InterviewSolution
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InterviewSolution
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\(\bigg[1+cos\fracπ8\bigg]\)\(\bigg[1+cos\frac{3π}8\bigg]\)\(\bigg[1+cos\frac{5π}8\bigg]\)\(\bigg[1+cos\frac{7π}8\bigg]\) is equal to(a) \(\frac18\) (b) \(\frac12\)(c) \(\frac{1+\sqrt2}{2\sqrt2}\)(d) cos\(\fracπ8\) |
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Answer» (a) \(\frac18\) \(\bigg(1+cos\fracπ8\bigg)\)\(\bigg(1+cos\frac{3π}8\bigg)\)\(\bigg(1+cos\frac{5π}8\bigg)\)\(\bigg(1+cos\frac{7π}8\bigg)\) = \(\bigg(1+cos\fracπ8\bigg)\)\(\bigg(1+cos\frac{3π}8\bigg)\)\(\bigg(1+cos\bigg(π-cos\frac{3π}8\bigg)\bigg)\)\(\bigg(1+cos\bigg(π-\frac{π}8\bigg)\bigg)\) = \(\bigg(1+cos\fracπ8\bigg)\)\(\bigg(1+cos\frac{3π}8\bigg)\)\(\bigg(1-cos\frac{3π}8\bigg)\)\(\bigg(1-cos\frac{π}8\bigg)\) (∵cos (π – θ) = – cos θ) = \(\bigg(1-cos^2\fracπ8\bigg)\)\(\bigg(1-cos^2\frac{3π}8\bigg)\) = sin2 \(\fracπ8\) sin2 \(\frac{3π}8\) = sin2 \(\fracπ8\) cos2 \(\frac{π}8\) \(\bigg(∵ sin\frac{3π}{8}= sin\bigg(\fracπ2-\fracπ8\bigg)=cos\fracπ8\bigg)\) = \(\frac14\bigg(4\,sin^2\fracπ8cos^2\fracπ8\bigg)\) = \(\frac14\bigg(2\,sin\fracπ8\,cos\fracπ8\bigg)^2\) = \(\frac14\big(sin\fracπ4\big)^2\,\frac14\big(\frac1{\sqrt2}\big)^2=\frac18.\) (Using, 2 sin θ cos θ = sin 2θ) |
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