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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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The value of $\cos^248^{\large\circ}-\sin^212^{\large\circ}$ is

$\begin{array}{1 1}(A)\;\large\frac{\sqrt 5+1}{8}&(B)\;\large\frac{-\sqrt 5-1}{8}\\(C)\;\large\frac{\sqrt 5+1}{5}&(D)\;\large\frac{\sqrt 5+1}{2\sqrt 2}\end{array} $

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1 Answer

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Toolbox:
  • $\cos^2A-\sin^2B=\cos(A+B).\cos (A-B)$
  • $\cos 36^{\large\circ}=\large\frac{\sqrt 5+1}{4}$
$\cos^248^{\large\circ}-\sin^212^{\large\circ}$
$\cos^2A-\sin^2B=\cos(A+B).\cos (A-B)$
$\Rightarrow \cos(48+12).\cos (48-12)$
$\Rightarrow \cos 60^{\large\circ}.\cos 36^{\large\circ}$
$\cos 60^{\large\circ}=\large\frac{1}{2}$
$\cos 36^{\large\circ}=\large\frac{\sqrt 5+1}{4}$
$\Rightarrow \large\frac{1}{2}\times \frac{\sqrt 5+1}{4}$
$\Rightarrow \large \frac{\sqrt 5+1}{8}$
Hence (A) is the correct answer.
answered Apr 21, 2014 by sreemathi.v
 

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