Cos 5pi/4 - Exact Value -√2/2 and How to Find It

#Trigonometry
TL;DR
The value of cos 5pi/4 is exactly −22-\dfrac{\sqrt{2}}{2}, about −0.7071-0.7071. The angle 5π4\dfrac{5\pi}{4} lands in Quadrant III with a reference angle of π4\dfrac{\pi}{4}, where cosine is negative, so this article shows the reference-angle method, the unit circle proof, a standard-angle table, worked examples, and the common mistakes.
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Bhanzu TeamLast updated on August 11, 20265 min read

What Does Cos 5pi/4 Mean?

On the unit circle, the cosine of an angle is the xx-coordinate of the point where the angle's radius meets the circle. A quadrant is one of the four regions the axes cut the plane into, numbered anticlockwise from the top right.

The angle 5π4\dfrac{5\pi}{4} is measured anticlockwise from the positive xx-axis and stops in Quadrant III, the bottom-left region. Its terminal point is (−22,−22)\left(-\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2}\right), so the xx-coordinate, and therefore the cosine, is −22-\dfrac{\sqrt{2}}{2}. Points in Quadrant III have both coordinates negative, which is why cosine and sine are both negative here.

Where Does Cos 5pi/4 Show Up?

The angle 5π4\dfrac{5\pi}{4} points to 225∘225^\circ, which is 45∘45^\circ below the negative xx-axis, straight into the bottom-left. A vector pointing southwest, a force acting down and to the left, or a compass bearing in that direction all have a horizontal component read from cos⁡5π4=−22\cos\dfrac{5\pi}{4} = -\dfrac{\sqrt{2}}{2}. The value is negative because the point has swung past the vertical into the left half of the circle.

Standard-Angle Reference Table

The value 22\dfrac{\sqrt{2}}{2} and its negative appear at four angles that share the reference angle π4\dfrac{\pi}{4}. The quadrant decides the sign.

Angle (radians)

Angle (degrees)

Quadrant

cos⁡θ\cos\theta

π4\dfrac{\pi}{4}

45∘45^\circ

I

22\dfrac{\sqrt{2}}{2}

3π4\dfrac{3\pi}{4}

135∘135^\circ

II

−22-\dfrac{\sqrt{2}}{2}

5π4\dfrac{5\pi}{4}

225∘225^\circ

III

−22-\dfrac{\sqrt{2}}{2}

7π4\dfrac{7\pi}{4}

315∘315^\circ

IV

22\dfrac{\sqrt{2}}{2}

All four have the same reference angle, so their cosines share the magnitude 22\dfrac{\sqrt{2}}{2}. Since 5π4\dfrac{5\pi}{4} sits in Quadrant III, its cosine is negative, matching cos 135 degrees in value while sitting one quadrant further round.

How Do You Find The Exact Value Of Cos 5pi/4?

Three routes all give −22-\dfrac{\sqrt{2}}{2}.

Method 1: Reference angle and quadrant sign.

A reference angle is the acute angle between the terminal side and the xx-axis. For a Quadrant III angle, subtract π\pi:

5π4−π=5π−4π4=π4\frac{5\pi}{4} - \pi = \frac{5\pi - 4\pi}{4} = \frac{\pi}{4}

Cosine is negative in Quadrant III, so:

cos⁡5π4=−cos⁡π4=−22\cos\frac{5\pi}{4} = -\cos\frac{\pi}{4} = -\frac{\sqrt{2}}{2}

Method 2: The unit circle.

Convert with the radians-to-degrees rule: 5π4=225∘\dfrac{5\pi}{4} = 225^\circ. The terminal point at 225∘225^\circ is (−22,−22)\left(-\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2}\right), so

cos⁡5π4=x-coordinate=−22\cos\frac{5\pi}{4} = x\text{-coordinate} = -\frac{\sqrt{2}}{2}

Method 3: From the reference-angle twin.

The magnitude comes straight from cos π/4, which is 22\dfrac{\sqrt{2}}{2} or equivalently 12\dfrac{1}{\sqrt{2}}. The rationalised form 22\dfrac{\sqrt{2}}{2} is the standard way to write it, and Quadrant III attaches the minus sign:

cos⁡5π4=−12=−22\cos\frac{5\pi}{4} = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}

Examples Of Cos 5pi/4

Example 1

Evaluate 42,cos⁡5π44\sqrt{2},\cos\dfrac{5\pi}{4}.

42×(−22)=−4×22=−44\sqrt{2} \times \left(-\frac{\sqrt{2}}{2}\right) = -\frac{4 \times 2}{2} = -4

Example 2

Find cos⁡5π4\cos\dfrac{5\pi}{4} using the reference angle.

Wrong attempt. A student finds the reference angle π4\dfrac{\pi}{4}, reads cos⁡π4=22\cos\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}, and writes cos⁡5π4=22\cos\dfrac{5\pi}{4} = \dfrac{\sqrt{2}}{2}.

That skips the quadrant step. The reference angle only fixes the size of the value, not its sign, and 5π4\dfrac{5\pi}{4} is in Quadrant III where the xx-coordinate is negative.

Correct. Apply the Quadrant III sign to the magnitude: cos⁡5π4=−22\cos\dfrac{5\pi}{4} = -\dfrac{\sqrt{2}}{2}.

Example 3

Compare cos⁡5π4\cos\dfrac{5\pi}{4} with cos⁡7π4\cos\dfrac{7\pi}{4}.

