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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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Find the value of $\sin 67\large\frac{1}{2}^{\large\circ}$ & $\cos 67\large\frac{1}{2}^{\large\circ}$

$\begin{array}{1 1}(a)\;\large\frac{1}{4}\normalsize[\sqrt{4+2\sqrt 2}+\sqrt{4-2\sqrt 2}],\large\frac{1}{4}\normalsize[\sqrt{4+2\sqrt 2}-\sqrt{4-2\sqrt 2}]\\(b)\;\large\frac{1}{2}\normalsize[\sqrt{4+2\sqrt 2}+\sqrt{4-2\sqrt 2}],\large\frac{1}{2}\normalsize[\sqrt{4+2\sqrt 2}-\sqrt{4-2\sqrt 2}]\\(c)\;\large\frac{1}{3}\normalsize[\sqrt{4+2\sqrt 2}+\sqrt{4-2\sqrt 2}],\large\frac{1}{4}\normalsize[\sqrt{4+2\sqrt 2}-\sqrt{4-2\sqrt 2}]\\(d)\;\large\frac{1}{4}\normalsize[\sqrt{4+2\sqrt 2}-\sqrt{4-2\sqrt 2}],\large\frac{1}{4}\normalsize[\sqrt{4+2\sqrt 2}+\sqrt{4-2\sqrt 2}]\end{array}$

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1 Answer

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$\sin 67\large\frac{1}{2}^{\large\circ}+$$\cos 67\large\frac{1}{2}^{\large\circ}$$=\sqrt{1+\sin 135^{\large\circ}}$
$\qquad\qquad\qquad\quad\;\;=\sqrt{1+\large\frac{1}{\sqrt 2}}$
$\qquad\qquad\qquad\quad\;\;=\large\frac{1}{2}$$\sqrt{4+2\sqrt 2}$------(1)
$\sin 67\large\frac{1}{2}^{\large\circ}-$$\cos 67\large\frac{1}{2}^{\large\circ}$$=\sqrt{1-\sin 135^{\large\circ}}$
Because $45^{\large\circ}<67\large\frac{1}{2}^{\large\circ}$$\leq 135^{\large\circ}$
$\qquad\qquad\qquad\quad\;\;=\large\frac{1}{2}$$\sqrt{4-2\sqrt 2}$------(2)
From equ(1) and equ(2)
$\sin 67\large\frac{1}{2}^{\large\circ}=\large\frac{1}{4}\normalsize[\sqrt{4+2\sqrt 2}+\sqrt{4-2\sqrt 2}]$
$\cos 67\large\frac{1}{2}^{\large\circ}=\large\frac{1}{4}\normalsize[\sqrt{4+2\sqrt 2}-\sqrt{4-2\sqrt 2}]$
Hence (a) is the correct answer.
answered Oct 14, 2013 by sreemathi.v
 

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