*Trigonometric Identities*
cos2 y − sin2 y = 1 − 2sin2 y
its cos^2y - sin^2y= 1 - 2sin^2y
try using the identity sin^2 y + cos^2 y = 1
so change cos^2?
sin^y = 1 - cos^2y
isnt it better to change cos so ill have all sin?
replace cos^2 y by 1 - sin^2 y on the left side
so its (1 - sin^2y) - sin^2y = 1- 2sin^2y
yes
1- 2sin^2y = 1-2siny
easy one this
huh?
- your last post was incorrect the previous one completed the proof
but shouldnt both sides be the same thing?o.O
No you have taken the left side and by using known trig identity converted it to the right side
- therefore the identity is true
1 - sin^2y) - sin^2y = 1- 2sin^2y
from here I expanded the brackets and got 1 - sin^2y - sin^2y = 1- 2sin^2y
- sin^2y - sin^2y = - 2 sin^2y
right??
cos^2y - sin^2y = 1 - 2sin^2y now cos^2 y = 1 - sin^2 y so substituting 1 - sin^2 y - sin^2 y = 1 - 2 sin^2 y = RHS
yes
RHS = right hand side
so then wouldnt it be 1 - 2sin^2y = 1 - 2sin^2y??
no - no need to write that usually with these , you pick the more complicated side of the identity and try to convert it to the other side . Once that is done you have confirmed the identity.
so wat will the answer be then? o.O
just write what i have put or if you like at the end write 'the above proves the original identity'.
- thats what this is - a Proof.
so this is the answer? 1 - sin^2 y - sin^2 y = 1 - 2 sin^2 y = RHS
No there are 2 lines before that - check 6 posts back.
the answer is the whole process of proving the identity
cos^2y - sin^2y = 1 - 2sin^2y now cos^2 y = 1 - sin^2 y so substituting 1 - sin^2 y - sin^2 y = 1 - 2 sin^2 y = RHS The above proves the original identity
you could also make a note to clarify that RHS = right hand side
gotta go - hope this has clarified things
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