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cos1 * cos2 * cos3 * ........* cos149 * cos150 =?

Note by Eraz Ahmed 5 years, 8 months ago

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Assuming that the angles are in degrees, the result is zero since \(\cos (90)=0\).

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Here's a similar problem that I believe most of you have seen.

Simplify \( (x-a)(x-b)(x-c) \cdots (x-z) \).

0

if the angles are in degrees then answer is obviously 0. but i don't think they meant that. i think angles are in radians , then the answer would be approx: -1.14893 * 10^-46

i have used degree values here don't use radians....

haroun M 's answer is correct,think easily because it's so much easy

Yes but one hour later than I posted...

approx -1.14893 * 10^-46

how???

maybe he take it in radian.

hello, the result is very ugly \(-1.14893012274526133709121721740219091292398251438476437266\times10^{-46}\)

Sonnhard.

how?please explain

it is 0

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## Comments

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TopNewestAssuming that the angles are in degrees, the result is zero since \(\cos (90)=0\).

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Here's a similar problem that I believe most of you have seen.

Simplify \( (x-a)(x-b)(x-c) \cdots (x-z) \).

Log in to reply

0

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if the angles are in degrees then answer is obviously 0. but i don't think they meant that. i think angles are in radians , then the answer would be approx: -1.14893 * 10^-46

Log in to reply

i have used degree values here don't use radians....

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haroun M 's answer is correct,think easily because it's so much easy

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Yes but one hour later than I posted...

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0

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0

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approx -1.14893 * 10^-46

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how???

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maybe he take it in radian.

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0

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hello, the result is very ugly \(-1.14893012274526133709121721740219091292398251438476437266\times10^{-46}\)

Sonnhard.

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how?please explain

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it is 0

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