# Pythagorean identity: why sin² + cos² = 1

> The identity sin²θ + cos²θ = 1 is the Pythagorean theorem on the unit circle. Tangent and secant are lengths in the same picture and give two more.

Series: Trigonometry (https://www.monocurl.com/monocurl-essays/trigonometry/)
Canonical: https://www.monocurl.com/monocurl-essays/trigonometry/pythagorean-identity/
Keywords: Pythagorean identity, sin squared plus cos squared equals 1, 1 + tan squared = sec squared, trig identities proof, tangent and secant on the unit circle, find cos from sin

[Figure 1, animation: As a point travels around the unit circle, the squares of its two coordinates share out a bar of length 1 between them. Whatever \cos^2\theta gives up, \sin^2\theta takes.]

Ask a calculator for $\sin 37°$ and $\cos 37°$. It gives $0.6018$ and $0.7986$, two numbers with no obvious connection. Now square each and add: $0.3622 + 0.6378 = 1.0000$. Try an angle from the far side of the circle, 200°: the sine is $-0.3420$, the cosine is $-0.9397$, and the squares are $0.1170$ and $0.8830$, which again add to exactly 1.

It happens for every angle you try. Two functions with complicated decimal values are bound together by one of the simplest equations there is. The reason is a theorem that is at least two and a half thousand years old.

## Why the sum is 1

Recall what the two numbers are. On [the unit circle](@trigonometry/unit-circle), the circle of radius 1 centered at the origin, start at the point $(1, 0)$ and turn about the origin through the angle $\theta$, counterclockwise. The point you reach has coordinates $(\cos\theta, \sin\theta)$: cosine is how far across it is, sine is how far up.

**Pause and try.** The point $(\cos\theta, \sin\theta)$ is on a circle of radius 1 around the origin. For an angle in the first quadrant, draw the right triangle formed by the radius to the point, the vertical line from the point down to the $x$ axis, and the piece of the axis between. What are its three side lengths, and what does the Pythagorean theorem say about them?

**Answer.** The horizontal leg runs from the origin to the foot of the vertical line, so its length is the point's $x$ coordinate, $\cos\theta$. The vertical leg has length $\sin\theta$. The hypotenuse is a radius, of length 1. The [Pythagorean theorem](@geometry/pythagorean-theorem) says the squares of the legs add up to the square of the hypotenuse: $(\cos\theta)^2 + (\sin\theta)^2 = 1^2 = 1$.

That is the whole secret for an acute angle. A good instance to carry through the lesson is the point $(0.8, 0.6)$, which lies on the unit circle because $0.64 + 0.36 = 1$. Its triangle is the 3-4-5 triangle shrunk to hypotenuse 1, with angle $\theta \approx 36.87°$, and for this angle $\cos\theta = 0.8$ and $\sin\theta = 0.6$ exactly.

You might object that the argument used a triangle, and the opening example at 200° has no such triangle with legs $\cos 200°$ and $\sin 200°$: both numbers are negative, and a side cannot have negative length.

[Figure 2, image: The point of the unit circle at 200° lies in the third quadrant. The triangle under it has legs of length 0.940 and 0.342, the sizes of the two negative coordinates, and their squares still add to 1.]

The triangle is still there. Its legs have lengths $|\cos\theta|$ and $|\sin\theta|$, the distances of the point from the two axes, and squaring a number gives the same result as squaring its absolute value. So the signs disappear exactly where they would have caused trouble. Only on the axes is there no triangle at all, and there the equation can be checked directly: at the top of the circle the point is $(0, 1)$ and $0^2 + 1^2 = 1$.

There is a cleaner way to say all of this at once. The Pythagorean theorem written in coordinates is the distance formula: the distance from the origin to a point $(x, y)$ is $\sqrt{x^2 + y^2}$. It holds for every point, in every quadrant and on the axes. The identity is the statement that $(\cos\theta, \sin\theta)$ is at distance 1 from the origin.

A word on notation before the formal statement. By long custom $\sin^2\theta$ means $(\sin\theta)^2$, the square of the sine. It does not mean $\sin(\theta^2)$, the sine of the squared angle.

