Benchmark Angles
By the end of this lesson, you’ll be able to:
- Recall exact sine, cosine, and tangent values for key benchmark angles.
- Use special right triangles to derive benchmark trig values.
- Connect benchmark angles in degrees and radians.
- Evaluate trig expressions exactly without a calculator.
- Recognize when tangent is undefined.
Key Ideas
Certain angles appear repeatedly in trigonometry.
The most important first-quadrant benchmark angles are:
\[ 0^\circ,\quad 30^\circ,\quad 45^\circ,\quad 60^\circ,\quad 90^\circ \]
In radians:
\[ 0,\quad \frac{\pi}{6},\quad \frac{\pi}{4},\quad \frac{\pi}{3},\quad \frac{\pi}{2} \]
Their exact trig values are:
| Angle | Radians | \(\sin\theta\) | \(\cos\theta\) | \(\tan\theta\) |
|---|---|---|---|---|
| \(0^\circ\) | \(0\) | \(0\) | \(1\) | \(0\) |
| \(30^\circ\) | \(\frac{\pi}{6}\) | \(\frac12\) | \(\frac{\sqrt3}{2}\) | \(\frac{\sqrt3}{3}\) |
| \(45^\circ\) | \(\frac{\pi}{4}\) | \(\frac{\sqrt2}{2}\) | \(\frac{\sqrt2}{2}\) | \(1\) |
| \(60^\circ\) | \(\frac{\pi}{3}\) | \(\frac{\sqrt3}{2}\) | \(\frac12\) | \(\sqrt3\) |
| \(90^\circ\) | \(\frac{\pi}{2}\) | \(1\) | \(0\) | undefined |
The values for \(30^\circ\), \(45^\circ\), and \(60^\circ\) come directly from the two special right triangles.
The \(45^\circ\)–\(45^\circ\)–\(90^\circ\) Triangle
The side ratio is:
\[ 1:1:\sqrt2 \]
Using either \(45^\circ\) angle:
\[ \sin45^\circ = \frac{1}{\sqrt2} = \frac{\sqrt2}{2} \]
Similarly:
\[ \cos45^\circ = \frac{1}{\sqrt2} = \frac{\sqrt2}{2} \]
and:
\[ \tan45^\circ = \frac11 = 1 \]
Therefore:
\[ \boxed{ \sin45^\circ=\cos45^\circ=\frac{\sqrt2}{2} } \]
and:
\[ \boxed{\tan45^\circ=1} \]
The \(30^\circ\)–\(60^\circ\)–\(90^\circ\) Triangle
The side ratio is:
\[ 1:\sqrt3:2 \]
Remember which side is which:
- Short leg \(=1\) → opposite \(30^\circ\)
- Long leg \(=\sqrt3\) → opposite \(60^\circ\)
- Hypotenuse \(=2\)
For the \(30^\circ\) angle:
\[ \sin30^\circ = \frac12 \]
\[ \cos30^\circ = \frac{\sqrt3}{2} \]
\[ \tan30^\circ = \frac{1}{\sqrt3} = \frac{\sqrt3}{3} \]
For the \(60^\circ\) angle:
\[ \sin60^\circ = \frac{\sqrt3}{2} \]
\[ \cos60^\circ = \frac12 \]
\[ \tan60^\circ = \sqrt3 \]
Because \(30^\circ\) and \(60^\circ\) are complementary:
\[ \sin30^\circ=\cos60^\circ=\frac12 \]
and:
\[ \cos30^\circ=\sin60^\circ=\frac{\sqrt3}{2} \]
This is the complementary-angle identity from the previous lesson.
What About \(0^\circ\) and \(90^\circ\)?
These values are easiest to see from the unit circle.
At:
\[ 0^\circ \]
the unit-circle point is:
\[ (1,0) \]
so:
\[ \cos0^\circ=1 \]
and:
\[ \sin0^\circ=0 \]
At:
\[ 90^\circ \]
the point is:
\[ (0,1) \]
so:
\[ \cos90^\circ=0 \]
and:
\[ \sin90^\circ=1 \]
Since:
\[ \tan\theta=\frac{\sin\theta}{\cos\theta} \]
we have:
\[ \tan90^\circ = \frac{1}{0} \]
Division by zero is undefined.
