Sine-cosine square identity

One trigonometric identity states that for any angle \theta:

\sin^2(\theta) + \cos^2(\theta) \equiv 1

Proof

For a unit triangle (one where the hypotenuse has length 1), the lengths of the two other sides correspond to the sine and cosine of the angle \theta.

In other words, the side opposite to angle \theta has length \sin(\theta), and the side adjacent to angle \theta has length \cos(\theta).

From the pythagorean theorem, we know that for any right triangle with sides of lengths a and b, and hypotenuse of length c:

a^2 + b^2 \equiv c^2

In this case:

• a = \sin(\theta)
• b = \cos(\theta)
• c = 1

So we can substitute these values into the pythagorean theorem!

(\sin(\theta))^2 + (\cos(\theta))^2 \equiv 1^2

which is otherwise written as:

\sin^2(\theta) + \cos^2(\theta) \equiv 1

Finding missing values

Given that \sin(\theta) = 0.6, find \cos(\theta)

• \sin^2(\theta) + \cos^2(\theta) \equiv 1
• (0.6)^2 + \cos^2(\theta) = 1
• 0.36 + \cos^2(\theta) = 1
• \cos^2(\theta) = 1 - 0.36
• \cos^2(\theta) = 0.64
• \cos(\theta) = \pm\sqrt{0.64}
• \cos(\theta) = \pm0.8

Given that \cos(\theta) = -0.5, find \sin(\theta)

• \sin^2(\theta) + \cos^2(\theta) \equiv 1
• \sin^2(\theta) + (-0.5)^2 = 1
• \sin^2(\theta) + 0.25 = 1
• \sin^2(\theta) = 1 - 0.25
• \sin^2(\theta) = 0.75
• \sin(\theta) = \pm\sqrt{0.75}
• \sin(\theta) = \pm\frac{\sqrt{3}}{2}

Given that \sin(\theta) = -\frac{3}{5}, find \cos(\theta)

• \sin^2(\theta) + \cos^2(\theta) \equiv 1
• \left(-\frac{3}{5}\right)^2 + \cos^2(\theta) = 1
• \frac{9}{25} + \cos^2(\theta) = 1
• \cos^2(\theta) = 1 - \frac{9}{25}
• \cos^2(\theta) = \frac{16}{25}
• \cos(\theta) = \pm\sqrt{\frac{16}{25}}
• \cos(\theta) = \pm\frac{4}{5}

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