Sin 7pi/12: Exact Value (√6 + √2)/4 And Proof

What Does $\sin \frac{7\pi}{12}$ Mean?

$\sin \frac{7\pi}{12}$ is the sine of an angle of $\frac{7\pi}{12}$ radians, measured counterclockwise from the positive $x$-axis. On the unit circle, a circle of radius $1$ centred at the origin, sine is the $y$-coordinate of the point where the angle's terminal side meets the circle.

The terms that carry this article, defined before they are used:

• Quadrant: one of the four regions cut by the axes, numbered Q1 to Q4 counterclockwise from the top right. Angles between $\frac{\pi}{2}$ and $\pi$ end in quadrant Q2, where sine is positive and cosine is negative.

• Reference angle: the acute angle between the terminal side and the $x$-axis. For $\frac{7\pi}{12}$ it is $\pi - \frac{7\pi}{12} = \frac{5\pi}{12}$, as the reference angle rule gives.

• Angle-sum identity: $\sin(A + B) = \sin A \cos B + \cos A \sin B$, one of the sum and difference identities that build new exact values from known ones.

No right triangle has a $105^\circ$ angle, so the unit circle is the home definition here, and the triangle only supplies the reference angle. Proving and using the addition formulas is CCSS-M standard F-TF.C.9, and NCERT Class 11 Mathematics, Chapter 3 (Trigonometric Functions), works the same idea in its exercise on $\sin 75^\circ$.

Standard-Angle Sine Reference Table

Sine has exact values at every multiple of $15^\circ$ ($\frac{\pi}{12}$), and $\frac{7\pi}{12}$ is one of the less familiar ones. The table runs from $0^\circ$ to $180^\circ$ in $15^\circ$ steps.

The column is symmetric around $90^\circ$. Each angle on the way down from $90^\circ$ to $180^\circ$ repeats the sine of its mirror on the way up, which is why $\frac{7\pi}{12}$ and $\frac{5\pi}{12}$ share a row value.

Where Does Sin 7pi/12 Show Up, And Why Does It Matter?

Sin 7pi/12 shows up wherever a $105^\circ$ angle appears, and $105^\circ$ is the third angle of the $30^\circ$-$45^\circ$-$105^\circ$ triangle that turns up in surveying and textbook problems. It is also the standard first test of the angle-sum formula, because $\frac{7\pi}{12}$ is the simplest obtuse angle that is not on the usual list.

• Triangle sides: by the law of sines, a triangle with angles $30^\circ$, $45^\circ$ and $105^\circ$ has its longest side in proportion to $\sin 105^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$.

• Forces at an angle: a force pulling at $105^\circ$ from the horizontal has vertical component $F \sin \frac{7\pi}{12} \approx 0.9659F$, almost all of the force.

• Area with an obtuse angle: two sides $a$ and $b$ with a $105^\circ$ angle between them enclose an area of $\frac{1}{2}ab \sin \frac{7\pi}{12}$.

What Is 7pi/12 In Degrees?

$\frac{7\pi}{12}$ radians is $105^\circ$. Multiply by $\frac{180^\circ}{\pi}$, the rule from converting degrees to radians run in reverse:

$$\frac{7\pi}{12} \times \frac{180^\circ}{\pi} = \frac{7 \times 180^\circ}{12} = 105^\circ$$

So sin 7pi/12 and $\sin 105^\circ$ are the same number. As a decimal, $\frac{7\pi}{12} \approx 1.8326$ radians.

What Is Sin 7pi/12 In Fraction Form?

In fraction form, sin 7pi/12 is $\frac{\sqrt{6} + \sqrt{2}}{4}$. A second correct form, $\frac{\sqrt{2 + \sqrt{3}}}{2}$, comes out of the half-angle route, and the two are equal.

Square the first form to see it:

$$\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)^2 = \frac{6 + 2\sqrt{12} + 2}{16}$$

$$= \frac{8 + 4\sqrt{3}}{16} = \frac{2 + \sqrt{3}}{4}$$

Both forms are positive and have the same square, $\frac{2 + \sqrt{3}}{4}$, so they are the same number. Textbooks usually prefer $\frac{\sqrt{6} + \sqrt{2}}{4}$ because it has no root inside a root (see simplifying radical expressions).

How Do You Find The Exact Value Of Sin 7pi/12?

