How do you use the sum and difference formula to simplify $cos (\frac{π}{12}) cos (\frac{5 π}{12}) + cos (\frac{π}{12}) sin (\frac{5 π}{12})$ $cos(pi/12)cos((5pi)/12)+cos(pi/12)sin((5pi)/12)$ ?

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How do you use the sum and difference formula to simplify $cos (\frac{π}{12}) cos (\frac{5 π}{12}) + cos (\frac{π}{12}) sin (\frac{5 π}{12})$ $cos(pi/12)cos((5pi)/12)+cos(pi/12)sin((5pi)/12)$ ?

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Answer 1 · 2016-10-20T08:43:06

$cos (\frac{π}{12}) cos (\frac{5 π}{12}) + cos (\frac{π}{12}) sin (\frac{5 π}{12}) = \frac{3 + \sqrt{3}}{4}$ $cos(pi/12)cos((5pi)/12)+cos(pi/12)sin((5pi)/12)=(3+sqrt3)/4$

Explanation:

$cos (\frac{π}{12}) cos (\frac{5 π}{12}) + cos (\frac{π}{12}) sin (\frac{5 π}{12})$ $cos(pi/12)cos((5pi)/12)+cos(pi/12)sin((5pi)/12)$

= $cos (\frac{π}{12}) [cos (\frac{5 π}{12}) + sin (\frac{5 π}{12})]$ $cos(pi/12)[cos((5pi)/12)+sin((5pi)/12)]$

Now as $sin (\frac{π}{4}) = cos (\frac{π}{4}) = \frac{1}{\sqrt{2}}$ $sin(pi/4)=cos(pi/4)=1/sqrt2$ above can be written as

$\sqrt{2} cos (\frac{π}{12}) [sin (\frac{π}{4}) cos (\frac{5 π}{12}) + cos (\frac{π}{4}) sin (\frac{5 π}{12})]$ $sqrt2cos(pi/12)[sin(pi/4)cos((5pi)/12)+cos(pi/4)sin((5pi)/12)]$

using $sin (A + B) = sin A cos B + cos A sin B$ $sin(A+B)=sinAcosB+cosAsinB$ , this becomes

$\sqrt{2} cos (\frac{π}{12}) [sin {(\frac{π}{4}) + (\frac{5 π}{12})}$ $sqrt2cos(pi/12)[sin{(pi/4)+((5pi)/12)}$

= $\sqrt{2} cos (\frac{π}{12}) sin (\frac{8 π}{12}) = \sqrt{2} sin (\frac{8 π}{12}) cos (\frac{π}{12})$ $sqrt2cos(pi/12)sin((8pi)/12)=sqrt2sin((8pi)/12)cos(pi/12)$

Now using $sin A cos B = \frac{1}{2} {sin (A - B) + sin (A + B)}$ $sinAcosB=1/2{sin(A-B)+sin(A+B)}$

$\sqrt{2} sin (\frac{8 π}{12}) cos (\frac{π}{12}) = \frac{\sqrt{2}}{2} (sin (\frac{8 π}{12} - \frac{π}{12}) + sin (\frac{8 π}{12} + \frac{π}{12}))$ $sqrt2sin((8pi)/12)cos(pi/12)=sqrt2/2(sin((8pi)/12-pi/12)+sin((8pi)/12+pi/12))$

= $\frac{1}{\sqrt{2}} (sin (\frac{7 π}{12}) + sin (\frac{9 π}{12}))$ $1/sqrt2(sin((7pi)/12)+sin((9pi)/12))$

= $\frac{1}{\sqrt{2}} (sin (\frac{3 π}{12} + \frac{4 π}{12}) + sin (\frac{3 π}{4}))$ $1/sqrt2(sin((3pi)/12+(4pi)/12)+sin((3pi)/4))$

= $\frac{1}{\sqrt{2}} (sin (\frac{3 π}{12} + \frac{4 π}{12}) + \frac{1}{\sqrt{2}})$ $1/sqrt2(sin((3pi)/12+(4pi)/12)+1/sqrt2)$

= $\frac{1}{\sqrt{2}} (sin (\frac{3 π}{12}) cos (\frac{4 π}{12}) + cos (\frac{3 π}{12}) sin (\frac{4 π}{12})) + \frac{1}{2}$ $1/sqrt2(sin((3pi)/12)cos((4pi)/12)+cos((3pi)/12)sin((4pi)/12))+1/2$

= $\frac{1}{\sqrt{2}} (sin (\frac{π}{4}) cos (\frac{π}{3}) + cos (\frac{π}{4}) sin (\frac{π}{3})) + \frac{1}{2}$ $1/sqrt2(sin(pi/4)cos(pi/3)+cos(pi/4)sin(pi/3))+1/2$

= $\frac{1}{\sqrt{2}} (\frac{1}{\sqrt{2}} \times \frac{1}{2} + \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}) + \frac{1}{2}$ $1/sqrt2(1/sqrt2xx1/2+1/sqrt2xxsqrt3/2)+1/2$

= $\frac{\sqrt{3} + 1}{4} + \frac{1}{2}$ $(sqrt3+1)/4+1/2$

= $\frac{3 + \sqrt{3}}{4}$ $(3+sqrt3)/4$

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How do you use the sum and difference formula to simplify $cos (\frac{π}{12}) cos (\frac{5 π}{12}) + cos (\frac{π}{12}) sin (\frac{5 π}{12})$ $cos(pi/12)cos((5pi)/12)+cos(pi/12)sin((5pi)/12)$ ?

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How do you use the sum and difference formula to simplify $cos (\frac{π}{12}) cos (\frac{5 π}{12}) + cos (\frac{π}{12}) sin (\frac{5 π}{12})$ $cos(pi/12)cos((5pi)/12)+cos(pi/12)sin((5pi)/12)$ ?