How do you express $cos (\frac{3 π}{2}) \cdot cos (\frac{5 π}{4})$ $cos( (3 pi)/ 2 ) * cos (( 5 pi) /4 )$ without using products of trigonometric functions?

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How do you express $cos (\frac{3 π}{2}) \cdot cos (\frac{5 π}{4})$ $cos( (3 pi)/ 2 ) * cos (( 5 pi) /4 )$ without using products of trigonometric functions?

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Answer 1 · 2016-02-26T10:26:48

0

Explanation:

From knowledge of the graph of cosx , we know that

$cos (\frac{3 π}{2}) = 0$ $cos((3pi)/2) = 0$

and $cos (\frac{5 π}{4}) = - cos (\frac{π}{4}) = - \frac{1}{\sqrt{2}}$ $cos((5pi)/4) = -cos(pi/4) = -1/sqrt2$

$\Rightarrow cos (\frac{3 π}{2}) . cos (\frac{5 π}{4}) = 0 . (- \frac{1}{\sqrt{2}}) = 0$ $rArr cos((3pi)/2) . cos((5pi)/4) = 0.(-1/sqrt2) = 0$

Answer 2 · 2016-02-26T10:40:49

It is equivalent to $0$ $0$ .

Explanation:

To express $cos (3 \frac{π}{2}) \cdot cos (5 \frac{π}{4})$ $cos(3pi/2)*cos(5pi/4)$ , without using trigonometric functions, we should first find value of $cos (3 \frac{π}{2})$ $cos(3pi/2)$ and $cos (5 \frac{π}{4})$ $cos(5pi/4)$ separately.

$cos (3 \frac{π}{2})$ $cos(3pi/2)$ is equal $cos (\frac{3 π}{2} - 2 π)$ $cos((3pi)/2-2pi)$ or $cos (- \frac{π}{2})$ $cos(-pi/2)$ , which is equal to $cos (\frac{π}{2})$ $cos(pi/2)$ . But as latter is equal to zero,

$cos (\frac{3 π}{2}) = 0$ $cos((3pi)/2)=0$

Although, $cos (\frac{5 π}{4}) = cos (2 π - \frac{5 π}{4}) = cos (\frac{3 π}{4}) = - cos (\frac{π}{4}) = (- \frac{1}{\sqrt{2}})$ $cos((5pi)/4)=cos(2pi-(5pi)/4)=cos((3pi)/4)=-cos(pi/4)=(-1/sqrt2)$

$cos (3 \frac{π}{2}) \cdot cos (5 \frac{π}{4})$ $cos(3pi/2)*cos(5pi/4)$ will still be $0$ $0$ , as $cos (\frac{3 π}{2}) = 0$ $cos((3pi)/2)=0$

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How do you express $cos (\frac{3 π}{2}) \cdot cos (\frac{5 π}{4})$ $cos( (3 pi)/ 2 ) * cos (( 5 pi) /4 )$ without using products of trigonometric functions?

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Explanation:

Explanation:

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How do you express $cos (\frac{3 π}{2}) \cdot cos (\frac{5 π}{4})$ $cos( (3 pi)/ 2 ) * cos (( 5 pi) /4 )$ without using products of trigonometric functions?