Please help! Reduce (sin65degrees)/(1+cos65degrees) to a single function of one angle. a. tan32degrees 30' b. tan130degrees c. cot32degrees 30' d. cot130degrees
We have \[\frac{\sin 65}{1+\cos 65}\] @IloveCharlie what's cos 0?
@IloveCharlie are you here?
Oh yea, just a second
np
1?
Good so, I'll rewrite it as \[\frac{\sin 65}{\cos 0+\cos 65}\] Do you know what's \(\large \cos A+\cos B\)?
@IloveCharlie ???
No, I don't :(
NP \[\large \cos A+\cos B=2\cos (\frac A2-\frac B2)\cos (\frac A 2+\frac B2)\] so we get now \[\large \frac{\sin 65}{2\cos (\frac 02-\frac {65}2)\cos (\frac 0 2+\frac {65}2)}\] Do you get this?
Oh, yea, makes sense
good...would you tell me what's \(\sin 2\theta\) in terms of \(\sin \theta\ and\ \cos \theta\)?
@IloveCharlie ??
Uhhhh, ? No :(
it's given as \[\sin 2\theta=2 \sin \theta \times \cos \theta\] using this formula. we'll get \[\large \frac{2\sin (\frac{65}{2})\times\cos(\frac{65}{2})}{2\cos (-\frac {65}2)\cos (\frac {65}2)}\] what would we get now?
@IloveCharlie ??
K sorry, this keyboard isn't working. Just a second
cool:)
.9063/1.422 ? I'm thinking I did that wrong :/
don't compute the values, just simplify
sin(65)/1+cos(65) ?
nope, check again. I think you have written the question again:)
:O I'm not sure :/
\[\large\large \frac{\cancel 2\sin (\frac{65}{2})\times\cancel{\cos(\frac{65}{2})}}{\cancel 2\cos (-\frac {65}2)\cancel{\cos (\frac {65}2)}}\] what would you get now?
sin(65/2)/cos(-65/2)
@IloveCharlie what's \[\large \cos (-x)\] ???
cos(x)?
good so we'll get now \[\large \frac{\sin \frac{65}{2}}{\cos \frac{65}{2}}\] now what would you get?
tan65/2?
yeah so it's \[\tan 32.5 degrees\] convert the 0.5 degrees into minutes
So a.! THANK YOU SO MUCH ASH!
you're welcome, did you understand?
Yes, thanks for explaining each step
np:D
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