Therefore:
\[ \boxed{\tan90^\circ\text{ is undefined}} \]
Common Problem Types
1. Evaluating Sine Exactly
For example:
\[ \sin30^\circ \]
From the \(30^\circ\)–\(60^\circ\)–\(90^\circ\) triangle:
\[ \boxed{\sin30^\circ=\frac12} \]
2. Evaluating Cosine Exactly
For example:
\[ \cos60^\circ \]
Using the special triangle:
\[ \boxed{\cos60^\circ=\frac12} \]
3. Evaluating Tangent Exactly
For example:
\[ \tan45^\circ \]
Since the opposite and adjacent legs are equal:
\[ \tan45^\circ = \frac11 \]
Therefore:
\[ \boxed{\tan45^\circ=1} \]
4. Recognizing Angles Written in Radians
Benchmark angles may be given in radians instead of degrees.
For example:
\[ \frac{\pi}{4}=45^\circ \]
Therefore:
\[ \cos\frac{\pi}{4} = \cos45^\circ \]
so:
\[ \boxed{ \cos\frac{\pi}{4}=\frac{\sqrt2}{2} } \]
Similarly:
\[ \frac{\pi}{6}=30^\circ \]
and:
\[ \frac{\pi}{3}=60^\circ \]
5. Recognizing Undefined Tangent
Remember:
\[ \tan\theta = \frac{\sin\theta}{\cos\theta} \]
At \(90^\circ\):
\[ \cos90^\circ=0 \]
so tangent would require division by zero.
Therefore:
\[ \boxed{\tan90^\circ\text{ is undefined}} \]
6. Using Complementary Angles
Since:
\[ 30^\circ+60^\circ=90^\circ \]
sine and cosine switch:
\[ \sin30^\circ=\cos60^\circ \]
and:
\[ \cos30^\circ=\sin60^\circ \]
This can help you remember the benchmark values instead of memorizing each one separately.
Strategies
- Memorize the side ratios of the two special right triangles:
\[ 1:1:\sqrt2 \]
and:
\[ 1:\sqrt3:2 \]
- If you forget a trig value, redraw the appropriate special triangle.
- Remember:
\[ 30^\circ=\frac{\pi}{6} \]
\[ 45^\circ=\frac{\pi}{4} \]
\[ 60^\circ=\frac{\pi}{3} \]
\[ 90^\circ=\frac{\pi}{2} \]
- Use SOH–CAH–TOA to reconstruct values instead of relying only on memorization.
- Use complementary angles to connect sine and cosine.
- Remember:
\[ \tan\theta=\frac{\sin\theta}{\cos\theta} \]
- Give exact values with fractions and radicals unless a decimal is specifically requested.
Worked Examples
Example 1 — Evaluate Cosine
Find:
\[ \cos60^\circ \]
From the \(30^\circ\)–\(60^\circ\)–\(90^\circ\) triangle, relative to \(60^\circ\):
- adjacent side \(=1\)
- hypotenuse \(=2\)
Using CAH:
\[ \cos60^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} \]
\[ = \frac12 \]
Therefore:
\[ \boxed{\cos60^\circ=\frac12} \]
Example 2 — Evaluate Tangent
Find:
\[ \tan45^\circ \]
A \(45^\circ\)–\(45^\circ\)–\(90^\circ\) triangle has equal legs.
Therefore:
\[ \tan45^\circ = \frac{\text{opposite}}{\text{adjacent}} \]
\[ = \frac11 \]
\[ \boxed{\tan45^\circ=1} \]
Example 3 — Evaluate an Angle in Radians
Find:
\[ \sin\frac{\pi}{3} \]
Recognize:
\[ \frac{\pi}{3}=60^\circ \]
Therefore:
\[ \sin\frac{\pi}{3} = \sin60^\circ \]
Using the \(30^\circ\)–\(60^\circ\)–\(90^\circ\) triangle:
\[ \boxed{ \sin\frac{\pi}{3}=\frac{\sqrt3}{2} } \]
Example 4 — Evaluate an Expression
Evaluate exactly:
\[ 2\cos60^\circ+\sin30^\circ \]
Use the benchmark values:
\[ \cos60^\circ=\frac12 \]
and:
\[ \sin30^\circ=\frac12 \]
Substitute:
\[ 2\left(\frac12\right)+\frac12 \]
\[ =1+\frac12 \]
\[ =\frac32 \]
Therefore:
\[ \boxed{\frac32} \]
Example 5 — Recognize an Undefined Value
Evaluate:
\[ \tan\frac{\pi}{2} \]
Since:
\[ \frac{\pi}{2}=90^\circ \]
and:
\[ \cos90^\circ=0 \]
we have:
\[ \tan90^\circ = \frac{\sin90^\circ}{\cos90^\circ} = \frac10 \]
Division by zero is undefined.
