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sin(2*pi) - not pie - is the same as sin(0) = 0

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12y ago
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Q: What is the sin of 2 pie?
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Related questions

What is the sin of pie over 2?

1


What is sin pie divided by six equals one half?

Sin(pi/6) = 1/2 is a true statement [not pie].


What is the sin of negative pie over 2?

-1


What is sine of 3 pie divided by 4?

0.75


What is the sin of pie over 9?

1/3


What does sin pie over three equal?

one half


How do you prove this trigonometric relationship sin3A equals 3sinA cos 2 A - sin 3 A?

sin(3A) = sin(2A + A) = sin(2A)*cos(A) + cos(2A)*sin(A)= sin(A+A)*cos(A) + cos(A+A)*sin(A) = 2*sin(A)*cos(A)*cos(A) + {cos^2(A) - sin^2(A)}*sin(A) = 2*sin(A)*cos^2(A) + sin(a)*cos^2(A) - sin^3(A) = 3*sin(A)*cos^2(A) - sin^3(A)


What is the sin of pi divided by 2?

Do you mean Sin(pi/2) = 1 or [Sin(pi)] /2 = 0.0274....


Verify that sin minus cos plus 1 divided by sin plus cos subtract 1 equals sin plus 1 divided by cos?

[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,


How do you show that 2 sin squared x minus 1 divided by sin x minus cos x equals sin x plus cos x?

(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x


Is 2 plus 2 equals Pie?

2 plus 2 does not equal pie


How do you simplify this expression... cot2x over csc2x-cscx... please note that cot2x and csc2x are really just raised to the second power?

I assume the expression is cot^2 x / ( csc^2 x - csc x) express it in terms of sin x and cos x: =(cos^2 x / sin^2 x) / (1/sin^2 x - 1/sin x) =(cos^2 x / sin^2 x) / [(1 - sin x)/sin^2 x] =cos^2 x / (1 - sin x) = (1 - sin^2 x) / (1 - sin x) = (1 + sin x)(1 - sin x) / (1 - sin x) = 1 + sin x