Split $\frac{7\pi}{12}$ into two angles whose sines and cosines you already know, then apply the angle-sum identity. Every method below lands on $\frac{\sqrt{6} + \sqrt{2}}{4}$; the first is the one most teachers expect.

Method 1: The Sum Identity, $\sin\left(\frac{\pi}{4} + \frac{\pi}{3}\right)$

Since $\frac{\pi}{4} + \frac{\pi}{3} = \frac{3\pi}{12} + \frac{4\pi}{12} = \frac{7\pi}{12}$, apply the sin(A + B) formula with $A = \frac{\pi}{4}$ and $B = \frac{\pi}{3}$:

$$\sin \frac{7\pi}{12} = \sin \frac{\pi}{4} \cos \frac{\pi}{3} + \cos \frac{\pi}{4} \sin \frac{\pi}{3}$$

$$= \frac{\sqrt{2}}{2} \times \frac{1}{2} + \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2}$$

$$= \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$$

Method 2: The Difference Identity, $\sin\left(\frac{3\pi}{4} - \frac{\pi}{6}\right)$

The same angle is $\frac{9\pi}{12} - \frac{2\pi}{12}$, so the sin(A − B) formula works too. Watch the sign of $\cos \frac{3\pi}{4}$, which is negative in Q2.

$$\sin \frac{7\pi}{12} = \sin \frac{3\pi}{4} \cos \frac{\pi}{6} - \cos \frac{3\pi}{4} \sin \frac{\pi}{6}$$

$$= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} - \left(-\frac{\sqrt{2}}{2}\right) \times \frac{1}{2}$$

$$= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$$

Method 3: The Reference Angle, $\sin\left(\pi - \frac{5\pi}{12}\right)$

In Q2, $\sin(\pi - \theta) = \sin\theta$, so $\sin \frac{7\pi}{12} = \sin \frac{5\pi}{12}$. That is the value of sin 75 degrees, the degree-first sibling of this page.

$$\sin \frac{5\pi}{12} = \sin\left(\frac{\pi}{4} + \frac{\pi}{6}\right) = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$$

Method 4: The Half-Angle Formula

$\frac{7\pi}{12}$ is half of $\frac{7\pi}{6}$, and $\cos \frac{7\pi}{6} = -\frac{\sqrt{3}}{2}$. The half angle formula gives

$$\sin \frac{7\pi}{12} = \sqrt{\frac{1 - \cos \frac{7\pi}{6}}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}}$$

$$= \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2}$$

The positive root is chosen because $\frac{7\pi}{12}$ is in Q2, where sine is positive. My first instinct here was to take the $\pm$ at face value and keep both signs; the quadrant settles it before any arithmetic is needed.

Method 5: The Co-Function Identity

The cofunction identities say $\sin\theta = \cos\left(\theta - \frac{\pi}{2}\right)$. Here $\frac{7\pi}{12} - \frac{\pi}{2} = \frac{\pi}{12}$, so

$$\sin \frac{7\pi}{12} = \cos \frac{\pi}{12} = \cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$$

Method 6: Calculator Check

Set radian mode and enter $\sin(7\pi \div 12)$ with the brackets. The screen shows $0.96592582\ldots$, and $\frac{\sqrt{6} + \sqrt{2}}{4} = \frac{2.44949 + 1.41421}{4} = 0.96593$, which matches.

Why Is Sin 7pi/12 Equal To Sin 5pi/12?

Sin 7pi/12 equals $\sin \frac{5\pi}{12}$ because the two angles are mirror images across the $y$-axis. Reflecting a point across the $y$-axis changes its $x$-coordinate's sign and leaves its height alone, and sine is the height.

$$\frac{5\pi}{12} + \frac{7\pi}{12} = \pi$$

Angles that add to $\pi$ ($180^\circ$) are supplementary, and supplementary angles always share a sine. Their cosines are opposites: $\cos \frac{5\pi}{12} = \frac{\sqrt{6} - \sqrt{2}}{4}$ while $\cos \frac{7\pi}{12} = \frac{\sqrt{2} - \sqrt{6}}{4}$.

What Are All Six Trig Values At 7pi/12?

The unit-circle point at $\frac{7\pi}{12}$ is $\left(\frac{\sqrt{2} - \sqrt{6}}{4}, \frac{\sqrt{6} + \sqrt{2}}{4}\right)$. The six values follow from it.