**Theorem.** **(Pythagorean identity.)** For every real number $\theta$,

$$
\sin^2\theta + \cos^2\theta = 1.
$$

**Proof.** Let $O$ be the origin and $P = (\cos\theta, \sin\theta)$. By the definition of sine and cosine, $P$ is the point obtained by rotating $(1, 0)$ about $O$ through the angle $\theta$. A rotation about $O$ does not change distances from $O$, and $(1, 0)$ is at distance 1, so $OP = 1$. By the distance formula,

$$
OP = \sqrt{(\cos\theta - 0)^2 + (\sin\theta - 0)^2} = \sqrt{\cos^2\theta + \sin^2\theta}.
$$

Therefore $\sqrt{\cos^2\theta + \sin^2\theta} = 1$, and squaring both sides gives the claim.

The proof leans on one earlier result, the distance formula, which is the Pythagorean theorem applied to the horizontal and vertical separations of two points and which remains true when one of the separations is zero. "For every real number" is meant literally: negative angles and angles beyond a full turn are included, in degrees or radians.

An equation that holds for every value of its variable is called an **identity**. That separates it from an equation such as $\sin\theta = \frac{1}{2}$, which is true for some angles and false for others and which one solves. An identity is used differently: as a rule for replacing one expression by another.

[Figure 3, image: The graphs of \cos^2\theta and \sin^2\theta over one full turn. At every angle the two heights add to 1, as the stacked bars at 30°, 135° and 250° show.]

## One coordinate nearly determines the other

The first use of the identity is to recover one of the two functions from the other without ever finding the angle.

**Worked example.** **Given $\sin\theta = \frac{3}{5}$, find $\cos\theta$.**

Substitute into the identity: $\frac{9}{25} + \cos^2\theta = 1$, so $\cos^2\theta = \frac{16}{25}$. Two numbers have that square, so

$$
\cos\theta = \frac{4}{5} \quad \text{or} \quad \cos\theta = -\frac{4}{5}.
$$

Both really occur. The points of the unit circle at height $\frac{3}{5}$ are $\left(\frac{4}{5}, \frac{3}{5}\right)$ in the first quadrant and $\left(-\frac{4}{5}, \frac{3}{5}\right)$ in the second. The sine alone cannot tell them apart. If we are also told the quadrant, the sign is settled: cosine is positive on the right half of the circle and negative on the left.

[Figure 4, image: The horizontal line at height \frac{3}{5} meets the unit circle in two points. They have the same sine and opposite cosines, \frac{4}{5} and -\frac{4}{5}, which is why the identity gives \cos\theta only up to sign.]

**Pause and try.** Suppose $\cos\theta = -\frac{5}{13}$ and the point at angle $\theta$ is in the third quadrant. Find $\sin\theta$.

**Answer.** $\sin^2\theta = 1 - \cos^2\theta = 1 - \frac{25}{169} = \frac{144}{169}$, so $\sin\theta = \pm\frac{12}{13}$. In the third quadrant the point is below the $x$ axis, so the sine is negative: $\sin\theta = -\frac{12}{13}$. Check: $\frac{25}{169} + \frac{144}{169} = 1$.

## Tangent and secant as lengths

Sine and cosine are lengths in the unit circle picture. Four more functions are built from them by division, and the two met most often are the **tangent**, $\tan\theta = \frac{\sin\theta}{\cos\theta}$, and the **secant**, $\sec\theta = \frac{1}{\cos\theta}$. As formulas they look like bookkeeping. Compute them at our running point, though, where $\cos\theta = 0.8$ and $\sin\theta = 0.6$:

$$
\tan\theta = \frac{0.6}{0.8} = 0.75, \qquad \sec\theta = \frac{1}{0.8} = 1.25.
$$

The numbers $1$, $0.75$ and $1.25$ are $4$, $3$ and $5$ divided by 4. They are the sides of another right triangle: $1^2 + 0.75^2 = 1 + 0.5625 = 1.5625 = 1.25^2$.

