anonymous
  • anonymous
5(cos^2)60 + 4(sec^2)30 - (tan^2)45 = x Find x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
freckles
  • freckles
well cos(60) and sec(30) and tan(45) can all be evaluated using the unit circle
freckles
  • freckles
http://buildingaunitcircle.weebly.com/uploads/6/2/2/3/622393/9409299.gif
anonymous
  • anonymous
Use the trig table........

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freckles
  • freckles
do you need help using that chart above ?
anonymous
  • anonymous
.... I need to the values like cos = 1/2 and all.......
freckles
  • freckles
cos(60)=1/2 that is correct so squaring both sides gives cos^2(60)=1/4
freckles
  • freckles
sec(30) means you need to find cos(30) and flip whatever number that is
freckles
  • freckles
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
freckles
  • freckles
\[5 \cos^2(60)+4\sec^2(30)-\tan^2(45) \\ 5(\frac{1}{4})+4(\frac{1}{\cos(30)})^2-1\]
freckles
  • freckles
so you still need to find cos(30)
anonymous
  • anonymous
4/3 = sec^2(30)
freckles
  • freckles
\[\cos(30)=\frac{\sqrt{3}}{2} \\ \cos^2(30)=\frac{3}{4} \\ \frac{1}{\cos^2(30)}=\frac{4}{3} \\ \sec^2(30)=\frac{4}{3}\] that is right so far you have this now: \[5(\frac{1}{4})+4(\frac{4}{3})-1\]
freckles
  • freckles
that can be simplify just be doing a few multiplications and additions(subtractions)
anonymous
  • anonymous
\[\frac{ 5 }{ 4 } +\frac{ 16 }{ 3 } - 1\]
freckles
  • freckles
right so far
anonymous
  • anonymous
shall I take LCM?
freckles
  • freckles
yes which the lcm(4,3,1)=12
freckles
  • freckles
\[\frac{5}{4}+\frac{16}{3}-\frac{1}{1} \\ \frac{5(3)}{4(3)}+\frac{16(4)}{3(4)}-\frac{1(12)}{1(12)} \\ \frac{5(3)}{12}+\frac{16(4)}{12}-\frac{12}{12} \\ \frac{5(3)+16(4)-12}{12}\]
anonymous
  • anonymous
67/12
freckles
  • freckles
looks awesome!
freckles
  • freckles
and totally correct :)
anonymous
  • anonymous
Thanks! Please help me a bit more?
freckles
  • freckles
Sure. I can take a stab at it.
triciaal
  • triciaal
@freckles would using identities just take longer?
freckles
  • freckles
@triciaal if you want to try something with identities you can
anonymous
  • anonymous
\[\frac{ cotA-1 }{ 2 -\sec^2(A) } = \frac{ \cot A }{ 1 + tanA }\]
freckles
  • freckles
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
anonymous
  • anonymous
Help me?
freckles
  • freckles
\[\frac{\frac{\cos(a)}{\sin(a)}-1}{2-\frac{1}{\cos^2(a)}} =\frac{\frac{\cos(a)}{\sin(a)}}{1+\frac{\sin(a)}{\cos(a)}}\] so this is what I would do as a first step second step would be to remove the compound fraction action going on
anonymous
  • anonymous
I'm suppose to use LHS and get RHS
freckles
  • freckles
ok then get rid of the compound fraction action on the the left hand side
anonymous
  • anonymous
\[\frac{ \cos(a) - \sin(a) }{ \sin(a) }\]
anonymous
  • anonymous
and for the denominator 2sin^(a)/cos^2(a)
freckles
  • freckles
\[\frac{\frac{\cos(a)}{\sin(a)}-1}{2-\frac{1}{\cos^2(a)}} \cdot \frac{\sin(a) \cos^2(a)}{\sin(a)\cos^2(a)} \\ \frac{\cos^3(a)-\cos^2(a)\sin(a)}{2 \cos^2(a)\sin(a)-\sin(a)} \\ \text{ so doing a bit of factoring } \\ \frac{\cos^2(a)(\cos(a)-\sin(a))}{\sin(a)(2 \cos^2(a)-1)} \\ \\ \text{ now recall on the other side we wanted } \cot(a) \\ \text{ on \top } \\ \text{ well we notice } \frac{\cos(a)}{\sin(a)}=\cot(a) \\ \cot(a) \frac{\cos(a)(\cos(a)-\sin(a))}{2 \cos^2(a)-1} \\ \] 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
freckles
  • freckles
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)
freckles
  • freckles
that is I used \[\cos(2a)=\cos^2(a)-\sin^2(a) \\ \text{ and also } \cos(2a)=2\cos^2(a)-1\]
freckles
  • freckles
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)) )
freckles
  • freckles
so the key in our proof was the use of double angle identity for cosine
anonymous
  • anonymous
Nice.. Thanks... I need help in quadratic equations too...
freckles
  • freckles
ok let's see it
freckles
  • freckles
hey @triciaal did you come up with a way using identities for the first one ?
freckles
  • freckles
I honestly haven't given it a try
anonymous
  • anonymous
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
freckles
  • freckles
wait what does that mean what is the angle of elevation 60 or 30?
freckles
  • freckles
and what is 30?
anonymous
  • anonymous
from 60 to 30 in secs from the smae point
freckles
  • freckles
oh the angle of elevation changed from 60 to 30 in 10 sec gotcha
freckles
  • freckles
|dw:1436680212196:dw|
freckles
  • freckles
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...
freckles
  • freckles
|dw:1436680529581:dw|
freckles
  • freckles
wait my drawing is off
freckles
  • freckles
I will just make two separate drawings |dw:1436680577539:dw| |dw:1436680738212:dw|
freckles
  • freckles
We need to find the distance the plane has flown for both pictures
freckles
  • freckles
notice that is x for the first drawing and B for the second drawing
freckles
  • freckles
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
anonymous
  • anonymous
How to find the dist?
freckles
  • freckles
you can use a trig ratio
freckles
  • freckles
for example you are given the height which is 1km (the opposite side of the angle of elevation)
freckles
  • freckles
and you are wanting to find the length that airplane has flown for both (which is the adjacent side )
freckles
  • freckles
opp/adj is the tangent ratio
freckles
  • freckles
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) \[\text{ the speed over that 10 sec change }=\frac{\text{ distance traveled in that 10 sec }}{ 10 \sec} \\ \frac{|x-B|}{10 \sec}\]
anonymous
  • anonymous
Thanks. I will manage the rest...
freckles
  • freckles
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
freckles
  • freckles
oh your question says to put it an hours
freckles
  • freckles
lol
freckles
  • freckles
just let me know if you need any further help with this one
freckles
  • freckles
or if you want me to check your answer

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