Therefore:
\[ \boxed{\tan\frac{\pi}{2}\text{ is undefined}} \]
- Mixing up the \(30^\circ\) and \(60^\circ\) sine and cosine values.
- Forgetting which side is \(1\) and which is \(\sqrt3\) in a \(30^\circ\)–\(60^\circ\)–\(90^\circ\) triangle.
- Forgetting radicals in exact answers.
- Writing a decimal when an exact value is requested.
- Forgetting that \(\tan90^\circ\) is undefined.
- Confusing degree and radian versions of the same benchmark angle.
- Using the wrong side as the hypotenuse when reconstructing a value.
Practice Problems
- Compute exactly:
\[ \sin45^\circ \]
- Compute exactly:
\[ \tan30^\circ \]
- Evaluate:
\[ \cos\frac{\pi}{4} \]
- Evaluate:
\[ \sin\frac{\pi}{6} \]
- Evaluate exactly:
\[ 2\sin60^\circ \]
- Evaluate:
\[ \tan\frac{\pi}{2} \]
1. A \(45^\circ\)–\(45^\circ\)–\(90^\circ\) triangle has side ratio:
\[ 1:1:\sqrt2 \]
Using SOH:
\[ \sin45^\circ = \frac{1}{\sqrt2} \]
Rationalize:
\[ \frac{1}{\sqrt2} = \frac{\sqrt2}{2} \]
Therefore:
\[ \boxed{\frac{\sqrt2}{2}} \]
2. In a \(30^\circ\)–\(60^\circ\)–\(90^\circ\) triangle, relative to \(30^\circ\):
\[ \text{opposite}=1 \]
and:
\[ \text{adjacent}=\sqrt3 \]
Therefore:
\[ \tan30^\circ = \frac{1}{\sqrt3} \]
Rationalize:
\[ \frac{1}{\sqrt3} = \frac{\sqrt3}{3} \]
Therefore:
\[ \boxed{\frac{\sqrt3}{3}} \]
3. Recognize:
\[ \frac{\pi}{4}=45^\circ \]
Therefore:
\[ \cos\frac{\pi}{4} = \cos45^\circ \]
Using the \(45^\circ\)–\(45^\circ\)–\(90^\circ\) triangle:
\[ \boxed{\frac{\sqrt2}{2}} \]
4. Recognize:
\[ \frac{\pi}{6}=30^\circ \]
Therefore:
\[ \sin\frac{\pi}{6} = \sin30^\circ \]
So:
\[ \boxed{\frac12} \]
5. Use:
\[ \sin60^\circ=\frac{\sqrt3}{2} \]
Then:
\[ 2\sin60^\circ = 2\left(\frac{\sqrt3}{2}\right) \]
\[ =\sqrt3 \]
Therefore:
\[ \boxed{\sqrt3} \]
6. Recognize:
\[ \frac{\pi}{2}=90^\circ \]
Since:
\[ \tan\theta = \frac{\sin\theta}{\cos\theta} \]
and:
\[ \cos90^\circ=0 \]
we would have division by zero.
Therefore:
\[ \boxed{\text{undefined}} \]
Summary
The most important first-quadrant benchmark angles are:
\[ 0^\circ,\quad30^\circ,\quad45^\circ,\quad60^\circ,\quad90^\circ \]
with radian equivalents:
\[ 0,\quad \frac{\pi}{6},\quad \frac{\pi}{4},\quad \frac{\pi}{3},\quad \frac{\pi}{2} \]
The \(30^\circ\), \(45^\circ\), and \(60^\circ\) exact values come from:
\[ 1:\sqrt3:2 \]
and:
\[ 1:1:\sqrt2 \]
Use these triangles and SOH–CAH–TOA to reconstruct exact trig values whenever needed.
- \(30^\circ\) → think \(1:\sqrt3:2\).
- \(45^\circ\) → think \(1:1:\sqrt2\).
- \(30^\circ=\frac{\pi}{6}\).
- \(45^\circ=\frac{\pi}{4}\).
- \(60^\circ=\frac{\pi}{3}\).
- Sine and cosine swap between \(30^\circ\) and \(60^\circ\).
- \(\tan45^\circ=1\).
- \(\tan90^\circ\) is undefined.
- Exact answer requested → keep fractions and radicals.