Question

PLEASE HELP

d = dCosB/CosA + yCosB

I need to prove that it equals

yCosACosB/CosA - CosB

Just give me help on how to get it like this, step by step.

Any help is greatly appreciated :)

Answer 1

can you post a quick screenshot of the material?

Answer 2

so we can see what you mean

Answer 3

ok wait a sec

Answer 4

@jdoe0001 thank you for having a look :)

Answer 5

hmmm I'm assuming there's a context to it
was wondering what "d" and "y" were =)

Answer 6

d is the distance, don't know what y is. would you like me to give you the original question?

Answer 7

ok

Answer 8

the question

Answer 9

working so far

Answer 10

does that help? @jdoe0001

Answer 11

yes

Answer 12

yay :)

Answer 13

@whpalmer4

Answer 14

@whpalmer4 @jdoe0001 are you guys having any luck?

Answer 15

hehe, not yet

Answer 16

Okay, it's not so bad.

\[\cos (\alpha )=\frac{d}{x}\]\[x = \frac{d}{\cos(\alpha)}\]\[d = x\cos(\alpha)\]
\[\cos (\beta )=\frac{d}{x+y}\]\[d = (x+y)\cos(\beta)\]
\[x\cos(\alpha) = (x+y)\cos(\beta)\]\[x\cos(\alpha)=x\cos(\beta) + y\cos(\beta)\]\[x(\cos(\alpha)-\cos(\beta)) = y\cos(\beta)\]
Now divide both sides by the difference of cosines term from the left, and substitute what we got for \(x\) in the second step. Then multiply by \(cos(\alpha)\) and Bob's yer uncle...

Answer 17

@whpalmer4 you are a lifesaver thank you so so much!! could i just show you a photo of the working out ot make sure its correct?? :)
and my teacher substituted x=d/cosA into it, but you substiuted d=xcosA, would that mke any difference?

Answer 18

\[x = \frac{d}{\cos(\alpha)}\]\[x*\cos(\alpha) = \frac{d}{\cancel{\cos(\alpha)}}*\cancel{\cos(\alpha)}=d\]

Answer 19

sure, show me a picture

Answer 20

ok, um i just don't get when you wrote
xcosA = xcosB +ycosB
xcosA -cosB = ycosB

Answer 21

\(\bf cos(\alpha)=\cfrac{d}{x}\implies xcos(\alpha)={\color{red}{ d}}\qquad {\color{blue}{ x}}=\cfrac{d}{cos(\alpha)}
\\ \quad \\
cos(\beta)=\cfrac{d}{x+y}\implies xcos(\beta)+ycos(\beta)={\color{red}{ d}}
\\ \quad \\
xcos(\alpha)=xcos(\beta)+ycos(\beta)\implies xcos(\alpha)-xcos(\beta)=ycos(\beta)
\\ \quad \\
x[cos(\alpha)-cos(\beta)]=ycos(\beta)
\\ \quad \\
{\color{blue}{ \cfrac{d}{cos(\alpha)}}}[cos(\alpha)-cos(\beta)]=ycos(\beta)
\implies
d=\cfrac{ycos(\beta)}{cos(\alpha)-cos(\beta)}\)

Answer 22

@whpalmer4

Answer 23

Look at the penultimate line of @jdoe0001 's writeup to see what I did in the "mystery step" :-)

Answer 24

@rhiannon1 you dropped the \(x\) off the \(x\cos(\beta)\) term when you moved it across the = sign.

Answer 25

\(\bf {\color{blue}{ \cfrac{d}{cos(\alpha)}}}[cos(\alpha)-cos(\beta)]=ycos(\beta)

\\ \quad \\
\implies \cfrac{d[cos(\alpha)-cos(\beta)]}{cos(\alpha)}=ycos(\beta)
\\ \quad \\

\implies d=\cfrac{ycos(\beta)}{cos(\alpha)-cos(\beta)}\)

Answer 26

yeah, i didnt get it but now i see that you have to take the xcosB across the equals sign then factorise, right?

Answer 27

common factor, yes

Answer 28

@jdoe0001 i dont get what you wrote in the message before the most recent one

Answer 29

ohh just an expanded version of the last line
in case you didn't see the simplification :)

Answer 30

yes, that's what I combined into 1 step, moving the term and factoring out the \(x\).

Answer 31

but its different ot the answer @jdoe0001

Answer 32

to*

Answer 33

hmmmm

Answer 34

he forgot to put in the \(\cos(\alpha)\) that he multiplied both sides by

\[\bf {\color{blue}{ \cfrac{d}{cos(\alpha)}}}[cos(\alpha)-cos(\beta)]=ycos(\beta)

\\ \quad \\
\implies \cfrac{d[cos(\alpha)-cos(\beta)]}{cos(\alpha)}=ycos(\beta)
\\ \quad \\

\implies d=\cfrac{y\cos(\alpha)cos(\beta)}{cos(\alpha)-cos(\beta)}\]

Answer 35

ohh shoot... ahemm yea =)

Answer 36

I'm still waiting for the instantaneous thought->LaTeX converter to arrive :-)

Answer 37

\(\bf \bf cos(\alpha)=\cfrac{d}{x}\implies xcos(\alpha)={\color{red}{ d}}\qquad {\color{blue}{ x}}=\cfrac{d}{cos(\alpha)}
\\ \quad \\
cos(\beta)=\cfrac{d}{x+y}\implies xcos(\beta)+ycos(\beta)={\color{red}{ d}}
\\ \quad \\
xcos(\alpha)=xcos(\beta)+ycos(\beta)\implies xcos(\alpha)-xcos(\beta)=ycos(\beta)
\\ \quad \\
x[cos(\alpha)-cos(\beta)]=ycos(\beta)
\\ \quad \\
{\color{blue}{ \cfrac{d}{cos(\alpha)}}}[cos(\alpha)-cos(\beta)]=ycos(\beta)
\implies
d=\cfrac{y{\color{blue}{ cos(\alpha)}}cos(\beta)}{cos(\alpha)-cos(\beta)}\)

Answer 38

hehe

Answer 39

rhiannon1 just had a typo =)

Answer 40

oh ok, but how did you get the denominator like that?

