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Question: How do you find the exact value of \(\cot \left( \dfrac{5\pi }{4} \right)\) ?...

How do you find the exact value of cot(5π4)\cot \left( \dfrac{5\pi }{4} \right) ?

Explanation

Solution

To solve the above question use the property of the trigonometric functions that they periodic functions and have a period of 2π2\pi . Write the given angle as the sum of 2π2\pi and some other angle and proceed further.

Complete step by step solution:
Let us first understand the procedure to find the exact value of any trigonometric ratios.We know that the basic trigonometric ratios are sine and cosine and all the other trigonometric ratios depend on these two ratios. We also know that functions of sine and cosine are periodic functions (meaning their values repeat after equal intervals). The sine and cosine functions have a period of 2π2\pi . And similarly, all the other trigonometric functions have a period of 2π2\pi . This means that any trigonometric function will repeat itself after every equal interval of 2π2\pi units. If we plot the given trigonometric function, we can see this property more clearly.

Suppose, we have a trigonometric function, say sin(x) and let the value of the function sin(x), at x=x0x={{x}_{0}} has some value, say y. Then the function will have the same value (i.e. y) at x=x0+2πx={{x}_{0}}+2\pi , since the function has a period of 2π2\pi .This means that sin(2π+x0)=sin(x0)\sin (2\pi +{{x}_{0}})=\sin ({{x}_{0}}).This relation is applicable for all the trigonometric ratios.In the question, the given function is cot(x).The value 5π4\dfrac{5\pi }{4} can be written as 5π4=2π3π4\dfrac{5\pi }{4}=2\pi -\dfrac{3\pi }{4}.
Then, this means that cot(5π4)=cot(2π3π4)\cot \left( \dfrac{5\pi }{4} \right)=\cot \left( 2\pi -\dfrac{3\pi }{4} \right).
And we know that cot(2π3π4)=cot(3π4)\cot \left( 2\pi -\dfrac{3\pi }{4} \right)=\cot \left( -\dfrac{3\pi }{4} \right).
We also know that cot(3π4)=1\cot \left( -\dfrac{3\pi }{4} \right)=1

Therefore, the exact value of cot(5π4)\cot \left( \dfrac{5\pi }{4} \right) is 1.

Note: To solve the given problem we can also use the relation cot(π+θ)=cot(θ)\cot (\pi +\theta )=\cot (\theta ). However, this relation is only valid for the trigonometric functions – tan(x) and cot(x). For sine and cosine functions,
sin(π+θ)=sin(θ)\sin (\pi +\theta )=-\sin (\theta )
And
cos(π+θ)=cos(θ)\cos (\pi +\theta )=-\cos (\theta )