Math Detours · Grade 11 · Sharpen · 3

sin(A + B), from a picture

It is not \(\sin A + \sin B\). One diagram of two stacked right triangles gives the true formula, and the double-angle formulas drop out.

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Tonight's idea

Sine does not split over a sum: \(\sin 30^\circ + \sin 60^\circ \approx 1.37\), yet \(\sin 90^\circ = 1\). The true rule comes from one picture.

OQR ABA 1 cos B sin B cos A cos B sin A cos Bcos A sin B cos(A + B)sin A sin B sin(A + B)
Two right triangles stacked, inside a rectangle. Opposite sides of the rectangle are equal.

The blue triangle has hypotenuse \(OQ = 1\) and angle \(B\) at \(O\), so \(OR = \cos B\) and \(RQ = \sin B\). It is tilted up by an angle \(A\) and boxed in.

The bottom triangle has hypotenuse \(\cos B\) and angle \(A\), so its sides are \(\cos A\cos B\) and \(\sin A\cos B\).

The top-right triangle has hypotenuse \(\sin B\). Its angle at \(R\) is also \(A\): the angles along the straight side at \(R\) are \(90^\circ - A\), a right angle, and what is left, and the three make \(180^\circ\). Its sides are \(\cos A\sin B\) and \(\sin A\sin B\).

\(OQ\) makes the angle \(A + B\) with the base, so \(Q\) is \(\sin(A + B)\) up and \(\cos(A + B)\) across. Compare opposite sides of the rectangle:

\(\sin(A + B) = \sin A\cos B + \cos A\sin B\)
\(\cos(A + B) = \cos A\cos B - \sin A\sin B\)

Put \(B = A\) for the double-angle formulas: \(\sin 2A = 2\sin A\cos A\) and \(\cos 2A = \cos^2 A - \sin^2 A\).

Watch one: \(\sin 75^\circ\) exactly

75° is not one of the exact-value angles, but it is \(45^\circ + 30^\circ\), and both of those are.

\[\sin 75^\circ = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ = \frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} + \frac{\sqrt2}{2}\cdot\frac12 = \frac{\sqrt6 + \sqrt2}{4}.\]

That is about 0.966, and a calculator agrees. The tempting wrong answer, \(\sin 45^\circ + \sin 30^\circ \approx 1.21\), is bigger than 1. No sine can be.

Your turn

  1. Use the formula to work out \(\sin(30^\circ + 60^\circ)\) from exact values. Do you get \(\sin 90^\circ\)?
  2. Find \(\cos 75^\circ\) exactly.
  3. \(A\) is an acute angle with \(\sin A = \tfrac35\). Find \(\cos A\), then \(\sin 2A\) and \(\cos 2A\). Is \(2A\) more or less than \(90^\circ\)?
  4. The formulas hold for every angle, including negative ones. Replace \(B\) by \(-B\), using \(\cos(-B) = \cos B\) and \(\sin(-B) = -\sin B\), to get a formula for \(\sin(A - B)\). Use it to find \(\sin 15^\circ\) exactly. Which earlier answer does it match, and why?

Stretch: triple angle

Write \(3A = 2A + A\) and use the formulas to express \(\sin 3A\) using only \(\sin A\). You will need \(\cos^2 A = 1 - \sin^2 A\). Check your answer at \(A = 30^\circ\).

Answers

  1. \(\sin 30^\circ\cos 60^\circ + \cos 30^\circ\sin 60^\circ = \tfrac12\cdot\tfrac12 + \tfrac{\sqrt3}{2}\cdot\tfrac{\sqrt3}{2} = \tfrac14 + \tfrac34 = 1\). Yes: \(\sin 90^\circ = 1\).
  2. \(\cos 75^\circ = \cos 45^\circ\cos 30^\circ - \sin 45^\circ\sin 30^\circ = \dfrac{\sqrt6 - \sqrt2}{4}\), about 0.259.
  3. \(\cos A = \tfrac45\) (a 3-4-5 triangle). \(\sin 2A = 2\cdot\tfrac35\cdot\tfrac45 = \tfrac{24}{25}\). \(\cos 2A = \tfrac{16}{25} - \tfrac{9}{25} = \tfrac{7}{25}\). The cosine is positive, so \(2A\) is less than \(90^\circ\) (it is about \(73.7^\circ\)).
  4. \(\sin(A - B) = \sin A\cos B - \cos A\sin B\). Then \(\sin 15^\circ = \sin(45^\circ - 30^\circ) = \tfrac{\sqrt2}{2}\cdot\tfrac{\sqrt3}{2} - \tfrac{\sqrt2}{2}\cdot\tfrac12 = \dfrac{\sqrt6 - \sqrt2}{4}\). It matches \(\cos 75^\circ\): 15° and 75° are the two acute angles of one right triangle, so the sine of one is the cosine of the other.

Stretch. \(\sin 3A = \sin 2A\cos A + \cos 2A\sin A = 2\sin A\cos^2 A + (\cos^2 A - \sin^2 A)\sin A = 3\sin A\cos^2 A - \sin^3 A\). Replace \(\cos^2 A\) by \(1 - \sin^2 A\): \(\sin 3A = 3\sin A - 4\sin^3 A\). At \(A = 30^\circ\): \(3\cdot\tfrac12 - 4\cdot\tfrac18 = 1 = \sin 90^\circ\).

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