Equivalent Trig Expressions

The same trig value can be written in many different ways: $\sin\dfrac{\pi}{6}$, $\cos\dfrac{\pi}{3}$ and $\sin\dfrac{5\pi}{6}$ are all equal to $\dfrac{1}{2}$. In this lesson you’ll see why, using right triangles, the unit circle, and graph transformations, and you’ll collect a toolkit of equivalent expressions that you’ll use to simplify, prove identities, and solve equations for the rest of the unit. All angles are in radians (remember $\pi = 180^\circ$, so $\dfrac{\pi}{2} = 90^\circ$).

Key ideas

Equivalent expressions

Two trig expressions are equivalent if they give the same value for every value of $x$ (where both are defined). Their graphs are then exactly the same curve. For example, $\cos\left(x + \dfrac{\pi}{2}\right)$ and $-\sin x$ are equivalent.

Agreeing at one or two values of $x$ is not enough. But finding just one value where they disagree is enough to show they are not equivalent.

Cofunction identities (from a right triangle)

In a right triangle, the two acute angles add to $\dfrac{\pi}{2}$. So if one acute angle is $x$, the other is $\dfrac{\pi}{2} - x$.

The side opposite one acute angle is the side adjacent to the other.

Side $a$ is opposite angle $x$ but adjacent to angle $\dfrac{\pi}{2} - x$. So

$$\sin x = \frac{a}{c} = \cos\left(\frac{\pi}{2} - x\right) \qquad\text{and}\qquad \cos x = \frac{b}{c} = \sin\left(\frac{\pi}{2} - x\right)$$

This is where the name “cosine” comes from: it’s the sine of the complementary angle. The same idea works for the other pairs:

Cofunction identities
$\sin\left(\dfrac{\pi}{2} - x\right) = \cos x$	$\cos\left(\dfrac{\pi}{2} - x\right) = \sin x$
$\tan\left(\dfrac{\pi}{2} - x\right) = \cot x$	$\cot\left(\dfrac{\pi}{2} - x\right) = \tan x$
$\csc\left(\dfrac{\pi}{2} - x\right) = \sec x$	$\sec\left(\dfrac{\pi}{2} - x\right) = \csc x$

The triangle only shows this for acute $x$, but the identities are true for every $x$ (you’ll prove that with the compound angle formulas).

Even and odd identities (from the unit circle)

The point for angle $-x$ on the unit circle is the reflection of the point for angle $x$ in the $x$-axis: $(\cos x, \sin x)$ becomes $(\cos x, -\sin x)$. So

$$\cos(-x) = \cos x \qquad \sin(-x) = -\sin x \qquad \tan(-x) = -\tan x$$

Cosine is an even function (its graph is symmetric in the $y$-axis), and sine and tangent are odd functions (their graphs have rotational symmetry about the origin). See even and odd functions for more.

Related-angle identities (from the unit circle)

Reflecting the point for $x$ in the $y$-axis gives the angle $\pi - x$; rotating it by half a turn gives $\pi + x$; reflecting in the $x$-axis gives $2\pi - x$. The coordinates only change sign, which gives these identities. The signs match the CAST rule if you picture $x$ as a small first-quadrant angle.

Angle	Quadrant (for small $x$)	sine	cosine	tangent
$\pi - x$	2	$\sin x$	$-\cos x$	$-\tan x$
$\pi + x$	3	$-\sin x$	$-\cos x$	$\tan x$
$2\pi - x$	4	$-\sin x$	$\cos x$	$-\tan x$

Adding a full turn changes nothing: $\sin(x + 2\pi) = \sin x$ and $\cos(x + 2\pi) = \cos x$, because the period is $2\pi$.

Identities from transformations

Shifting a sine or cosine graph by a quarter period ($\dfrac{\pi}{2}$) turns it into another sine or cosine graph. For example, $y = \cos\left(x + \dfrac{\pi}{2}\right)$ is the graph of $y = \cos x$ shifted $\dfrac{\pi}{2}$ to the left. The maximum at $(0, 1)$ moves to $\left(-\dfrac{\pi}{2}, 1\right)$, and the new graph starts at $0$ and goes down: that’s the graph of $y = -\sin x$.

Shifting $y = \cos x$ left by $\dfrac{\pi}{2}$ gives the graph of $y = -\sin x$.