**Pause and try.** Is that a coincidence of the 3-4-5 triangle? Test another angle: compute $1 + \tan^2 60°$ and $\sec^2 60°$ exactly, using $\sin 60° = \frac{\sqrt 3}{2}$ and $\cos 60° = \frac{1}{2}$.

**Answer.** $\tan 60° = \frac{\sqrt 3 / 2}{1/2} = \sqrt 3$, so $1 + \tan^2 60° = 1 + 3 = 4$. And $\sec 60° = \frac{1}{1/2} = 2$, so $\sec^2 60° = 4$. They agree again, so $1$, $\tan\theta$ and $\sec\theta$ look like the sides of a right triangle for every angle.

If there is a right triangle with sides $1$, $\tan\theta$ and $\sec\theta$, it ought to be somewhere in the picture. Step through the figure to find it, along with a third triangle for the remaining two functions, the **cotangent** $\cot\theta = \frac{\cos\theta}{\sin\theta}$ and the **cosecant** $\csc\theta = \frac{1}{\sin\theta}$.

[Figure 5, steps: All six trigonometric functions as lengths, for the angle whose point on the unit circle is (0.8, 0.6). Tangent and secant lie on the line touching the circle at (1, 0); cotangent and cosecant lie on the line touching it at (0, 1).]

  - step: The point P at angle θ on the unit circle is (cos θ, sin θ), here (0.8, 0.6). The two coordinates and the radius form a right triangle with hypotenuse 1.
  - step: Draw the line that touches the circle at A = (1, 0), and extend the radius through P until it meets that line at T. The height AT is tan θ = 0.75.
  - step: The distance from the center O out to T is sec θ = 1.25, measured here just below the line OT. The triangle OAT has sides 1, tan θ and sec θ.
  - step: Now the line that touches the circle at B = (0, 1). The same radius, extended further, meets it at C. The length BC is cot θ ≈ 1.33.
  - step: The distance from O out to C is csc θ ≈ 1.67. The triangle OBC has sides 1, cot θ and csc θ.
The names come from this picture. A line that touches a circle at one point is a tangent line, and $\tan\theta$ is a length measured along the tangent line at $A$. The line through $O$ and $T$ cuts across the circle, and "secant" comes from the Latin for cutting.

Why are those lengths the quotients we defined? Let $F$ be the foot of the vertical line from $P$. The triangles $OFP$ and $OAT$ both have a right angle (at $F$ and at $A$) and share the angle $\theta$ at $O$. [Triangles with the same angles are similar](@geometry/similar-triangles): one is a scaled copy of the other. The horizontal leg grows from $OF = \cos\theta$ to $OA = 1$, so the scale factor is $\frac{1}{\cos\theta}$, and the other two sides grow by the same factor:

$$
AT = \frac{\sin\theta}{\cos\theta} = \tan\theta, \qquad OT = \frac{1}{\cos\theta} = \sec\theta.
$$

The third triangle, $OBC$, has its right angle at $B$. The line $BC$ is parallel to the $x$ axis, so the angle at $C$ equals the angle $\theta$ that the line $OC$ makes with the axis. It is again similar to $OFP$, this time with the side opposite $\theta$ growing from $\sin\theta$ to $OB = 1$. The scale factor is $\frac{1}{\sin\theta}$, which gives $BC = \frac{\cos\theta}{\sin\theta} = \cot\theta$ and $OC = \frac{1}{\sin\theta} = \csc\theta$.

[Figure 6, image: The three right triangles of the diagram drawn side by side at the same scale. Each has the angle \theta, so they are similar, and the Pythagorean theorem applied to each one gives one identity.]

Three similar right triangles give three instances of the Pythagorean theorem. The first is the identity we started with. The second says $1 + \tan^2\theta = \sec^2\theta$, which is the pattern in the numbers $1$, $0.75$, $1.25$. The third says $\cot^2\theta + 1 = \csc^2\theta$.

The picture only covers angles between 0° and 90°, where all six lengths exist as drawn. The algebra has no such restriction. Dividing the sides of a triangle by $\cos\theta$ corresponds to dividing the equation $\sin^2\theta + \cos^2\theta = 1$ by $\cos^2\theta$, and that can be done for any angle at which $\cos\theta$ is not zero.

