) 1 (d) 100

Answer. A number plus its reciprocal is 2 only when the number is 1. So $\tan\theta = 1$ and $\cot\theta = 1$, and any power of 1 is 1. The answer is 1 + 1 = (b) 2.

Question 8

If $\tan\theta + \cot\theta = 5$, what is the value of $\tan^2\theta + \cot^2\theta$? (a) 25 (b) 27 (c) 21 (d) 23

Answer. Squaring, $\tan^2\theta + 2\tan\theta\cot\theta + \cot^2\theta = 25$. The middle term is 2, because $\tan\theta\cot\theta = 1$. So $\tan^2\theta + \cot^2\theta = 25 - 2$ = (d) 23.

Question 9

If $\sin\theta + \sin^2\theta = 1$, what is the value of $\tan^4\theta + \tan^2\theta$? (a) 1 (b) 0 (c) 2 (d) $\frac{1}{2}$

Answer. The condition says $\sin\theta = 1 - \sin^2\theta = \cos^2\theta$. So $\tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta} = \frac{\sin^2\theta}{\sin\theta} = \sin\theta$, and then $\tan^4\theta = \sin^2\theta$. The expression is $\sin^2\theta + \sin\theta$, which is the given condition, so it equals (a) 1.

Question 10

If $\cos\theta + \cos^2\theta = 1$, what is the value of $\sin^{12}\theta + 3\sin^{10}\theta + 3\sin^8\theta + \sin^6\theta + 2\sin^4\theta + 2\sin^2\theta - 2$? (a) 0 (b) 2 (c) 1 (d) 3

Answer. The condition gives $\cos\theta = 1 - \cos^2\theta = \sin^2\theta$, so $\sin^2\theta + \sin^4\theta = \cos\theta + \cos^2\theta = 1$. The first four terms have coefficients 1, 3, 3, 1, so they make the cube $(\sin^4\theta + \sin^2\theta)^3 = 1$. The next two are $2(\sin^4\theta + \sin^2\theta) = 2$. So the value is 1 + 2 − 2 = (c) 1.

Question 11

If $3\tan\theta = 4$, what is the value of $\frac{3\sin\theta - 2\cos\theta}{3\sin\theta + \cos\theta}$? (a) $\frac{2}{3}$ (b) $\frac{2}{5}$ (c) $\frac{1}{5}$ (d) $\frac{5}{2}$

Answer. Every term has one sin or cos, so I divide the top and the bottom by $\cos\theta$: $\frac{3\tan\theta - 2}{3\tan\theta + 1} = \frac{4 - 2}{4 + 1}$ = (b) $\frac{2}{5}$. I never needed $\sin\theta$ and $\cos\theta$ themselves.

Question 12

If $\frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = 5$, what is the value of $\sin^2\theta - \cos^2\theta$? (a) $\frac{12}{13}$ (b) $\frac{3}{2}$ (c) $\frac{13}{5}$ (d) $\frac{5}{13}$

Answer. Dividing the top and the bottom by $\cos\theta$: $\frac{\tan\theta + 1}{\tan\theta - 1} = 5$, so $\tan\theta + 1 = 5\tan\theta - 5$ and $\tan\theta = \frac{3}{2}$. Now I write $\sin^2\theta - \cos^2\theta$ over $\sin^2\theta + \cos^2\theta$ (which is 1) and divide by $\cos^2\theta$: $\frac{\tan^2\theta - 1}{\tan^2\theta + 1} = \frac{\frac{9}{4} - 1}{\frac{9}{4} + 1} = \frac{5}{13}$. The answer is (d) $\frac{5}{13}$.

Question 13

If $\sin\theta - \cos\theta = \sqrt{2}\cos\theta$, then $\sin\theta + \cos\theta$ is equal to (a) $\sqrt{2}\sin\theta$ (b) $\sqrt{2}\cos\theta$ (c) $-\sqrt{2}\sin\theta$ (d) $\sqrt{2}\tan\theta$

Answer. From the condition, $\sin\theta = (\sqrt{2} + 1)\cos\theta$. I multiply both sides by $\sqrt{2} - 1$, using $(\sqrt{2} + 1)(\sqrt{2} - 1) = 1$: $(\sqrt{2} - 1)\sin\theta = \cos\theta$. So $\cos\theta = \sqrt{2}\sin\theta - \sin\theta$, which rearranges to $\sin\theta + \cos\theta$ = (a) $\sqrt{2}\sin\theta$.

