Finding arctan(1/sqrt3): Solving the Puzzle

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In summary, the speaker asks about finding arctan(1/sqrt3) and notes that the coordinates of 7pi/6 equal 1/sqrt3 on a pie chart. However, they found the answer to be pi/6 when using a right triangle. The speaker questions why y/x for the coordinates of pi/6 gives sqrt3 instead of 1/sqrt3. They then mention drawing a right triangle with pi/6 and finding the opposite (y) and adjacent (x) sides to be 1/2 and sqrt(3)/2 respectively, resulting in y/x=1/sqrt(3). They are seeking an explanation for this discrepancy.
  • #1
fk378
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This is a general question.
I was given a problem where you need to find arctan(1/sqrt3). Referring to a pie chart, I see that if you compute y/x, you can see that the coordinates of 7pi/6 equal 1/sqrt3.

However, I found the answer is pi/6 and I know the proof is in drawing a right triangle. But if you do y/x for the coordinates of pi/6, you get sqrt3 instead of 1/sqrt3. Can anyone tell me why?
 
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  • #2
If I draw a right triangle with an angle of pi/6, I see an opposite side (y) with length 1/2 and and adjacent side (x) with length sqrt(3)/2. y/x=1/sqrt(3). What's the problem?
 
  • #3


I would approach this question by first clarifying the notation being used. In mathematics, the notation "arctan" typically refers to the inverse trigonometric function, also known as the arctangent function. This function takes in a ratio of two sides of a right triangle and outputs the angle opposite the side with the given ratio.

In this case, the given ratio is 1/sqrt3, which represents the tangent of an angle in a right triangle. However, it is important to note that the tangent function is not the same as the arctangent function. The tangent function takes in an angle and outputs the ratio of the opposite side to the adjacent side in a right triangle.

In order to find the correct angle for arctan(1/sqrt3), we need to use the inverse of the tangent function, which is the arctangent function. This means we need to use the inverse property of trigonometric functions, which states that the inverse of a trigonometric function "undoes" the original function. In this case, we need to use the inverse property to "undo" the tangent function and find the angle that produces a ratio of 1/sqrt3.

To do this, we can use a right triangle with an angle of pi/6 (30 degrees) and sides of 1 and sqrt3. This triangle satisfies the given ratio of 1/sqrt3, and we can see that the angle opposite the side with a length of 1 is indeed pi/6. Therefore, the correct answer for arctan(1/sqrt3) is pi/6.

In summary, the confusion may arise from mixing up the tangent and arctangent functions, which are inverse operations. It is important to use the correct function in order to find the correct angle. Additionally, using a visual representation, such as a right triangle, can help to understand and solve the problem.
 

Related to Finding arctan(1/sqrt3): Solving the Puzzle

1. What is arctan(1/sqrt3)?

Arctan(1/sqrt3) is an inverse trigonometric function that represents the angle whose tangent is equal to 1/sqrt3. In other words, it is the angle in a right triangle whose opposite side is equal to 1 and whose adjacent side is equal to sqrt3.

2. Why is finding arctan(1/sqrt3) important?

Finding arctan(1/sqrt3) is important because it allows us to solve for the missing angle in a right triangle when we know the length of its sides. This can be useful in many real-world applications, such as engineering and navigation.

3. What is the process for finding arctan(1/sqrt3)?

The process for finding arctan(1/sqrt3) involves using a calculator or a trigonometric table to determine the inverse tangent of 1/sqrt3. This will give you the angle in radians, which can be converted to degrees by multiplying by 180/pi.

4. Can arctan(1/sqrt3) be simplified?

Yes, arctan(1/sqrt3) can be simplified to pi/6 or 30 degrees. This is because 1/sqrt3 is equivalent to sin(pi/6) or cos(pi/6), and the inverse tangent of these values will give us pi/6 or 30 degrees.

5. How can I use arctan(1/sqrt3) to solve a puzzle?

Arctan(1/sqrt3) can be used to solve a puzzle by providing the missing angle in a given right triangle. It can also be used in conjunction with other trigonometric functions to find the missing side lengths of a triangle. Additionally, it can be used in geometry puzzles that involve angles and triangles.

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