The four quarter-period shifts are worth knowing:

$$\cos\left(x + \frac{\pi}{2}\right) = -\sin x \qquad \cos\left(x - \frac{\pi}{2}\right) = \sin x \qquad \sin\left(x + \frac{\pi}{2}\right) = \cos x \qquad \sin\left(x - \frac{\pi}{2}\right) = -\cos x$$

Checking with a graph

To test whether two expressions are equivalent, graph both on graphing technology (such as Desmos, in radian mode) over at least one full period. If the graphs are identical, the expressions are very likely equivalent (a proof uses algebra, as in proving trig identities). If the graphs differ anywhere, read off a value of $x$ where they disagree: that’s your counterexample.

Worked examples

Example 1: Cofunctions

(a) Write $\cos\dfrac{3\pi}{10}$ as the sine of an angle.

(b) Write $\tan\dfrac{\pi}{8}$ as the cotangent of an angle.

Solution.

(a) Use $\cos x = \sin\left(\dfrac{\pi}{2} - x\right)$:

$$\cos\frac{3\pi}{10} = \sin\left(\frac{\pi}{2} - \frac{3\pi}{10}\right) = \sin\left(\frac{5\pi}{10} - \frac{3\pi}{10}\right) = \sin\frac{2\pi}{10} = \sin\frac{\pi}{5}$$

Check on a calculator in radian mode: both are about $0.5878$. ✓

(b) Use $\tan x = \cot\left(\dfrac{\pi}{2} - x\right)$:

$$\tan\frac{\pi}{8} = \cot\left(\frac{4\pi}{8} - \frac{\pi}{8}\right) = \cot\frac{3\pi}{8}$$

The two angles in each answer add to $\dfrac{\pi}{2}$, which is a quick way to check a cofunction.

Example 2: Simplifying with related angles

Simplify each expression.

(a) $\cos(-x) - \cos(\pi - x)$

(b) $\tan(\pi + x)\tan(-x)$

Solution.

(a) $\cos(-x) = \cos x$ (cosine is even) and $\cos(\pi - x) = -\cos x$ (quadrant 2, where cosine is negative):

$$\cos(-x) - \cos(\pi - x) = \cos x - (-\cos x) = 2\cos x$$

(b) $\tan(\pi + x) = \tan x$ (quadrant 3, where tangent is positive) and $\tan(-x) = -\tan x$ (tangent is odd):

$$\tan(\pi + x)\tan(-x) = \tan x \cdot (-\tan x) = -\tan^2 x$$

Example 3: Two ways to see an identity

Show that $\sin\left(x - \dfrac{\pi}{2}\right) = -\cos x$.

Solution.

With a transformation. $y = \sin\left(x - \dfrac{\pi}{2}\right)$ is the graph of $y = \sin x$ shifted $\dfrac{\pi}{2}$ to the right. The minimum of $y = \sin x$ at $\left(-\dfrac{\pi}{2}, -1\right)$ moves to $(0, -1)$, so the new graph starts at its minimum and rises. That’s the graph of $y = -\cos x$.

With identities. Sine is odd, so $\sin(-\theta) = -\sin\theta$. Write $x - \dfrac{\pi}{2} = -\left(\dfrac{\pi}{2} - x\right)$:

$$\sin\left(x - \frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2} - x\right) = -\cos x \qquad \text{(cofunction identity)}$$

Check with $x = \dfrac{\pi}{3}$: $\sin\left(\dfrac{\pi}{3} - \dfrac{\pi}{2}\right) = \sin\left(-\dfrac{\pi}{6}\right) = -\dfrac{1}{2}$, and $-\cos\dfrac{\pi}{3} = -\dfrac{1}{2}$. ✓

Example 4: Not equivalent

Is $\cos(x - \pi)$ equivalent to $\cos x$? If not, find an expression that it is equivalent to.

Solution. Try $x = 0$: $\cos(0 - \pi) = \cos(-\pi) = -1$, but $\cos 0 = 1$. One counterexample is enough: they are not equivalent.

Shifting $y = \cos x$ right by $\pi$ (half a period) moves each maximum to where a minimum was, which flips the graph upside down. So $\cos(x - \pi) = -\cos x$. You can also see this from the table: $\cos(x - \pi) = \cos(\pi - x)$ because cosine is even, and $\cos(\pi - x) = -\cos x$.

