How to Find Sine, Cosine, and Tangent of 15° and 75° (without using a Trig Identity)

SINE, COSINE, AND TANGENT OF 15° AND 75°

It is possible to find the sine, cosine, and tangent of 15° and also 75° without using a trig identity (and without using a calculator).

The trick is to begin with a 75°-75°-30° isosceles triangle, as shown above. Let AB = AC = 2. These are the sides opposite to the 75° angles. The remaining side, BC, is unknown at this point.

With side AB serving as the base, draw an altitude down from point C, as shown below. This divides the isosceles triangle into two right triangles. Triangle ACD is a 30°-60°-90° triangle and triangle BCD is a 75°-15°-90° triangle.

It is well-known that the sides of a 30°-60-°90° triangle come in the ratio 1:sqrt(3):2, with 1 opposite to 30°, sqrt(3) opposite to 60°, and 2 as the hypotenuse. Since AC = 2 is the hypotenuse, this means that CD = 1 and AD = sqrt(3).

Now we know that the altitude of triangle ABC is CD = 1, which is one side of triangle BCD. We can find side BD by subtracting AD from AB. Since AB = 2 and AD = sqrt(3), we get BD = 2 – sqrt(3).

The Pythagorean theorem can then be used to determine BC, which is the hypotenuse of triangle BCD.

Recall that BD = 2 – sqrt(3) and that CD = 1. After applying the Pythagorean theorem, apply the “foil” method from algebra: (x – y)² = x² – 2xy + y². Recall that (sqrt(3))² = sqrt(3) × sqrt(3) = 3.

To solve for BC, take the square root of both sides of the equation. Factor out the 4. Recall that sqrt(ab) = sqrt(a) × sqrt(b). Note that sqrt(4) = 2.

The above answer has one square root inside of another. There is a clever way to rewrite this without using a nested square root. Rewrite 8 as 6 + 2. The reason behind this is that (sqrt(6))² = 6 and (sqrt(2))² = 2. If you “foil” out (sqrt(6) – sqrt(2))², you get 6 – 2sqrt(12) + 2 = 8 – 4sqrt(3) because sqrt(12) = sqrt(4) × sqrt(3) = 2 sqrt(3) such that 2sqrt(12) = 2(2)sqrt(3) = 4 sqrt(3). If you’re still not convinced, note that sqrt(8 – 4sqrt(3)) is approximately 1.03527618 on a calculator, and that sqrt(6) – sqrt(2) is also approximately 1.03527618 on a calculator.

Since BC = 2 sqrt(2 – sqrt(3)) and BC = sqrt(6) – sqrt(2), it follows from the transitive property that 2 sqrt(2 – sqrt(3)) = sqrt(6) – sqrt(2). Divide both sides by 2 to get the following:

The following forms for BC are equivalent, but the right expression is considered to be “standard form.”

Now that we know BC, BD, and CD, we can easily find the sine, cosine, and tangent of 15°. In triangle BCD, note that BD is opposite to 15°, CD is adjacent to 15°, and BC is the hypotenuse. For the sine of 15°, note that (2 – sqrt(3)) / sqrt(2 – sqrt(3)) = sqrt(2 – sqrt(3)) for the same reason that x/sqrt(x) = sqrt(x). For the cosine of 15°, the answer 1/(sqrt(6) – sqrt(2)) is equivalent to the answer (sqrt(6) + sqrt(2))/4. Both answers approximately equal 0.965925826 on a calculator, but the rightmost expression is considered to be “standard form” because it has a rational denominator. We multiplied the numerator and denominator each by (sqrt(6) + sqrt(2)) in order to rationalize the denominator. This is called “multiplying by the conjugate.” The conjugate of sqrt(6) – sqrt(2) is sqrt(6) + sqrt(2) because the product of these two conjugate expressions is rational. When applying the “foil” method, the irrational terms cancel out.

Note that the sine of 15° is equivalent to the cosine of 75°, and that the cosine of 15° is equivalent to the sine of 75°. What is “opposite” for 15° is “adjacent” for 75°, and vice-versa. For the tangent of 75°, we again multiplied by the conjugate in order to rationalize the denominator.

Below is a summary of our final answers.

CHRIS MCMULLEN, PH.D.

Copyright © 2021 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks

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Memory Tip for Sine, Cosine, and Tangent of Special Angles (Trigonometry)

Trig Table

TRIGONOMETRY MEMORY TIP

There is a simple way to remember the sine, cosine, and tangent of special trigonometry angles.

The special trig angles are 0º, 30º, 45º, 60º, and 90º. What makes these angles special? The 30º-60º-90º triangle is one-half of an equilateral triangle, while the 45º-45º-90º triangle is one-half of a square. In both cases, the trig functions (sine, cosine, and tangent) can be expressed as simple ratios.

Here is the trick for quickly working out the sine, cosine, and tangent of 0º, 30º, 45º, 60º, and 90º.

STEP 1: Special angles.

Write the special angles in order.

Trig Table top

STEP 2: Integers.

Write the integers 0 thru 4 in order.

Trig Table numbers

STEP 3: Squareroots.

Squareroot each number.

Trig Table roots

STEP 4: Find the sine of theta.

Divide each number by 2.

Trig Table sine

These are the sine of 0º, 30º, 45º, 60º, and 90º.

It’s that simple. Here’s a recap:

  • Write the numbers 0 thru 4.
  • Squareroot each number.
  • Divide each number by 2.

STEP 5: Find the cosine of theta.

Just write the previous numbers in reverse order.

Trig Table cos

Why does this work? Because the sine of theta equals the cosine of the complement of theta: sin(θ)=cos(90º–θ). What’s opposite to theta is adjacent to its complement.

STEP 6: Find the tangent of theta.

Divide sine theta by cosine theta.

Trig Table tan

TRIG CHART

This chart shows all of the steps together.

Trig Table Tips

  • Write the special angles.
  • Write the integers 0 thru 4.
  • Squareroot each number.
  • Divide each number by 2. This gives you sine of theta.
  • Write the numbers in reverse order. This gives you cosine of theta.
  • Divide the previous two rows (sine over cosine). This gives you tangent theta.

NOTE

There are two different, yet equivalent ways, to write the above chart.

That’s because of the following properties of irrational numbers:

Roots

So, for example, there are alternative ways to express the following trig values:

Trig Values

The chart on this blog uses standard form. Most math courses use standard form, which means that there no irrational numbers (like root 2) in the denominator.

Sometimes you find the trig table in another nonstandard form. In that form, you see 1 over root 2 in place of root 2 over 2, and you see 1 over root 3 in place of root 3 over 3.

It’s important to realize that both forms are correct. The standard form, however, is expected in most math courses.

CHRIS MCMULLEN, PH.D.

Copyright © 2015 Chris McMullen, author of the Improve Your Math Fluency series of math workbooks

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