
Find the value of \[{{\cos }^{2}}\dfrac{\pi }{16}+{{\cos }^{2}}\dfrac{3\pi }{16}+{{\cos }^{2}}\dfrac{5\pi }{16}+{{\cos }^{2}}\dfrac{7\pi }{16}\]
Answer
610.2k+ views
Hint: We will first simplify all the terms in the expression mentioned in the question in multiples of 90 plus some angle and then with the help of cofunction identities we will convert these terms into simple function with standard angles and then we will use \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] a few times and after applying this we will solve this question.
Complete step-by-step answer:
The given expression in the question is \[{{\cos }^{2}}\dfrac{\pi }{16}+{{\cos }^{2}}\dfrac{3\pi }{16}+{{\cos }^{2}}\dfrac{5\pi }{16}+{{\cos }^{2}}\dfrac{7\pi }{16}.......(1)\]
Now breaking \[\dfrac{5\pi }{16}\] and \[\dfrac{7\pi }{16}\] in terms of \[\dfrac{\pi }{2}\] in equation (1) and simplifying we get,
\[\Rightarrow {{\cos }^{2}}\dfrac{\pi }{16}+{{\cos }^{2}}\dfrac{3\pi }{16}+{{\cos }^{2}}\left( \dfrac{\pi }{2}-\dfrac{3\pi }{16} \right)+{{\cos }^{2}}\left( \dfrac{\pi }{2}-\dfrac{\pi }{16} \right).......(2)\]
We know that \[\cos (90-\theta )=\sin \theta \] and when we square both sides we get \[{{\cos }^{2}}(90-\theta )={{\sin }^{2}}\theta \] and hence applying this in equation (2) we get,
\[\Rightarrow {{\cos }^{2}}\dfrac{\pi }{16}+{{\cos }^{2}}\dfrac{3\pi }{16}+{{\sin }^{2}}\dfrac{3\pi }{16}+{{\sin }^{2}}\dfrac{\pi }{16}.......(3)\]
Now rearranging the terms in equation (3) we get,
\[\Rightarrow \left( {{\sin }^{2}}\dfrac{\pi }{16}+{{\cos }^{2}}\dfrac{\pi }{16} \right)+\left( {{\sin }^{2}}\dfrac{3\pi }{16}+{{\cos }^{2}}\dfrac{3\pi }{16} \right).......(4)\]
We also know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] and hence applying this in equation (4) we get,
\[\Rightarrow 1+1=2\]
Hence we get 2 as the answer to the expression given in the question.
Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We in a hurry can make a mistake in applying the cofunction identities as we can write cos in place of sin and sin in place of cos in equation (3). Also if we don’t remember the formula then we can get confused about how to proceed further after equation (2).
Complete step-by-step answer:
The given expression in the question is \[{{\cos }^{2}}\dfrac{\pi }{16}+{{\cos }^{2}}\dfrac{3\pi }{16}+{{\cos }^{2}}\dfrac{5\pi }{16}+{{\cos }^{2}}\dfrac{7\pi }{16}.......(1)\]
Now breaking \[\dfrac{5\pi }{16}\] and \[\dfrac{7\pi }{16}\] in terms of \[\dfrac{\pi }{2}\] in equation (1) and simplifying we get,
\[\Rightarrow {{\cos }^{2}}\dfrac{\pi }{16}+{{\cos }^{2}}\dfrac{3\pi }{16}+{{\cos }^{2}}\left( \dfrac{\pi }{2}-\dfrac{3\pi }{16} \right)+{{\cos }^{2}}\left( \dfrac{\pi }{2}-\dfrac{\pi }{16} \right).......(2)\]
We know that \[\cos (90-\theta )=\sin \theta \] and when we square both sides we get \[{{\cos }^{2}}(90-\theta )={{\sin }^{2}}\theta \] and hence applying this in equation (2) we get,
\[\Rightarrow {{\cos }^{2}}\dfrac{\pi }{16}+{{\cos }^{2}}\dfrac{3\pi }{16}+{{\sin }^{2}}\dfrac{3\pi }{16}+{{\sin }^{2}}\dfrac{\pi }{16}.......(3)\]
Now rearranging the terms in equation (3) we get,
\[\Rightarrow \left( {{\sin }^{2}}\dfrac{\pi }{16}+{{\cos }^{2}}\dfrac{\pi }{16} \right)+\left( {{\sin }^{2}}\dfrac{3\pi }{16}+{{\cos }^{2}}\dfrac{3\pi }{16} \right).......(4)\]
We also know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] and hence applying this in equation (4) we get,
\[\Rightarrow 1+1=2\]
Hence we get 2 as the answer to the expression given in the question.
Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We in a hurry can make a mistake in applying the cofunction identities as we can write cos in place of sin and sin in place of cos in equation (3). Also if we don’t remember the formula then we can get confused about how to proceed further after equation (2).
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