Outside the first quadrant the same two tangent lines still carry the four functions, as signed positions in place of lengths. The line through $O$ and $P$ consists of the points $(t\cos\theta, t\sin\theta)$, where $t$ is the position along the line measured from $O$, positive on the side of $P$. If $\cos\theta \ne 0$, this line meets the vertical line $x = 1$ where $t\cos\theta = 1$, that is at $t = \sec\theta$, and the point there is $T = (1, \tan\theta)$. So $\tan\theta$ is the height of $T$, negative when $T$ is below the axis, and $\sec\theta$ is the position of $T$ along the line, negative when $T$ and $P$ lie on opposite sides of $O$. At 200°, for example, $P$ is in the third quadrant and the line has to be extended backward through $O$ to reach $x = 1$. It arrives above the axis, at height $\tan 200° \approx 0.364$, and $\sec 200° \approx -1.064$ is negative. In the same way, if $\sin\theta \ne 0$ the line meets $y = 1$ at $C = (\cot\theta, 1)$, at position $t = \csc\theta$. The distances themselves are $OT = |\sec\theta|$ and $OC = |\csc\theta|$.

**Definition.** For a real number $\theta$,

$$
\tan\theta = \frac{\sin\theta}{\cos\theta}, \quad \sec\theta = \frac{1}{\cos\theta} \quad (\text{defined when } \cos\theta \ne 0),
$$

$$
\cot\theta = \frac{\cos\theta}{\sin\theta}, \quad \csc\theta = \frac{1}{\sin\theta} \quad (\text{defined when } \sin\theta \ne 0).
$$

**Theorem.** For every real number $\theta$ with $\cos\theta \ne 0$,

$$
1 + \tan^2\theta = \sec^2\theta.
$$

For every real number $\theta$ with $\sin\theta \ne 0$,

$$
1 + \cot^2\theta = \csc^2\theta.
$$

**Proof.** Start from $\sin^2\theta + \cos^2\theta = 1$, which holds for every $\theta$. If $\cos\theta \ne 0$, we may divide both sides by $\cos^2\theta$:

$$
\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}, \qquad \text{that is,} \qquad \tan^2\theta + 1 = \sec^2\theta.
$$

If $\sin\theta \ne 0$, divide both sides by $\sin^2\theta$ instead:

$$
\frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \frac{1}{\sin^2\theta}, \qquad \text{that is,} \qquad 1 + \cot^2\theta = \csc^2\theta.
$$

**Key idea.** $\sin^2\theta + \cos^2\theta = 1$ is the Pythagorean theorem for the triangle inside the unit circle. Dividing it by $\cos^2\theta$ or by $\sin^2\theta$ gives $1 + \tan^2\theta = \sec^2\theta$ and $1 + \cot^2\theta = \csc^2\theta$, the same theorem for two enlarged copies of that triangle.

[Figure 7, animation: The angle sweeps between 32° and 58° and the six lengths change with it. Tangent and secant grow as the angle opens while cotangent and cosecant shrink, yet 1 + \tan^2\theta always equals \sec^2\theta and \cot^2\theta + 1 always equals \csc^2\theta.]

## Working with the identities

**Worked example.** **Given $\tan\theta = -2$ with $\theta$ in the second quadrant, find $\cos\theta$ and $\sin\theta$.**

The identity that contains the tangent is $1 + \tan^2\theta = \sec^2\theta$, so start there: $\sec^2\theta = 1 + 4 = 5$, and $\sec\theta = \pm\sqrt 5$. In the second quadrant the cosine is negative, and $\sec\theta = \frac{1}{\cos\theta}$ has the same sign as the cosine, so $\sec\theta = -\sqrt 5$ and

$$
\cos\theta = -\frac{1}{\sqrt 5} \approx -0.447.
$$

For the sine, use the definition of tangent rather than another square root, since that avoids a second choice of sign: $\sin\theta = \tan\theta \cdot \cos\theta = (-2)\left(-\frac{1}{\sqrt 5}\right) = \frac{2}{\sqrt 5} \approx 0.894$.

