5(cos^2)60 + 4(sec^2)30 - (tan^2)45 = x Find x

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5(cos^2)60 + 4(sec^2)30 - (tan^2)45 = x Find x

Mathematics
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well cos(60) and sec(30) and tan(45) can all be evaluated using the unit circle
http://buildingaunitcircle.weebly.com/uploads/6/2/2/3/622393/9409299.gif
Use the trig table........

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do you need help using that chart above ?
.... I need to the values like cos = 1/2 and all.......
cos(60)=1/2 that is correct so squaring both sides gives cos^2(60)=1/4
sec(30) means you need to find cos(30) and flip whatever number that is
tan(45) means you need to find sin(45) and cos(45) but both sin(45) and cos(45) are the same so tan(45)=1
\[5 \cos^2(60)+4\sec^2(30)-\tan^2(45) \\ 5(\frac{1}{4})+4(\frac{1}{\cos(30)})^2-1\]
so you still need to find cos(30)
4/3 = sec^2(30)
\[\cos(30)=\frac{\sqrt{3}}{2} \\ \cos^2(30)=\frac{3}{4} \\ \frac{1}{\cos^2(30)}=\frac{4}{3} \\ \sec^2(30)=\frac{4}{3}\] that is right so far you have this now: \[5(\frac{1}{4})+4(\frac{4}{3})-1\]
that can be simplify just be doing a few multiplications and additions(subtractions)
\[\frac{ 5 }{ 4 } +\frac{ 16 }{ 3 } - 1\]
right so far
shall I take LCM?
yes which the lcm(4,3,1)=12
\[\frac{5}{4}+\frac{16}{3}-\frac{1}{1} \\ \frac{5(3)}{4(3)}+\frac{16(4)}{3(4)}-\frac{1(12)}{1(12)} \\ \frac{5(3)}{12}+\frac{16(4)}{12}-\frac{12}{12} \\ \frac{5(3)+16(4)-12}{12}\]
67/12
looks awesome!
and totally correct :)
Thanks! Please help me a bit more?
Sure. I can take a stab at it.
@freckles would using identities just take longer?
@triciaal if you want to try something with identities you can
\[\frac{ cotA-1 }{ 2 -\sec^2(A) } = \frac{ \cot A }{ 1 + tanA }\]
some people don't need to put in terms of sin and cos but I always do because I remember more identities with sin and cos
Help me?
\[\frac{\frac{\cos(a)}{\sin(a)}-1}{2-\frac{1}{\cos^2(a)}} =\frac{\frac{\cos(a)}{\sin(a)}}{1+\frac{\sin(a)}{\cos(a)}}\] so this is what I would do as a first step second step would be to remove the compound fraction action going on
I'm suppose to use LHS and get RHS
ok then get rid of the compound fraction action on the the left hand side
\[\frac{ \cos(a) - \sin(a) }{ \sin(a) }\]
and for the denominator 2sin^(a)/cos^2(a)
\[\frac{\frac{\cos(a)}{\sin(a)}-1}{2-\frac{1}{\cos^2(a)}} \cdot \frac{\sin(a) \cos^2(a)}{\sin(a)\cos^2(a)} \\ \frac{\cos^3(a)-\cos^2(a)\sin(a)}{2 \cos^2(a)\sin(a)-\sin(a)} \\ \text{ so doing a bit of factoring } \\ \frac{\cos^2(a)(\cos(a)-\sin(a))}{\sin(a)(2 \cos^2(a)-1)} \\ \\ \text{ now recall on the other side we wanted } \cot(a) \\ \text{ on \top } \\ \text{ well we notice } \frac{\cos(a)}{\sin(a)}=\cot(a) \\ \cot(a) \frac{\cos(a)(\cos(a)-\sin(a))}{2 \cos^2(a)-1} \\ \] now somehow we have to show that \[\frac{\cos(a)(\cos(a)-\sin(a))}{2 \cos^2(a)-1}=\frac{1}{1+\tan(a)}\] any ideas @BlackDranzer
if not read this if you want spoilers: \[\frac{\cos(a)(\cos(a)-\sin(a))}{2 \cos^2(a)-1} \cdot \frac{\cos(a)+\sin(a)}{\cos(a)+\sin(a)} \\ \frac{\cos(a)(\cos^2(a)-\sin^2(a))}{(2 \cos^2(a)-1)(\cos(a)+\sin(a))} \\ \text{ well recall double angle identity for } cosine \\ \frac{ \cos(a)}{\cos(a)+\sin(a)}\] now divide both top and bottom by cos(a)
that is I used \[\cos(2a)=\cos^2(a)-\sin^2(a) \\ \text{ and also } \cos(2a)=2\cos^2(a)-1\]
yep and we can bring down that cot(a) factor we had from earlier (though I normally write my fractions like this 1/(1+tan(a)) )
so the key in our proof was the use of double angle identity for cosine
Nice.. Thanks... I need help in quadratic equations too...
ok let's see it
hey @triciaal did you come up with a way using identities for the first one ?
I honestly haven't given it a try
An aero plane flying at 1 km horizontally above the ground make an angle of elevation 60 after 10 seconds its 30 find the speed of the aerplane in km/hr
wait what does that mean what is the angle of elevation 60 or 30?
and what is 30?
from 60 to 30 in secs from the smae point
oh the angle of elevation changed from 60 to 30 in 10 sec gotcha
|dw:1436680212196:dw|
so to find the speed of the airplane over that 10 sec we will need to find the distance it traveled for both angle of elevations...
|dw:1436680529581:dw|
wait my drawing is off
I will just make two separate drawings |dw:1436680577539:dw| |dw:1436680738212:dw|
We need to find the distance the plane has flown for both pictures
notice that is x for the first drawing and B for the second drawing
then once you find the distances for both pictures you will find the difference of those distances and that resulting distance/10 secs will be the speed of the airplane in that 10 secs you have that change
How to find the dist?
you can use a trig ratio
for example you are given the height which is 1km (the opposite side of the angle of elevation)
and you are wanting to find the length that airplane has flown for both (which is the adjacent side )
opp/adj is the tangent ratio
once you computer x and B (don't ask why I called them that; I wasn't thinking too much about the letters to use) \[\text{ the speed over that 10 sec change }=\frac{\text{ distance traveled in that 10 sec }}{ 10 \sec} \\ \frac{|x-B|}{10 \sec}\]
Thanks. I will manage the rest...
when you are done with that computation your units should by km/sec but if you don't like how small the number is next to those units you can always covert the sec part to either minutes or hours
oh your question says to put it an hours
lol
just let me know if you need any further help with this one
or if you want me to check your answer

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