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Question: Find the exact value of \( \cos \left( -\dfrac{5\pi }{4} \right) \) ....

Find the exact value of cos(5π4)\cos \left( -\dfrac{5\pi }{4} \right) .

Explanation

Solution

Hint : We have to find the value of cos(5π4)\cos \left( -\dfrac{5\pi }{4} \right) . We know that, cos(θ)=cosθ\cos \left( -\theta \right)=\cos \theta . Now, cos(5π4)\cos \left( -\dfrac{5\pi }{4} \right) is also in the form of cos(θ)\cos \left( -\theta \right) so we can use this trigonometric relation cos(θ)=cosθ\cos \left( -\theta \right)=\cos \theta . We know that 5π4\dfrac{5\pi }{4} lies in the third quadrant in which the value of cosine is negative so whatever the value of cos(5π4)\cos \left( -\dfrac{5\pi }{4} \right) , it comes with a negative sign. There is a trigonometric property which states that cos(π+θ)=cosθ\cos \left( \pi +\theta \right)=-\cos \theta so we can write cos(5π4)\cos \left( \dfrac{5\pi }{4} \right) as cos(π+π4)\cos \left( \pi +\dfrac{\pi }{4} \right) and then we can use this trigonometric property. Then substitute the value of cos(π4)\cos \left( \dfrac{\pi }{4} \right) to get the exact value.

Complete step-by-step answer :
We are asked to find the value of cos(5π4)\cos \left( -\dfrac{5\pi }{4} \right) .
We know that,
cos(θ)=cosθ\cos \left( -\theta \right)=\cos \theta
As in the above problem, the angle at which we have to find the cosine is also negative so we can use this relation in the given cos(5π4)\cos \left( -\dfrac{5\pi }{4} \right) .
cos(5π4)=cos(5π4)\cos \left( -\dfrac{5\pi }{4} \right)=\cos \left( \dfrac{5\pi }{4} \right)
So, we have reduced the given expression in cosine to:
cos(5π4)\cos \left( \dfrac{5\pi }{4} \right)
Now, the angle given above is 5π4\dfrac{5\pi }{4} and we can also write this angle as:
5π4=π+π4\dfrac{5\pi }{4}=\pi +\dfrac{\pi }{4}
Substituting the above angle in cos(5π4)\cos \left( \dfrac{5\pi }{4} \right) we get,
cos(5π4)=cos(π+π4)\cos \left( \dfrac{5\pi }{4} \right)=\cos \left( \pi +\dfrac{\pi }{4} \right)
As the above angle lies in the third quadrant so the value of cosine in the third quadrant is negative.
In the below figure, we have shown the four quadrants and have also shown the angle 5π4\dfrac{5\pi }{4} lies in the III Quadrant from the positive x axis.

We know that,
cos(π+θ)=cosθ\cos \left( \pi +\theta \right)=-\cos \theta
Using the above relation in cos(π+π4)\cos \left( \pi +\dfrac{\pi }{4} \right) we get,
cos(π+π4)=cos(π4)\cos \left( \pi +\dfrac{\pi }{4} \right)=-\cos \left( \dfrac{\pi }{4} \right) …………. Eq. (1)
From the trigonometric ratios we know that,
cos(π4)=12\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}
Substituting the above value in eq. (1) we get,
cos(π+π4)=12\cos \left( \pi +\dfrac{\pi }{4} \right)=-\dfrac{1}{\sqrt{2}}
From the above calculations, we have got the value of cos(5π4)\cos \left( -\dfrac{5\pi }{4} \right) as 12-\dfrac{1}{\sqrt{2}} .

Note : The plausible problem that could happen is that instead of writing cos(5π4)\cos \left( \dfrac{5\pi }{4} \right) as cos(π+π4)\cos \left( \pi +\dfrac{\pi }{4} \right) you might make a mistake of perceiving the angle 5π4\dfrac{5\pi }{4} as 3π4\dfrac{3\pi }{4} and write cos(5π4)\cos \left( \dfrac{5\pi }{4} \right) as cos(ππ4)\cos \left( \pi -\dfrac{\pi }{4} \right) .
You might think that how it will click in the mind that we can write 5π4\dfrac{5\pi }{4} as π+π4\pi +\dfrac{\pi }{4} . The answer is it is clearly visible on the prima facie that 5π4\dfrac{5\pi }{4} is greater than π\pi and we know the value of cosπ4\cos \dfrac{\pi }{4} so in the following line of logic you can write 5π4\dfrac{5\pi }{4} as π+π4\pi +\dfrac{\pi }{4} .