How do you express $cos (\frac{5 π}{6}) \cdot cos (\frac{π}{6})$ $cos( (5 pi)/6 ) * cos (( pi) /6 )$ without using products of trigonometric functions?

Question

How do you express $cos (\frac{5 π}{6}) \cdot cos (\frac{π}{6})$ $cos( (5 pi)/6 ) * cos (( pi) /6 )$ without using products of trigonometric functions?

1 Answer

Impact of this question

1854 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-12T23:41:27

Using Sum:
$cos (\frac{5 π}{6}) \cdot cos (\frac{π}{6}) = \frac{1}{2} [cos (\frac{5 π}{6} + \frac{π}{6}) + cos (\frac{5 π}{6} - \frac{π}{6})]$ $cos((5pi)/6)*cos(pi/6)=1/2[cos((5pi)/6+pi/6)+cos((5pi)/6-pi/6)]$
$cos (\frac{5 π}{6}) \cdot cos (\frac{π}{6}) = \frac{1}{2} [cos π + cos (\frac{2 π}{3})] = \frac{1}{2} (- 1 - \frac{1}{2}) = - \frac{3}{4}$ $cos((5pi)/6)*cos(pi/6)=1/2[cos pi+cos ((2pi)/3)]=1/2(-1-1/2)=-3/4$

Explanation:

$cos (x + y) = cos x cos y - sin x sin y$ $cos(x+y)=cos x cos y - sin x sin y" " " " " "$ 1st equation
$cos (x - y) = cos x cos y + sin x sin y$ $cos(x-y)=cos x cos y+sin x sin y" " " " " "$ 2nd equation

Add first and second equations

$cos (x + y) + cos (x - y) = 2 \cdot cos x cos y + 0$ $cos(x+y)+cos(x-y)=2*cos x cos y+0$
$cos (x + y) + cos (x - y) = 2 \cdot cos x cos y$ $cos(x+y)+cos(x-y)=2*cos x cos y$

and then

$cos x cos y = \frac{1}{2} [cos (x + y) + cos (x - y)]$ $cos x cos y=1/2[cos(x+y)+cos(x-y)]$

Let $x = \frac{5 π}{6}$ $x=(5pi)/6$ and $y = \frac{π}{6}$ $y=pi/6$

$cos (\frac{5 π}{6}) \cdot cos (\frac{π}{6}) = \frac{1}{2} [cos (\frac{5 π}{6} + \frac{π}{6}) + cos (\frac{5 π}{6} - \frac{π}{6})]$ $cos((5pi)/6)*cos(pi/6)=1/2[cos((5pi)/6+pi/6)+cos((5pi)/6-pi/6)]$
$cos (\frac{5 π}{6}) \cdot cos (\frac{π}{6}) = \frac{1}{2} [cos π + cos (\frac{2 π}{3})] = \frac{1}{2} (- 1 - \frac{1}{2}) = - \frac{3}{4}$ $cos((5pi)/6)*cos(pi/6)=1/2[cos pi+cos ((2pi)/3)]=1/2(-1-1/2)=-3/4$

God bless America ....

Science

Math

Humanities

... and beyond

How do you express $cos (\frac{5 π}{6}) \cdot cos (\frac{π}{6})$ $cos( (5 pi)/6 ) * cos (( pi) /6 )$ without using products of trigonometric functions?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you express cos( (5 pi)/6 ) * cos (( pi) /6 ) cos(5π6)⋅cos(π6)cos( (5 pi)/6 ) * cos (( pi) /6 ) without using products of trigonometric functions?

1 Answer

Explanation:

Related questions

Impact of this question

How do you express $cos (\frac{5 π}{6}) \cdot cos (\frac{π}{6})$ $cos( (5 pi)/6 ) * cos (( pi) /6 )$ without using products of trigonometric functions?