Carter is going to invest $79000 and leave it in an account for 12 years. Assuming the interest is compounded continuously, what
1 answer:
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That'd be true only if the value of "s" is the exact same one for both
namely if sec(s) = cos(s)
then solving for "s"
thus
![\bf sec(s)=cos(s)\qquad but\implies sec(\theta)=\cfrac{1}{cos(\theta)} \\\\\\ thus\cfrac{1}{cos(s)}=cos(s)\implies 1=cos^2(s)\implies \pm \sqrt{1}=cos(s) \\\\\\ \pm 1=cos(s)\impliedby \textit{now taking }cos^{-1}\textit{ to both sides} \\\\\\ cos^{-1}(\pm 1)=cos^{-1}[cos(s)]\implies cos^{-1}(\pm 1)=\measuredangle s](https://tex.z-dn.net/?f=%5Cbf%20sec%28s%29%3Dcos%28s%29%5Cqquad%20but%5Cimplies%20sec%28%5Ctheta%29%3D%5Ccfrac%7B1%7D%7Bcos%28%5Ctheta%29%7D%0A%5C%5C%5C%5C%5C%5C%0Athus%5Ccfrac%7B1%7D%7Bcos%28s%29%7D%3Dcos%28s%29%5Cimplies%201%3Dcos%5E2%28s%29%5Cimplies%20%5Cpm%20%5Csqrt%7B1%7D%3Dcos%28s%29%0A%5C%5C%5C%5C%5C%5C%0A%5Cpm%201%3Dcos%28s%29%5Cimpliedby%20%5Ctextit%7Bnow%20taking%20%7Dcos%5E%7B-1%7D%5Ctextit%7B%20to%20both%20sides%7D%0A%5C%5C%5C%5C%5C%5C%0Acos%5E%7B-1%7D%28%5Cpm%201%29%3Dcos%5E%7B-1%7D%5Bcos%28s%29%5D%5Cimplies%20cos%5E%7B-1%7D%28%5Cpm%201%29%3D%5Cmeasuredangle%20s)
namely if sec(s) = cos(s)
then solving for "s"
thus
Answer:
Step-by-step explanation:
Using the addition formulae for cosine
cos(x ± y) = cosxcosy ∓ sinxsiny
---------------------------------------------------------------
cos(120 + x) = cos120cosx - sin120sinx
= - cos60cosx - sin60sinx
= - cosx -
sinx
squaring to obtain cos² (120 + x)
= cos²x +
sinxcosx +
sin²x
--------------------------------------------------------------------
cos(120 - x) = cos120cosx + sin120sinx
= -cos60cosx + sin60sinx
= - cosx +
sinx
squaring to obtain cos²(120 - x)
= cos²x -
sinxcosx +
sin²x
--------------------------------------------------------------------------
Putting it all together
cos²x + cos²x +
sinxcosx +
sin²x +
cos²x -
sinxcosx +
sin²x
= cos²x + cos²x +
sin²x
= cos²x +
sin²x
= (cos²x + sin²x) =