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Question: What is \(\arctan \left( \cos \pi \right)\)?...

What is arctan⁡(cos⁡π)\arctan \left( \cos \pi \right)?

Explanation

Solution

For solving this question we can use the concept of trigonometric inverse functions. In this question we will take the value of cos⁡π\cos \pi and then we will take the inverse of tan⁡\tan regarding that value. And then you will get some angle and that will be the solution for this question. And all questions of this type should be solved by this method.

Complete step by step solution:
According to our question we have to calculate the value of arctan⁡(cos⁡π)\arctan \left( \cos \pi \right). For solving this question, we will use the trigonometric function concept and first we will solve the value which is given in the bracket and then we will solve the next step. So, as we can see that cos⁡π\cos \pi is given in the bracket with tan⁡−1{{\tan }^{-1}} of arctan⁡\arctan , so we will solve that first. We know that the value of cos⁡π=−1\cos \pi =-1. Now we will go to the next step and in that step, we have to take tan⁡−1{{\tan }^{-1}} of the value of cos⁡π\cos \pi . For taking the arctan⁡\arctan we will always take a value which is in the range of −∞-\infty to ∞\infty . It means that we can’t use the imaginary values in this.
Now, if we take arctan⁡\arctan of cos⁡π\cos \pi ,
Then, arctan⁡(cos⁡π)=arctan⁡(−1)\arctan \left( \cos \pi \right)=\arctan \left( -1 \right)
∵cos⁡π=−1⇒arctan⁡(cos⁡π)=−π4\because \cos \pi =-1 \Rightarrow \arctan \left( \cos \pi \right)=-\dfrac{\pi }{4}
Note that as a function the range of arctan⁡\arctan is limited to [−π2,+π2]\left[ -\dfrac{\pi }{2},+\dfrac{\pi }{2} \right]. Since arctan⁡(−1)\arctan \left( -1 \right) means that the oppositeadjacent\dfrac{\text{opposite}}{\text{adjacent}} for the angle must be (-1) or if we find it in form of xx and yy coordinates, then,
yx=−1⇒y=−x\dfrac{y}{x}=-1\Rightarrow y=-x
And we have an equilateral right angled triangle below the x-axis.
So, the value of arctan⁡(cos⁡π)\arctan \left( \cos \pi \right) is equal to −π4-\dfrac{\pi }{4}.

Note: During solving this question you should be careful about the values for which we are taking the arctan⁡\arctan or tan⁡−1{{\tan }^{-1}} because it is compulsory that they must be a real value. They can’t be imaginary values. And they must be in the range of arctan⁡(x)\arctan \left( x \right), otherwise the solution of this question will go wrong.