*Check.* $\sin^2\theta + \cos^2\theta = \frac{4}{5} + \frac{1}{5} = 1$, and a positive sine with a negative cosine is right for the second quadrant.

**Worked example.** **Show that $\tan\theta + \cot\theta = \sec\theta \csc\theta$ whenever $\sin\theta \ne 0$ and $\cos\theta \ne 0$.**

When an expression mixes several of the six functions, a dependable plan is to write everything in terms of sine and cosine and watch for $\sin^2\theta + \cos^2\theta$ to appear. Put the left side over a common denominator:

$$
\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}.
$$

The right side is $\frac{1}{\cos\theta} \cdot \frac{1}{\sin\theta}$, the same thing.

*Check* at the running point: $\tan\theta + \cot\theta = \frac{3}{4} + \frac{4}{3} = \frac{25}{12}$, and $\sec\theta\csc\theta = \frac{5}{4} \cdot \frac{5}{3} = \frac{25}{12}$.

## Where the identities break down

The first identity never fails. The other two have conditions, and the picture shows what the conditions mean. At $\theta = 90°$ the radius points straight up, parallel to the tangent line at $A$, and never meets it. There is no point $T$, and correspondingly $\cos 90° = 0$, so $\tan 90°$ and $\sec 90°$ are undefined. Just short of 90° the point $T$ is very far up the line: $\tan 89° \approx 57.3$ and $\tan 89.9° \approx 573$. Meanwhile $\sin^2 90° + \cos^2 90° = 1 + 0 = 1$ holds without complaint. Dividing by $\cos^2\theta$ produced a new identity but lost the angles where $\cos\theta = 0$.

The other limit concerns square roots.

**Pause and try.** A student reasons: "$\sin^2\theta + \cos^2\theta = 1$, so taking square roots, $\sin\theta + \cos\theta = 1$." Test the conclusion at 37° using the numbers from the start of the lesson. Then find every angle with $0° \le \theta < 360°$ for which $\sin\theta + \cos\theta = 1$ is true. (Square both sides and use the identity.)

**Answer.** At 37° the sum is $0.6018 + 0.7986 = 1.4004$, not 1. To find where it does hold, square: $(\sin\theta + \cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = 1 + 2\sin\theta\cos\theta$. If $\sin\theta + \cos\theta = 1$ then this equals 1, so $\sin\theta\cos\theta = 0$, which means $\sin\theta = 0$ or $\cos\theta = 0$. The candidates are 0°, 90°, 180° and 270°. Squaring can create false solutions, so test each: at 0° the sum is $0 + 1 = 1$ and at 90° it is $1 + 0 = 1$, but at 180° and 270° it is $-1$. So the equation holds only at 0° and 90°.

**Common mistake.** **Taking the square root term by term.** As the exercise shows, $\sin\theta + \cos\theta = 1$ is false for almost every angle. The square root of a sum is not the sum of the square roots. It is the same error as concluding $a + b = c$ from $a^2 + b^2 = c^2$ in a right triangle, where it would say that the hypotenuse is as long as the two legs together.

**Common mistake.** **Dropping the sign.** From the identity, $\cos^2\theta = 1 - \sin^2\theta$, and it is tempting to write $\cos\theta = \sqrt{1 - \sin^2\theta}$. The square root symbol means the non-negative root, so what is true is $|\cos\theta| = \sqrt{1 - \sin^2\theta}$. At $\theta = 120°$, for instance, $\sin\theta = \frac{\sqrt 3}{2}$ and $\sqrt{1 - \frac{3}{4}} = \frac{1}{2}$, but $\cos 120° = -\frac{1}{2}$. The identity gives the size of the other coordinate. Its sign must come from the quadrant.

## Looking back

The opening puzzle was two unrelated-looking decimals whose squares add to 1. The difficulty was only in seeing where to look: once sine and cosine are read as the coordinates of a point on a circle of radius 1, the relation between them is the Pythagorean theorem and nothing more. The step that needed care was the passage beyond the first quadrant, where the triangle's legs are $|\cos\theta|$ and $|\sin\theta|$ and the squares hide the signs.

