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Question: How do you evaluate \({\cos ^2}\left( {\dfrac{\pi }{8}} \right) - {\sin ^2}\left( {\dfrac{\pi }{8}} ...

How do you evaluate cos⁡2(π8)−sin⁡2(π8){\cos ^2}\left( {\dfrac{\pi }{8}} \right) - {\sin ^2}\left( {\dfrac{\pi }{8}} \right)?

Explanation

Solution

In order to evaluate the given question, we must first know the trigonometric ratios and the trigonometric identities. Every trigonometric function and formulae are designed on the basis of three primary ratios. Sine, Cosine and tangents are these ratios in trigonometry based on Perpendicular, Hypotenuse and Base of a right triangle . In order to calculate the angles sin , cos and tan functions . For this particular question we need to know the double angle formula through which on applying we can get our required solution.

Complete step-by-step answer:
For evaluating the given question cos⁡2(π8)−sin⁡2(π8){\cos ^2}\left( {\dfrac{\pi }{8}} \right) - {\sin ^2}\left( {\dfrac{\pi }{8}} \right) , we must recall the trigonometric identity related to this given question .
The Double – Angle Formula fits best to the question and we can apply that having the formula as follows –
⇒cos⁡2θ=cos⁡2θ−sin⁡2θ ⇒cos⁡2θ=1−2sin⁡2θ ⇒cos⁡2θ=2cos⁡2θ−1  \Rightarrow \cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta \\\ \Rightarrow \cos 2\theta = 1 - 2{\sin ^2}\theta \\\ \Rightarrow \cos 2\theta = 2{\cos ^2}\theta - 1 \\\
We can easily see from the above Double - Angle formulae that there is one formula which resembles our given question .
⇒cos⁡2θ=cos⁡2θ−sin⁡2θ\Rightarrow \cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta
Now , applying this formula will make our solution as follows -
⇒cos⁡2θ=cos⁡2θ−sin⁡2θ\Rightarrow \cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta =cos⁡2(π8)−sin⁡2(π8){\cos ^2}\left( {\dfrac{\pi }{8}} \right) - {\sin ^2}\left( {\dfrac{\pi }{8}} \right)
Here θ=π8\theta = \dfrac{\pi }{8}
⇒cos⁡2θ=\Rightarrow \cos 2\theta = cos⁡2(π8)−sin⁡2(π8){\cos ^2}\left( {\dfrac{\pi }{8}} \right) - {\sin ^2}\left( {\dfrac{\pi }{8}} \right)
⇒cos⁡2π8=\Rightarrow \cos 2\dfrac{\pi }{8} = cos⁡2(π8)−sin⁡2(π8){\cos ^2}\left( {\dfrac{\pi }{8}} \right) - {\sin ^2}\left( {\dfrac{\pi }{8}} \right)
⇒cos⁡π4=\Rightarrow \cos \dfrac{\pi }{4} = cos⁡2(π8)−sin⁡2(π8){\cos ^2}\left( {\dfrac{\pi }{8}} \right) - {\sin ^2}\left( {\dfrac{\pi }{8}} \right)
And as we know that cos⁡π4=\cos \dfrac{\pi }{4} = 12\dfrac{1}{{\sqrt 2 }}

Therefore , the final answer is 12\dfrac{1}{{\sqrt 2 }}.

Note: The sine function can be expressed as angle which is equal to the length of opposite side divided by the length of hypotenuse side and the formula is given , sin⁡θ=opp. sidehypotenuse side\sin \theta = \dfrac{{opp.\,side}}{{hypotenuse\,side}}
Learn the standard values of trigonometry angles by heart .
We know that sin⁡(−θ)=−sin⁡θ.cos⁡(−θ)=cos⁡θ and tan⁡(−θ)=−tan⁡θ\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta
Therefore, asin⁡θ\sin \theta nd tan⁡θ\tan \theta and their reciprocals,csc⁡θ\csc \theta and cot⁡θ\cot \theta are odd functions whereas cos⁡θ\cos \theta and its reciprocal sec⁡θ\sec \theta are even functions .
One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.