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Calculus1 8 Online
OpenStudy (anonymous):

How do I get cos(2x)-1=-2sin^2 x ?

OpenStudy (anonymous):

look at the attachment.

OpenStudy (anonymous):

wow thanks. Why is it that \[\cos 2x=\cos x \cos x-\sin x \sin x\] ?

OpenStudy (anonymous):

it is a general rule. cos(a+b)=cosacosb-sinasinb. if you write cos2x as cos(x+x) you can expand as to that rule.

OpenStudy (anonymous):

I dont know what this rule is called in your terminology. but it is calles as "half angle formula". \[\sin(a+b)=\sin(a) \cos(b)+\sin(b) \cos(a) and for cosinus \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\]

OpenStudy (anonymous):

..and if \[\cos ^{2}x=1-\sin^2x\] then \[\sin ^2x=1+\cos^2x\] right?

OpenStudy (anonymous):

nope \[\sin ^{2}x=1-\cos ^{2}x\]

OpenStudy (anonymous):

ok thanks a bunch

OpenStudy (anonymous):

your welcome :)

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