Both share the reference angle π4\dfrac{\pi}{4}, so both have magnitude 22\dfrac{\sqrt{2}}{2}. But 7π4\dfrac{7\pi}{4} is in Quadrant IV where cosine is positive:

cos⁡5π4=−22,cos⁡7π4=22\cos\frac{5\pi}{4} = -\frac{\sqrt{2}}{2}, \qquad \cos\frac{7\pi}{4} = \frac{\sqrt{2}}{2}

Example 4

Evaluate cos⁡5π4+sin⁡5π4\cos\dfrac{5\pi}{4} + \sin\dfrac{5\pi}{4}.

Both coordinates are −22-\dfrac{\sqrt{2}}{2} in Quadrant III:

−22+(−22)=−2≈−1.414-\frac{\sqrt{2}}{2} + \left(-\frac{\sqrt{2}}{2}\right) = -\sqrt{2} \approx -1.414

Example 5

Find sec⁡5π4\sec\dfrac{5\pi}{4}.

Secant is the reciprocal of cosine:

sec⁡5π4=1−22=−22=−2\sec\frac{5\pi}{4} = \frac{1}{-\frac{\sqrt{2}}{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}

Where Students Trip Up On Cos 5pi/4

Mistake 1: Forgetting the Quadrant III minus sign

Where it slips in: Finding the reference angle, reading the positive value, and stopping there.

Don't do this: Writing cos⁡5π4=22\cos\dfrac{5\pi}{4} = \dfrac{\sqrt{2}}{2}. The reference-angle value is always positive, and it is easy to hand in that number without applying the quadrant sign.

The correct way: In Quadrant III both coordinates are negative, so cosine is negative: cos⁡5π4=−22\cos\dfrac{5\pi}{4} = -\dfrac{\sqrt{2}}{2}. Always pair the reference angle with a quadrant-sign check.

Mistake 2: Using the wrong reference angle

Where it slips in: Subtracting from 2π2\pi or from π2\dfrac{\pi}{2} instead of from π\pi.

Don't do this: Computing 2π−5π4=3π42\pi - \dfrac{5\pi}{4} = \dfrac{3\pi}{4} and treating that as the reference angle.

The correct way: For a Quadrant III angle, the reference angle is the angle minus π\pi: 5π4−π=π4\dfrac{5\pi}{4} - \pi = \dfrac{\pi}{4}.

Mistake 3: Leaving the value unrationalised

Where it slips in: Stopping at −12-\dfrac{1}{\sqrt{2}} when a rationalised answer is expected.

Don't do this: Treating −12-\dfrac{1}{\sqrt{2}} and −22-\dfrac{\sqrt{2}}{2} as if only one is acceptable when the form matters.

The correct way: They are equal, but the rationalised −22-\dfrac{\sqrt{2}}{2} is the standard form. Bearings and direction vectors carry this same idea: 225∘225^\circ points southwest, and reporting the horizontal component as a clean −22-\dfrac{\sqrt{2}}{2} keeps later calculations consistent.

Key Takeaways

  • Cos 5pi/4 equals −22-\dfrac{\sqrt{2}}{2}, about −0.7071-0.7071 - negative because 5π4\dfrac{5\pi}{4} lies in Quadrant III.

  • The reference angle is π4\dfrac{\pi}{4}, so the magnitude matches cos⁡45∘=22\cos 45^\circ = \dfrac{\sqrt{2}}{2}; the quadrant supplies the minus sign.

  • In degrees, cos⁡5π4=cos⁡225∘=−22\cos\dfrac{5\pi}{4} = \cos 225^\circ = -\dfrac{\sqrt{2}}{2}, and the terminal point is (−22,−22)\left(-\dfrac{\sqrt{2}}{2}, -\dfrac{\sqrt{2}}{2}\right).

  • The most common slip is reporting the positive reference-angle value and forgetting the Quadrant III sign.

To master reference angles with a teacher, explore Bhanzu's trigonometry tutor sessions, a high school math tutor, or a math tutor for focused practice.

Practice These Before Moving On

  1. Evaluate 2cos⁡5π4+sec⁡5π42\cos\dfrac{5\pi}{4} + \sec\dfrac{5\pi}{4}.

  2. Without a calculator, decide the sign of cos⁡3π4\cos\dfrac{3\pi}{4} and give its value.

  3. Rewrite −12-\dfrac{1}{\sqrt{2}} in rationalised form and confirm it equals cos⁡5π4\cos\dfrac{5\pi}{4}.

Want a live Bhanzu trainer to walk through quadrant signs and rationalised forms? Book a free demo class.

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Frequently Asked Questions

Is cos 5pi/4 positive or negative?
Negative. The angle is in Quadrant III, where cosine (the xx-coordinate) is negative, so cos⁡5π4=−22\cos\dfrac{5\pi}{4} = -\dfrac{\sqrt{2}}{2}.
What is cos 5pi/4 in degrees?
5π4\dfrac{5\pi}{4} radians is 225∘225^\circ, and cos⁡225∘=−22\cos 225^\circ = -\dfrac{\sqrt{2}}{2}.
What is the reference angle for 5pi/4?
π4\dfrac{\pi}{4}, found by subtracting π\pi from 5π4\dfrac{5\pi}{4}. It is the same reference angle as 45∘45^\circ.
Is -1/√2 the same as -√2/2?
Yes. Multiplying −12-\dfrac{1}{\sqrt{2}} by 22\dfrac{\sqrt{2}}{\sqrt{2}} gives −22-\dfrac{\sqrt{2}}{2}, the rationalised standard form.
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