Trigonometric Identities

When students are first exposed to trigonometric identities, they are often given a list of formulas, which they are asked to memorize. Here is a way for you to remember many of these ideas.

Start by drawing a regular hexagon and connect each of the vertices to the center. Label the left-most vertex tan \tan , label the bottom left vertex sin \sin , and label the bottom right vertex cos \cos . Label the center 1. Now, to figure out how to label the remaining vertices, simply look at the diagonally opposite vertex and label it as the reciprocal. You should get the following diagram:

Hexagon of Trigonometric Functions Hexagon of Trigonometric Functions

(Note: 1cosθ=cscθ \frac {1}{ \cos \theta} = \csc \theta , which is sometimes denoted as cosec θ \theta . Either version is valid, and we will be using csc \csc in this post.)

Now, let's figure out some properties of this hexagon.

Property 1: How do the labels on the endpoints of a diameter relate? Recall that to establish the labels, we labelled the diagonally opposite vertex as the reciprocal. Hence, the product of the labels on a diameter is 1, which corresponds to the center vertex.

Property 2: How do we relate vertices that are connected by edges? Consider the central vertex 1. Moving directly to the left is equivalent to multiplying by tan \tan , moving to the lower right is equivalent to multiplying by cos \cos , moving to the lower left is equivalent to multiplying by sin \sin . This seems obvious when we're at vertex 1, and in fact holds true for any other vertex. For example, moving left from cos \cos brings us to sin \sin, which corresponds to cosθ×tanθ=sinθ \cos \theta \times \tan \theta = \sin \theta . There are similar statements for moving to the right, upper left and upper right.

Property 3: What property do we get when reading off the 3 vertices in an anti-clockwise manner? With our 3 starting vertices, we know that tanθ=sinθcosθ \tan \theta = \frac {\sin \theta} { \cos \theta} . This behavior holds true for the rest, that the first vertex is equal to the second divided by the third. For example, another set of 3 vertices is cos,cot,csc \cos, \cot, \csc , and you should be able to verify that cosθ=cotθcscθ \cos \theta = \frac { \cot \theta} { \csc \theta} .

Property 4: Recall the Pythagorean identity which states that sin2θ+cos2θ=12 \sin^2 \theta + \cos ^2 \theta = 1^2 . How is this expressed in the hexagon? If we look at the bottom upright triangle, we see it it has vertices of sin,cos \sin, \cos at the base, and 1 1 at the top. In fact, given any other upright triangle, a similar relation holds. For example, with the right upright triangle, we get 12+cot2θ=csc2θ 1^2 + \cot ^2 \theta = \csc ^2 \theta , and with the left upright triangle, we get tan2θ+1=sec2θ \tan^2 \theta + 1 = \sec^2 \theta .

Worked Examples

1. What is secθtanθ \frac {\sec \theta} { \tan \theta} ?

Solution: By property 3, we know that cscθ=secθtanθ \csc \theta = \frac { \sec \theta} { \tan \theta} .


2. Why does property 4 hold?

Solution: We already know that sin2θ+cos2θ=12 \sin ^2 \theta + \cos^2 \theta = 1^2 . Consider what happens when we shift this triangle to the upper right. We are simply multiplying each vertex by secθ \sec \theta . Hence, the equivalent identity is that sec2θ[sin2θ+cos2θ]=sec2θ12 \sec^2 \theta [ \sin ^2 \theta + \cos^2 \theta] = \sec^2 \theta \cdot 1^2 , which becomes tan2θ+12=sec2θ \tan^2 \theta + 1^2 = \sec^2 \theta .


3. What other properties are there?

Solution: This is an open ended question. There are as many properties as you can find for yourself. We will state one more, which is related to Property 3 (and actually is simply a different way of expressing the same idea):

Property 5: If we read a set of 3 vertices off in a clockwise-order, we get that the first vertex is equal to the second over the third. For example, sin,tan,sec \sin, \tan , \sec are 3 consecutive vertices in clockwise order, so this property gives us that sinθ=tanθsecθ \sin \theta = \frac { \tan \theta} { \sec \theta} .

Note by Arron Kau
7 years, 3 months ago

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Where's the "Test Yourself" section?

Hero Miles - 7 years, 1 month ago

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It's not present anymore, but you can always get lots of Trig Identity problems here:

Arron Kau Staff - 7 years, 1 month ago

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where is the challenge a friend option????

Ashwin Upadhyay - 6 years, 8 months ago

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Thanks a lot. A nice way to remember.

Niranjan Khanderia - 7 years ago

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Anand Raj - 6 years, 6 months ago

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this is really GOOD!!!Thanks but could you just explain more about properties of property 4?

Shouvonik Bos3 - 6 years, 4 months ago

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Prove that 1+cot^2 theta/1+cosec theta=1/sin theta

Sreemathi Gnanasekar - 6 years, 1 month ago

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