Question 14

If $5\sin\theta + 12\cos\theta = 13$, what is the value of $\tan\theta$? (a) $\frac{12}{5}$ (b) $\frac{5}{13}$ (c) $\frac{5}{12}$ (d) $\frac{13}{12}$

Answer. The partner expression is $5\cos\theta - 12\sin\theta$, and the squares of the two add to $5^2 + 12^2 = 169$. Since the given one is already 13, the partner is $\sqrt{169 - 169} = 0$. So $5\cos\theta = 12\sin\theta$, and $\tan\theta = \frac{\sin\theta}{\cos\theta}$ = (c) $\frac{5}{12}$. (13 is also the greatest value $5\sin\theta + 12\cos\theta$ can take.)

Question 15

If $2\cos\theta - \sin\theta = 1$ ($\theta$ acute), what is the value of $2\sin\theta + \cos\theta$? (a) 1 (b) 2 (c) $\sqrt{5}$ (d) 3

Answer. The two expressions pair up, so their squares add to $2^2 + 1^2 = 5$. That gives $(2\sin\theta + \cos\theta)^2 = 5 - 1 = 4$, and for an acute angle it is positive, so it equals (b) 2. A check: $\sin\theta = \frac{3}{5}$, $\cos\theta = \frac{4}{5}$ fits the condition ($\frac{8}{5} - \frac{3}{5} = 1$), and then $\frac{6}{5} + \frac{4}{5} = 2$.

Question 16

If $2\sin^2\theta + 5\cos^2\theta = 3$ ($\theta$ acute), what is the value of $\tan\theta$? (a) $\frac{1}{\sqrt{2}}$ (b) $\sqrt{3}$ (c) 2 (d) $\sqrt{2}$

Answer. Two of the five $\cos^2\theta$ join the two $\sin^2\theta$ to make 2, so the condition reads $2 + 3\cos^2\theta = 3$. That gives $\cos^2\theta = \frac{1}{3}$ and $\sin^2\theta = \frac{2}{3}$. So $\tan^2\theta = 2$ and $\tan\theta$ = (d) $\sqrt{2}$. Taking $\tan^2\theta$ the other way up gives $\frac{1}{\sqrt{2}}$, which is $\cot\theta$.

Question 17

[Tier 2] If $\operatorname{cosec}\theta = x + \frac{1}{4x}$ ($x > 0$), then $\operatorname{cosec}\theta + \cot\theta$ is equal to (a) $2x$ or $\frac{1}{2x}$ (b) $x$ or $\frac{1}{x}$ (c) $2x$ only (d) $\frac{1}{4x}$

Answer. I find $\cot^2\theta = \operatorname{cosec}^2\theta - 1 = x^2 + \frac{1}{2} + \frac{1}{16x^2} - 1 = x^2 - \frac{1}{2} + \frac{1}{16x^2} = \left(x - \frac{1}{4x}\right)^2$. So $\cot\theta = \pm\left(x - \frac{1}{4x}\right)$, the sign depending on whether x is more or less than $\frac{1}{2}$. Adding to $\operatorname{cosec}\theta$ gives either $2x$ or $\frac{1}{2x}$, so the answer is (a) $2x$ or $\frac{1}{2x}$. For x = 2, $\operatorname{cosec}\theta = \frac{17}{8}$, $\cot\theta = \frac{15}{8}$ and the sum is 4, which is 2x.

Question 18

If $2\cos^2\theta + 5\cos\theta - 3 = 0$ and $\theta$ is acute, what is $\theta$? (a) 30° (b) 45° (c) 60° (d) No acute angle satisfies it

Answer. It is a quadratic in $\cos\theta$: $(2\cos\theta - 1)(\cos\theta + 3) = 0$. $\cos\theta = -3$ is impossible because cosine never goes below −1, so $\cos\theta = \frac{1}{2}$ and $\theta$ is (c) 60°.