May Lord forgive us our sins.Wait, what about θ \theta ?

Level pending

A) sine is algebraically less than the cosine for any angle between 2 n π − 3 π 4 2n\pi -\frac { 3\pi }{ 4 } and 2 n π + π 4 2n\pi +\frac { \pi }{ 4 } ,where n is any integer.

Insufficient data Statement A is false Statement A is true. I don't like trigonometry

Soumo Mukherjee by Soumo Mukherjee , 25, India

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1 solution

Soumo Mukherjee
Oct 15, 2014

When A lies between − 3 π 4 -\frac { 3\pi }{ 4 } and − π 4 -\frac { \pi }{ 4 }

2 sin ⁡ A = − 1 + sin ⁡ 2 A − 1 − sin ⁡ 2 A 2 cos ⁡ A = − 1 + sin ⁡ 2 A + 1 − sin ⁡ 2 A ∴ sin ⁡ A < cos ⁡ A 2\sin { A } =\quad -\sqrt { 1+\sin { 2A } } -\sqrt { 1-\sin { 2A } } \\ 2\cos { A } =-\sqrt { 1+\sin { 2A } } +\sqrt { 1-\sin { 2A } } \\ \therefore \quad \sin { A } <\cos { A }

When A lies between − π 4 -\frac { \pi }{ 4 } and π 4 \frac { \pi }{ 4 }

2 sin ⁡ A = + 1 + sin ⁡ 2 A − 1 − sin ⁡ 2 A 2 cos ⁡ A = + 1 + sin ⁡ 2 A + 1 − sin ⁡ 2 A ∴ sin ⁡ A < cos ⁡ A 2\sin { A } =\quad +\sqrt { 1+\sin { 2A } } -\sqrt { 1-\sin { 2A } } \\ 2\cos { A } =+\sqrt { 1+\sin { 2A } } +\sqrt { 1-\sin { 2A } } \\ \therefore \quad \sin { A } <\cos { A }

Hence the statement is true.

Please Also Visit May Lord ... [Backstage]

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What does this have to do with sins and Lord??

John M. - 6 years, 8 months ago

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Probably nothing.Just playing with sins[ bad doings] and sins[trigo].Well I m not gud at presentations and naming things .. :p . hey @John Muradeli how to upload pics ?

Soumo Mukherjee - 6 years, 7 months ago

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! [anything, this is just for your checkmarking] (URL link of the image)

if you have a file on your computer you'll have to first upload it to the internet somehow and then get its URL link.

I do it via my google site, which I made on sites.google.com. It's real quick.

~Cheers

John M. - 6 years, 7 months ago

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@John M. – @John Muradeli Check my comments . Here

Soumo Mukherjee - 6 years, 7 months ago

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@Soumo Mukherjee – Yeah sorry for not replying. I gave you a vote up instead, I couldn't really say anything at the time.

Thanks again.

John M. - 6 years, 7 months ago

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