Question Video: Using the Unit Circle to Explain Symmetry of the Trigonometric Functions | Nagwa Question Video: Using the Unit Circle to Explain Symmetry of the Trigonometric Functions | Nagwa

Question Video: Using the Unit Circle to Explain Symmetry of the Trigonometric Functions Mathematics

Which symmetry of the unit circle proves that the tangent function is odd?

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Video Transcript

Which symmetry of the unit circle proves that the tangent function is odd?

Let’s start by reminding ourselves of the definition of an odd function. A function is odd if and only if 𝑓 of negative π‘₯ is equal to negative 𝑓 of π‘₯. So how do we use the unit circle to prove that the tangent function is odd?

Remember, the unit circle is a circle with a radius of one that allows us to evaluate sin, cos, and tan for specific values of πœƒ. At its most basic level, it can tell us whether the values of sin, cos, and tan πœƒ in each quadrant are positive or negative.

Let’s take the tangent function. We can construct a right-angled triangle in the first quadrant using an ordered pair π‘Ž and 𝑏, where both π‘Ž and 𝑏 are positive real numbers. The opposite side of this right-angled triangle, represented by 𝑏 in our ordered pair, is positive, and the adjacent, represented by π‘Ž, is also positive. Tan πœƒ is equal to 𝑏 over π‘Ž, and a positive divided by a positive gives us a positive solution. For this value of πœƒ, tan πœƒ is positive.

Remember, an odd function is one for which 𝑓 of negative π‘₯ is equal to negative 𝑓 of π‘₯. We’ve already worked out 𝑓 of π‘₯, so let’s look at a negative value of πœƒ. For this ordered pair, which is a reflection in the π‘₯-axis of the previous one, πœƒ is now negative, as we’re measuring it in a clockwise direction. But what about tan πœƒ?

Well, the opposite side is now negative, since the 𝑦 value in our ordered pair is negative 𝑏. The adjacent however is still positive, since π‘₯ in our ordered pair remains positive π‘Ž. In this case, tan πœƒ is equal to negative 𝑏 divided by positive π‘Ž, which is negative. By considering the ordered pair which we reflected in the π‘₯-axis, we’ve proven that 𝑓 of negative π‘₯ is equal to negative 𝑓 of π‘₯. The tangent function is odd. Therefore, symmetry about the π‘₯-axis proves that the tangent function is odd.

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