From tan theta = sqrt 3 , how do we arrive at 4pi/3? I did theta = inverse tan sqrt3 and did not get 4pi/3
The Correct Answer and Explanation is :
To solve for ( \theta ) when given that ( \tan(\theta) = \sqrt{3} ), the process involves understanding both the inverse tangent function and the nature of the tangent function. Let’s break this down step by step:
Step 1: Understand the given equation.
The equation you’re working with is:
[
\tan(\theta) = \sqrt{3}
]
We want to find ( \theta ). To do this, we can apply the inverse tangent function (also called ( \tan^{-1} ) or arctan) to both sides of the equation.
Step 2: Apply the inverse tangent function.
[
\theta = \tan^{-1}(\sqrt{3})
]
The arctangent function ( \tan^{-1}(\sqrt{3}) ) gives the principal value of ( \theta ), which is the value of ( \theta ) in the interval ( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) ).
Since ( \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} ), we know that:
[
\theta = \frac{\pi}{3}
]
However, this only gives one solution within the principal range.
Step 3: Consider the periodic nature of the tangent function.
The tangent function has a period of ( \pi ), meaning that it repeats every ( \pi ) radians. This implies that if ( \tan(\theta) = \sqrt{3} ) at ( \theta = \frac{\pi}{3} ), then ( \tan(\theta) ) will also be ( \sqrt{3} ) at ( \theta = \frac{\pi}{3} + n\pi ), where ( n ) is any integer.
Step 4: Find the general solution.
To find all possible values of ( \theta ) that satisfy ( \tan(\theta) = \sqrt{3} ), we write the general solution as:
[
\theta = \frac{\pi}{3} + n\pi
]
Step 5: Consider all possible solutions within a specific range.
If you’re asked to find the solution in a specific interval, such as ( 0 \leq \theta < 2\pi ), you would need to consider the values of ( n ) that satisfy this range.
- For ( n = 0 ), ( \theta = \frac{\pi}{3} )
- For ( n = 1 ), ( \theta = \frac{\pi}{3} + \pi = \frac{4\pi}{3} )
Thus, the two solutions in the interval ( 0 \leq \theta < 2\pi ) are:
[
\theta = \frac{\pi}{3}, \frac{4\pi}{3}
]
Step 6: Conclusion
The value ( \theta = \frac{4\pi}{3} ) comes from considering the periodic nature of the tangent function. While the inverse tangent gives you ( \frac{\pi}{3} ), the tangent function repeats every ( \pi ) radians, so ( \frac{4\pi}{3} ) is another valid solution in the interval ( 0 \leq \theta < 2\pi ).