The reusable idea is that one geometric fact, seen in similar triangles of different sizes, appears as several formulas that look unrelated on the page. When you meet a new trigonometric identity, ask which triangle it is the Pythagorean theorem for, or which known identity it is after a division. The next lesson, on [the angle addition formulas](@trigonometry/angle-addition), takes up a question the unit circle does not answer so quickly: what are the sine and cosine of a sum of two angles?

**Check yourself.** Suppose $\sin\theta = 0.6$ and the point at angle $\theta$ lies in the second quadrant. What is $\cos\theta$?
Answer: $-0.8$ $\cos^2\theta = 1 - 0.36 = 0.64$, so $\cos\theta = \pm 0.8$, and the second quadrant makes it negative.

**Recap.** - For every angle $\theta$, $\sin^2\theta + \cos^2\theta = 1$, because $(\cos\theta, \sin\theta)$ is a point at distance 1 from the origin. This is the Pythagorean theorem on the unit circle.
- Knowing one of $\sin\theta$, $\cos\theta$ gives the other up to sign: $\cos\theta = \pm\sqrt{1 - \sin^2\theta}$. The quadrant decides the sign.
- $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and $\sec\theta = \frac{1}{\cos\theta}$ are the vertical side and the hypotenuse of the triangle cut off by the tangent line at $(1, 0)$; $\cot\theta$ and $\csc\theta$ play the same roles on the tangent line at $(0, 1)$.
- $1 + \tan^2\theta = \sec^2\theta$ when $\cos\theta \ne 0$, and $1 + \cot^2\theta = \csc^2\theta$ when $\sin\theta \ne 0$. Each is the first identity divided by $\cos^2\theta$ or $\sin^2\theta$.
- To prove an identity, rewrite in terms of sine and cosine and look for $\sin^2\theta + \cos^2\theta$.
- $\sin\theta + \cos\theta = 1$ is not an identity; it holds only at 0° and 90° in each full turn.

## Problems

**Problem 1.** Given $\cos\theta = 0.28$ with $\theta$ in the fourth quadrant, find $\sin\theta$.

**Solution.** $\sin^2\theta = 1 - 0.28^2 = 1 - 0.0784 = 0.9216$, and $0.9216 = 0.96^2$, so $\sin\theta = \pm 0.96$. In the fourth quadrant the point is below the $x$ axis, so $\sin\theta = -0.96$. Check: $0.0784 + 0.9216 = 1$.

**Problem 2.** Given $\sin\theta = -\frac{5}{13}$ with $\theta$ in the third quadrant, find $\cos\theta$, $\tan\theta$ and $\sec\theta$.

**Solution.** $\cos^2\theta = 1 - \frac{25}{169} = \frac{144}{169}$, so $\cos\theta = \pm\frac{12}{13}$, and the third quadrant makes it negative: $\cos\theta = -\frac{12}{13}$. Then $\tan\theta = \frac{-5/13}{-12/13} = \frac{5}{12}$ and $\sec\theta = \frac{1}{\cos\theta} = -\frac{13}{12}$. Check with the second identity: $1 + \frac{25}{144} = \frac{169}{144} = \left(-\frac{13}{12}\right)^2$.

**Problem 3.** Given $\sec\theta = 3$ with $\theta$ in the first quadrant, find $\tan\theta$ and $\sin\theta$.

**Solution.** $\tan^2\theta = \sec^2\theta - 1 = 9 - 1 = 8$, and the tangent is positive in the first quadrant, so $\tan\theta = \sqrt 8 = 2\sqrt 2 \approx 2.828$. Since $\cos\theta = \frac{1}{\sec\theta} = \frac{1}{3}$, we get $\sin\theta = \tan\theta\cos\theta = \frac{2\sqrt 2}{3} \approx 0.943$. Check: $\sin^2\theta + \cos^2\theta = \frac{8}{9} + \frac{1}{9} = 1$.

