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

- anonymous

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

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- freckles

well cos(60) and sec(30) and tan(45) can all be evaluated using the unit circle

- freckles

http://buildingaunitcircle.weebly.com/uploads/6/2/2/3/622393/9409299.gif

- anonymous

Use the trig table........

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## More answers

- freckles

do you need help using that chart above ?

- anonymous

....
I need to the values
like cos = 1/2
and all.......

- freckles

cos(60)=1/2 that is correct
so squaring both sides gives
cos^2(60)=1/4

- freckles

sec(30) means you need to find cos(30) and flip whatever number that is

- 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

\[5 \cos^2(60)+4\sec^2(30)-\tan^2(45) \\ 5(\frac{1}{4})+4(\frac{1}{\cos(30)})^2-1\]

- freckles

so you still need to find cos(30)

- anonymous

4/3 = sec^2(30)

- 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

that can be simplify just be doing a few multiplications and additions(subtractions)

- anonymous

\[\frac{ 5 }{ 4 } +\frac{ 16 }{ 3 } - 1\]

- freckles

right so far

- anonymous

shall I take LCM?

- freckles

yes which the lcm(4,3,1)=12

- 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

67/12

- freckles

looks awesome!

- freckles

and totally correct :)

- anonymous

Thanks!
Please help me a bit more?

- freckles

Sure. I can take a stab at it.

- triciaal

@freckles would using identities just take longer?

- freckles

@triciaal if you want to try something with identities you can

- anonymous

\[\frac{ cotA-1 }{ 2 -\sec^2(A) } = \frac{ \cot A }{ 1 + tanA }\]

- 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

Help me?

- 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

I'm suppose to use LHS and get RHS

- freckles

ok then get rid of the compound fraction action on the the left hand side

- anonymous

\[\frac{ \cos(a) - \sin(a) }{ \sin(a) }\]

- anonymous

and for the denominator
2sin^(a)/cos^2(a)

- 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

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

that is I used \[\cos(2a)=\cos^2(a)-\sin^2(a) \\ \text{ and also } \cos(2a)=2\cos^2(a)-1\]

- 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

so the key in our proof was the use of double angle identity for cosine

- anonymous

Nice..
Thanks...
I need help in quadratic equations too...

- freckles

ok let's see it

- freckles

hey @triciaal did you come up with a way using identities for the first one ?

- freckles

I honestly haven't given it a try

- 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

wait what does that mean
what is the angle of elevation 60 or 30?

- freckles

and what is 30?

- anonymous

from 60 to 30 in secs
from the smae point

- freckles

oh the angle of elevation changed from 60 to 30 in 10 sec
gotcha

- freckles

|dw:1436680212196:dw|

- 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

|dw:1436680529581:dw|

- freckles

wait my drawing is off

- freckles

I will just make two separate drawings
|dw:1436680577539:dw|
|dw:1436680738212:dw|

- freckles

We need to find the distance the plane has flown for both pictures

- freckles

notice that is x for the first drawing and B for the second drawing

- 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

How to find the dist?

- freckles

you can use a trig ratio

- freckles

for example you are given the height which is 1km (the opposite side of the angle of elevation)

- freckles

and you are wanting to find the length that airplane has flown for both (which is the adjacent side )

- freckles

opp/adj is the tangent ratio

- 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

Thanks.
I will manage the rest...

- 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

oh your question says to put it an hours

- freckles

lol

- freckles

just let me know if you need any further help with this one

- freckles

or if you want me to check your answer

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