Check with $x = \dfrac{\pi}{3}$: $\cos\left(-\dfrac{2\pi}{3}\right) = -\dfrac{1}{2}$ and $-\cos\dfrac{\pi}{3} = -\dfrac{1}{2}$. ✓

Common mistakes

Mixing up even and odd. $\cos(-x) = \cos x$, but $\sin(-x) = -\sin x$. Picture the unit circle: going clockwise instead of counterclockwise flips the $y$-coordinate (sine) but not the $x$-coordinate (cosine).

Getting the sign wrong in a related angle. $\cos(\pi - x)$ is $-\cos x$, not $\cos x$. Picture $x$ as a small angle, decide which quadrant the new angle is in, and use CAST to get the sign.

Shifting the wrong way. $\cos\left(x + \dfrac{\pi}{2}\right)$ is a shift to the left, not the right. Check by substituting: at $x = \dfrac{\pi}{2}$ it gives $\cos\pi = -1$, which matches $-\sin\dfrac{\pi}{2} = -1$. A shift to the right, $\cos\left(x - \dfrac{\pi}{2}\right)$, would give $\cos 0 = 1$ there instead.

Deciding two expressions are equivalent because they agree at one value. $\cos 2x$ and $2\cos x - 1$ are both equal to $1$ at $x = 0$, but at $x = \pi$ they give $1$ and $-3$. Agreement at a few points proves nothing; disagreement at one point disproves.

Mixing degrees and radians. In this course the complementary angle is $\dfrac{\pi}{2} - x$, not $90 - x$. If your calculator check doesn’t match, make sure it’s in radian mode.

Practice

1. (Warm-up) Use a cofunction identity to complete each statement with an angle between $0$ and $\dfrac{\pi}{2}$.

• (a) $\sin\dfrac{\pi}{6} = \cos\ ?$
• (b) $\cos\dfrac{2\pi}{5} = \sin\ ?$

Solution

(a) $\dfrac{\pi}{2} - \dfrac{\pi}{6} = \dfrac{3\pi}{6} - \dfrac{\pi}{6} = \dfrac{\pi}{3}$, so $\sin\dfrac{\pi}{6} = \cos\dfrac{\pi}{3}$. (Both are $\dfrac{1}{2}$.)

(b) $\dfrac{\pi}{2} - \dfrac{2\pi}{5} = \dfrac{5\pi}{10} - \dfrac{4\pi}{10} = \dfrac{\pi}{10}$, so $\cos\dfrac{2\pi}{5} = \sin\dfrac{\pi}{10}$.

2. (Warm-up) Find the exact value of each.

• (a) $\cos\left(-\dfrac{\pi}{3}\right)$
• (b) $\sin\left(-\dfrac{\pi}{4}\right)$
• (c) $\tan\left(-\dfrac{\pi}{6}\right)$

Solution

(a) Cosine is even: $\cos\left(-\dfrac{\pi}{3}\right) = \cos\dfrac{\pi}{3} = \dfrac{1}{2}$.

(b) Sine is odd: $\sin\left(-\dfrac{\pi}{4}\right) = -\sin\dfrac{\pi}{4} = -\dfrac{\sqrt{2}}{2}$.

(c) Tangent is odd: $\tan\left(-\dfrac{\pi}{6}\right) = -\tan\dfrac{\pi}{6} = -\dfrac{1}{\sqrt{3}} = -\dfrac{\sqrt{3}}{3}$.

3. (Warm-up) Suppose $\sin x = 0.6$. Find the value of each.

• (a) $\sin(-x)$
• (b) $\sin(\pi - x)$
• (c) $\cos\left(\dfrac{\pi}{2} - x\right)$

Solution

(a) $\sin(-x) = -\sin x = -0.6$

(b) $\sin(\pi - x) = \sin x = 0.6$

(c) $\cos\left(\dfrac{\pi}{2} - x\right) = \sin x = 0.6$

4. (Core) Simplify each expression.

• (a) $\sin(\pi - x) - \cos\left(\dfrac{\pi}{2} - x\right)$
• (b) $\sin(\pi + x) + 3\sin(-x)$

Solution

(a) $\sin(\pi - x) = \sin x$ and $\cos\left(\dfrac{\pi}{2} - x\right) = \sin x$, so the expression is $\sin x - \sin x = 0$.

(b) $\sin(\pi + x) = -\sin x$ and $\sin(-x) = -\sin x$, so the expression is $-\sin x - 3\sin x = -4\sin x$.