**Problem 4.** Show that $(\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2 = 2$ for every $\theta$.

**Solution.** Expand both squares:

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

Adding, the middle terms cancel and we are left with $2\sin^2\theta + 2\cos^2\theta = 2(\sin^2\theta + \cos^2\theta) = 2$.

**Problem 5.** Show that $\sin^4\theta - \cos^4\theta = 1 - 2\cos^2\theta$ for every $\theta$.

**Solution.** The left side is a difference of two squares, $(\sin^2\theta)^2 - (\cos^2\theta)^2$, so it factors:

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

Now replace $\sin^2\theta$ by $1 - \cos^2\theta$: the result is $1 - \cos^2\theta - \cos^2\theta = 1 - 2\cos^2\theta$.

**Problem 6.** Show that $\dfrac{1}{1 - \sin\theta} + \dfrac{1}{1 + \sin\theta} = 2\sec^2\theta$ for every $\theta$ with $\cos\theta \ne 0$.

**Solution.** First note that $\cos\theta \ne 0$ guarantees $\sin\theta \ne \pm 1$ (if $\sin\theta = \pm 1$ the identity would force $\cos^2\theta = 0$), so neither denominator is zero. Use the common denominator $(1 - \sin\theta)(1 + \sin\theta) = 1 - \sin^2\theta = \cos^2\theta$:

$$
\frac{1}{1 - \sin\theta} + \frac{1}{1 + \sin\theta} = \frac{(1 + \sin\theta) + (1 - \sin\theta)}{\cos^2\theta} = \frac{2}{\cos^2\theta} = 2\sec^2\theta.
$$

**Problem 7.** (a) A point moves so that its coordinates at each moment are $x = 3\cos\theta$ and $y = 3\sin\theta$ for some angle $\theta$. Show that it stays on the circle $x^2 + y^2 = 9$. (b) Another point has $x = 5\cos\theta$ and $y = 2\sin\theta$. Find an equation in $x$ and $y$ alone that it always satisfies, and say where the curve crosses the axes.

**Solution.** (a) $x^2 + y^2 = 9\cos^2\theta + 9\sin^2\theta = 9(\cos^2\theta + \sin^2\theta) = 9$. So the point is always at distance 3 from the origin.

(b) Here squaring and adding directly gives $25\cos^2\theta + 4\sin^2\theta$, which does not simplify, so isolate the cosine and sine first: $\cos\theta = \frac{x}{5}$ and $\sin\theta = \frac{y}{2}$. The identity then says

$$
\frac{x^2}{25} + \frac{y^2}{4} = 1.
$$

On the $x$ axis $y = 0$, so $x^2 = 25$ and the curve crosses at $(\pm 5, 0)$. On the $y$ axis $x = 0$, so $y^2 = 4$ and it crosses at $(0, \pm 2)$. The curve is a circle stretched by 5 horizontally and by 2 vertically, an ellipse.

**Problem 8.** What is the largest value that $\sin\theta + \cos\theta$ can take, and at which angle between 0° and 360° does it take it? (Use the result of the fourth problem.)

**Solution.** By the fourth problem, $(\sin\theta + \cos\theta)^2 = 2 - (\sin\theta - \cos\theta)^2$. A square is never negative, so $(\sin\theta + \cos\theta)^2 \le 2$, and therefore $\sin\theta + \cos\theta \le \sqrt 2 \approx 1.414$. Equality requires $(\sin\theta - \cos\theta)^2 = 0$, that is, $\sin\theta = \cos\theta$. Then the identity gives $2\sin^2\theta = 1$, so $\sin\theta = \cos\theta = \pm\frac{\sqrt 2}{2}$. The positive choice is the point at 45°, where the sum is $\frac{\sqrt 2}{2} + \frac{\sqrt 2}{2} = \sqrt 2$. The negative choice, at 225°, gives the smallest value, $-\sqrt 2$. So the maximum is $\sqrt 2$, attained at $\theta = 45°$. This agrees with the lesson's observation that the sum is about $1.40$ at 37°: it exceeds 1 throughout the first quadrant and peaks at 45°.