What is the value of $cos [\frac{π}{3} - \frac{π}{4}]$ $cos[pi/3-pi/4]$ ?

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What is the value of $cos [\frac{π}{3} - \frac{π}{4}]$ $cos[pi/3-pi/4]$ ?

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Answer 1 · 2016-07-29T13:32:39

$\frac{\sqrt{2} + \sqrt{6}}{4}$ $(sqrt2 + sqrt6)/4$

Explanation:

Apply the trig identity:
cos (a - b) = cos a.cos b + sin a.sin b
$cos (\frac{π}{3} - \frac{π}{4}) = cos (\frac{π}{3}) . cos (\frac{π}{4}) + sin (\frac{π}{3}) . sin (\frac{π}{4}) =$ $cos (pi/3 - pi/4) = cos (pi/3).cos (pi/4) + sin (pi/3).sin (pi/4) =$
Trig table of special arcs gives:
$cos (\frac{π}{3}) = \frac{1}{2}$ $cos (pi/3) = 1/2$ , and $cos (\frac{π}{4}) = \frac{\sqrt{2}}{2}$ $cos (pi/4) = sqrt2/2$
$sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$ $sin (pi/3) = sqrt3/2$ , and $sin (\frac{π}{4}) = \frac{\sqrt{2}}{2}$ $sin (pi/4) = sqrt2/2$
There for:
$cos (\frac{π}{3} - \frac{π}{4}) = (\frac{1}{2}) (\frac{\sqrt{2}}{2}) + (\frac{\sqrt{3}}{2}) (\frac{\sqrt{2}}{2}) =$ $cos (pi/3 - pi/4) = (1/2)(sqrt2/2) + (sqrt3/2)(sqrt2/2) =$
$= \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} = \frac{\sqrt{2} + \sqrt{6}}{4}$ $= sqrt2/4 + sqrt6/4 = (sqrt2 + sqrt6)/4$

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What is the value of $cos [\frac{π}{3} - \frac{π}{4}]$ $cos[pi/3-pi/4]$ ?

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Explanation:

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What is the value of $cos [\frac{π}{3} - \frac{π}{4}]$ $cos[pi/3-pi/4]$ ?