5. (Core) Use the cofunction identities for sine and cosine, and the quotient identity, to show that $\tan\left(\dfrac{\pi}{2} - x\right) = \cot x$.

Solution

$$\tan\left(\frac{\pi}{2} - x\right) = \frac{\sin\left(\frac{\pi}{2} - x\right)}{\cos\left(\frac{\pi}{2} - x\right)} = \frac{\cos x}{\sin x} = \cot x$$

6. (Core) Use a transformation to explain why $\cos\left(x - \dfrac{\pi}{2}\right) = \sin x$. Then check it with $x = \dfrac{\pi}{6}$.

Solution

$y = \cos\left(x - \dfrac{\pi}{2}\right)$ is $y = \cos x$ shifted $\dfrac{\pi}{2}$ to the right. The maximum at $(0, 1)$ moves to $\left(\dfrac{\pi}{2}, 1\right)$, and the zero at $\left(-\dfrac{\pi}{2}, 0\right)$ (where cosine is rising) moves to $(0, 0)$. A graph that rises through the origin to a maximum at $\left(\dfrac{\pi}{2}, 1\right)$ is the graph of $y = \sin x$.

Check: $\cos\left(\dfrac{\pi}{6} - \dfrac{\pi}{2}\right) = \cos\left(-\dfrac{\pi}{3}\right) = \dfrac{1}{2}$, and $\sin\dfrac{\pi}{6} = \dfrac{1}{2}$. ✓

7. (Core) Given that $\cos\dfrac{\pi}{5} \approx 0.8090$, find each value without a calculator.

• (a) $\sin\dfrac{3\pi}{10}$
• (b) $\cos\dfrac{4\pi}{5}$
• (c) $\cos\dfrac{6\pi}{5}$

Solution

(a) $\dfrac{3\pi}{10} = \dfrac{\pi}{2} - \dfrac{\pi}{5}$, so $\sin\dfrac{3\pi}{10} = \cos\dfrac{\pi}{5} \approx 0.8090$.

(b) $\dfrac{4\pi}{5} = \pi - \dfrac{\pi}{5}$, so $\cos\dfrac{4\pi}{5} = -\cos\dfrac{\pi}{5} \approx -0.8090$.

(c) $\dfrac{6\pi}{5} = \pi + \dfrac{\pi}{5}$, so $\cos\dfrac{6\pi}{5} = -\cos\dfrac{\pi}{5} \approx -0.8090$.

8. (Challenge) Lena says that $\sin(x + \pi)$ and $\sin(\pi - x)$ are equivalent, because “adding is the same as subtracting when you go around a circle.” Show that she is wrong, and give the correct simplified form of each expression.

Solution

Try $x = \dfrac{\pi}{2}$: $\sin\left(\dfrac{\pi}{2} + \pi\right) = \sin\dfrac{3\pi}{2} = -1$, but $\sin\left(\pi - \dfrac{\pi}{2}\right) = \sin\dfrac{\pi}{2} = 1$. Since $-1 \ne 1$, they are not equivalent.

From the related-angle table, $\sin(x + \pi) = -\sin x$ and $\sin(\pi - x) = \sin x$. They are negatives of each other.

9. (Challenge) Write $y = \sin x$ as a cosine function in two different ways: once using a shift to the right and once using a shift to the left, with each shift between $0$ and $2\pi$. Explain why both work.

Solution

Shift right: $y = \cos\left(x - \dfrac{\pi}{2}\right)$. Shifting $y = \cos x$ right by $\dfrac{\pi}{2}$ moves its maximum from $0$ to $\dfrac{\pi}{2}$, where $y = \sin x$ has its maximum.

Shift left: $y = \cos\left(x + \dfrac{3\pi}{2}\right)$. Shifting left by $\dfrac{3\pi}{2}$ moves the maximum from $0$ to $-\dfrac{3\pi}{2}$. Since the period is $2\pi$, there is also a maximum at $-\dfrac{3\pi}{2} + 2\pi = \dfrac{\pi}{2}$, the same place as before.

The two shifts differ by $\dfrac{\pi}{2} + \dfrac{3\pi}{2} = 2\pi$, one full period, so they give the same graph.

Check with $x = 0$: $\cos\dfrac{3\pi}{2} = 0 = \sin 0$. ✓