Only sine and cosecant are positive, as expected in Q2. Sine can also be recovered from the others, for example $\sin\theta = \frac{1}{\csc\theta} = \frac{1}{\sqrt{6} - \sqrt{2}}$, which rationalises back to $\frac{\sqrt{6} + \sqrt{2}}{4}$.

Why Does A Calculator Show 0 For Sin 7pi/12?

A search-bar calculator fed "sin 7pi/12" with no brackets answers $0$. It reads the input as $\frac{\sin 7\pi}{12}$, and $\sin 7\pi = 0$, so the result is $0$.

The same parsing gives "cos 7pi/12" as $-0.0833$, which is $\frac{\cos 7\pi}{12} = -\frac{1}{12}$. Typing "sin(7pi/12)" with brackets fixes both.

Two other settings produce wrong decimals:

• Degree mode with a radian angle: $\sin(1.8326^\circ) \approx 0.0320$.

• Radian mode with a degree angle: typing "sin 105" returns $\sin(105 \text{ rad}) \approx -0.9705$, a value that is negative even though $105^\circ$ is in Q2.

Examples Of Sin 7pi/12

Example 1

Evaluate $4\sin \frac{7\pi}{12}$.

$$4\sin \frac{7\pi}{12} = 4 \times \frac{\sqrt{6} + \sqrt{2}}{4}$$

$$= \sqrt{6} + \sqrt{2} \approx 3.8637$$

Example 2

Find $\sin \frac{7\pi}{12}$ by splitting the angle as $\frac{\pi}{4} + \frac{\pi}{3}$.

A tempting first attempt adds the two sines:

$$\sin \frac{\pi}{4} + \sin \frac{\pi}{3} = \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \approx 1.5731$$

That cannot be right, because sine never exceeds $1$. Sine of a sum is not the sum of the sines; the angle-sum identity mixes sines and cosines:

$$\sin\left(\frac{\pi}{4} + \frac{\pi}{3}\right) = \sin \frac{\pi}{4} \cos \frac{\pi}{3} + \cos \frac{\pi}{4} \sin \frac{\pi}{3} = \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}$$

Final answer: $\frac{\sqrt{6} + \sqrt{2}}{4} \approx 0.9659$.

Example 3

Evaluate $\sin \frac{7\pi}{12} \cos \frac{\pi}{4} - \cos \frac{7\pi}{12} \sin \frac{\pi}{4}$.

The expression is the right-hand side of $\sin(A - B)$ with $A = \frac{7\pi}{12}$ and $B = \frac{\pi}{4}$.

$$\sin\left(\frac{7\pi}{12} - \frac{3\pi}{12}\right) = \sin \frac{4\pi}{12} = \sin \frac{\pi}{3}$$

$$= \frac{\sqrt{3}}{2}$$

Example 4

Find $\sin^2 \frac{7\pi}{12}$ in exact form.

$$\sin^2 \frac{7\pi}{12} = \frac{1 - \cos \frac{7\pi}{6}}{2} = \frac{1 + \frac{\sqrt{3}}{2}}{2}$$

$$= \frac{2 + \sqrt{3}}{4} \approx 0.9330$$

The squaring step in the fraction-form section gave the same $\frac{2 + \sqrt{3}}{4}$, which is a useful cross-check.

Example 5

Evaluate $\sin \frac{7\pi}{12} \cos \frac{\pi}{12}$.

Use the product-to-sum rule from the sin A cos B formula, $\sin A \cos B = \frac{1}{2}[\sin(A + B) + \sin(A - B)]$.

$$\sin \frac{7\pi}{12} \cos \frac{\pi}{12} = \frac{1}{2}\left[\sin \frac{8\pi}{12} + \sin \frac{6\pi}{12}\right]$$

$$= \frac{1}{2}\left[\frac{\sqrt{3}}{2} + 1\right] = \frac{2 + \sqrt{3}}{4}$$

It matches Example 4 for a reason: $\cos \frac{\pi}{12}$ is the same number as $\sin \frac{7\pi}{12}$ (Method 5), so this product is really $\sin^2 \frac{7\pi}{12}$. Students who memorise the product-to-sum rule but never test it against a known value tend to miss checks like this one.

Example 6

Two sides of a garden bed measure $6$ m and $8$ m, and the angle between them is $\frac{7\pi}{12}$ ($105^\circ$). Find the area of the triangular bed.

$$\text{Area} = \frac{1}{2} \times 6 \times 8 \times \sin \frac{7\pi}{12}$$

$$= 24 \times \frac{\sqrt{6} + \sqrt{2}}{4} = 6(\sqrt{6} + \sqrt{2})$$

$$\approx 23.18 \text{ m}^2$$

Where Do Students Trip Up On Sin 7pi/12?

Mistake 1: Splitting Sine Across A Sum

Where it slips in: The first step after writing $\frac{7\pi}{12} = \frac{\pi}{4} + \frac{\pi}{3}$.

Don't do this: Write $\sin \frac{\pi}{4} + \sin \frac{\pi}{3} \approx 1.5731$. A sine above $1$ is impossible.

The correct way: Use $\sin A \cos B + \cos A \sin B$. The rusher who has just used the distributive property in algebra applies it to sine as if sine were multiplication; checking that the answer is at most $1$ catches it in one glance.

Mistake 2: Mixing Up The Sine And Cosine Sum Formulas

Where it slips in: Writing the identity from memory under time pressure.

Don't do this: Use $\sin A \sin B$ products or a minus sign, as in the cosine formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$, and get $\frac{\sqrt{2} - \sqrt{6}}{4}$, which is actually $\cos \frac{7\pi}{12}$.

The correct way: Sine's sum formula mixes functions ($\sin \cos + \cos \sin$) and keeps the sign of the angle operation. A negative answer is also a red flag, since $\frac{7\pi}{12}$ is in Q2; students who check the quadrant sign before simplifying rarely carry this error to the end.

Mistake 3: Mismatching The Calculator Mode And The Angle

Where it slips in: Checking the exact answer on a calculator.

Don't do this: Type "sin 105" in radian mode and get $-0.9705$, or type $\frac{7\pi}{12}$ in degree mode and get $0.0320$.

The correct way: Match the mode to the angle: radians for $\frac{7\pi}{12}$, degrees for $105^\circ$, with brackets around the whole angle. Both correct entries return $0.9659$.

Where Can You Get Help With Sum And Difference Formulas?

Sin 7pi/12 is the point where many students first meet the angle-sum identity, and steady practice is what makes it reliable. A trigonometry tutor or a high school math tutor can walk through each split, and math classes online keep the identities in regular use.

Key Takeaways

• Sin 7pi/12 equals $\frac{\sqrt{6} + \sqrt{2}}{4}$, about $0.9659$, and the equivalent form is $\frac{\sqrt{2 + \sqrt{3}}}{2}$.

• $\frac{7\pi}{12} = 105^\circ$, an angle in Q2 with reference angle $\frac{5\pi}{12}$ ($75^\circ$), so its sine is positive.

• The standard proof is $\sin\left(\frac{\pi}{4} + \frac{\pi}{3}\right) = \sin \frac{\pi}{4} \cos \frac{\pi}{3} + \cos \frac{\pi}{4} \sin \frac{\pi}{3}$.

• Sin 7pi/12 equals $\sin \frac{5\pi}{12}$ (supplementary angles) and $\cos \frac{\pi}{12}$ (co-function).

• Sine does not split across a sum; any "sine" above $1$ signals that error.

• Calculator checks need brackets and the matching angle mode.

Practice These Before Moving On

1. Find $\cos \frac{7\pi}{12}$ exactly using $\cos\left(\frac{\pi}{4} + \frac{\pi}{3}\right)$.

2. Find $\sin\left(-\frac{7\pi}{12}\right)$.

3. Find $\sin \frac{19\pi}{12}$.

4. Evaluate $\sin \frac{7\pi}{12} - \sin \frac{\pi}{12}$.

Answers:

1. $\frac{\sqrt{2}}{2} \times \frac{1}{2} - \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{2} - \sqrt{6}}{4} \approx -0.2588$.

2. Sine is odd, so the value is $-\frac{\sqrt{6} + \sqrt{2}}{4} \approx -0.9659$.

3. $\frac{19\pi}{12} = \frac{7\pi}{12} + \pi$, and adding $\pi$ flips the sign of sine, so the value is $-\frac{\sqrt{6} + \sqrt{2}}{4}$.

4. $\frac{\sqrt{6} + \sqrt{2}}{4} - \frac{\sqrt{6} - \sqrt{2}}{4} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2}$.

If question 1 came out positive, revisit Mistake 2: the cosine sum formula has a minus sign, and the quadrant says the answer must be negative.

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