Answer 41

\(\bf cos(\alpha)=\cfrac{d}{x}\implies xcos(\alpha)={\color{red}{ d}}\qquad {\color{blue}{ x}}=\cfrac{d}{cos(\alpha)}
\\ \quad \\
----------------------\\
{\color{blue}{ \cfrac{d}{cos(\alpha)}}}[cos(\alpha)-cos(\beta)]=ycos(\beta)

\\ \quad \\
\implies \cfrac{{\color{blue}{ d}}[cos(\alpha)-cos(\beta)]}{{\color{blue}{ cos(\alpha)}}}=ycos(\beta)
\\ \quad \\

\implies {\color{blue}{ d}}=\cfrac{y{\color{blue}{ cos(\alpha)}}cos(\beta)}{cos(\alpha)-cos(\beta)}
\)

see it?

Answer 42

not really :(

Answer 43

\(\bf cos(\alpha)=\cfrac{d}{x}\implies xcos(\alpha)={\color{red}{ d}}\qquad {\color{blue}{ x}}=\cfrac{d}{cos(\alpha)}
\\ \quad \\
----------------------\\
{\color{blue}{ \cfrac{d}{cos(\alpha)}}}[cos(\alpha)-cos(\beta)]=ycos(\beta)

\\ \quad \\
\implies \cfrac{{\color{blue}{ d}}[cos(\alpha)-cos(\beta)]}{{\color{blue}{ cos(\alpha)}}}=ycos(\beta)
\\ \quad \\
\implies {\color{blue}{ d}}[cos(\alpha)-cos(\beta)]=y{\color{blue}{ cos(\alpha)}}cos(\beta)
\\ \quad \\

\implies {\color{blue}{ d}}=\cfrac{y{\color{blue}{ cos(\alpha)}}cos(\beta)}{cos(\alpha)-cos(\beta)}\)

Answer 44

up to what part you understand it?

Answer 45

i think i get it now that you did every step. so you time by cosa then divide by the part in brackets?

Answer 46

yes, more or less
you'd cross-multiply to get the \(cos(\alpha)\) over
then cross-multiply to get the bracket group down

Answer 47

@jdoe0001 @whpalmer4 do you think this is year 10 work? just wondering.

Answer 48

10th grade you mean?

Answer 49

yep

Answer 50

hmmmm might be ambiguous
depends on how much you've covered trig
and what has been over in the book

but maybe or maybe not

Answer 51

ok, don't know what ambiguous means, but ive heard of it.

Answer 52

ambiguous, it can go either way

Answer 53

when you started the working out, from the d/cosa (cosa - cosB) = ycosB i dont see how you did that (this is my last question) i mean where did you get it from?

Answer 54

oh wait, did you sub in the x = d/cosa into x(cosa-cosB) = ycosB ??@jdoe0001

Answer 55

yes

Answer 56

oh ok, thank you guys so much! i know who to ask for math problems haha
thank you for your time i understand a lot better thankyou!! :)

Answer 57

\[\cos (\alpha )=\frac{d}{x}\]\[x = \frac{d}{\cos(\alpha)}\]\[d = x\cos(\alpha)\]\[\cos (\beta )=\frac{d}{x+y}\]\[d = (x+y)\cos(\beta)\]
\[x\cos(\alpha) = (x+y)\cos(\beta)\]\[x\cos(\alpha)=x\cos(\beta) + y\cos(\beta)\]\[x\cos(\alpha)-x\cos(\beta) = y\cos(\beta)\]\[x(\cos(\alpha)-\cos(\beta)) = y\cos(\beta)\]Now substitute for \(x\)\[\frac{d}{\cos(\alpha)}*(\cos(\alpha)-\cos(\beta)) = y\cos(\beta)\]
\[\frac{d}{\cos(\alpha)}= \frac{y\cos(\beta)}{\cos(\alpha)-\cos(\beta)}\]Multiply both sides by \(\cos(\alpha)\)\[d = \frac{y\cos(\alpha)\cos(\beta)}{\cos(\alpha)-\cos(\beta)}\]

Answer 58

I think this is maybe on the slightly harder end of a high school trigonometry class, but not too much. We didn't have to do any real heavy lifting here with multiple trig identities or anything like that.

Answer 59

Mostly, it was just an exercise in algebra after you drew the picture, right?

Answer 60

yeah @whpalmer4

Answer 61

A tactic I occasionally find helpful (thought about doing it here, but didn't want to have to explain it while you were trying to understand the problem) is to replace all the trig functions with letters. We aren't doing any trig identity work here, so it isn't crucial that you be able to recognize them as such, and often the algebra becomes much simpler to see (and write/typeset!) I also do this with algebra where I have a bunch of square roots or other radical signs...

Answer 62

Then you undo the substitution after you've done all of the shuffling around and take another look to see if there is anything more you can do.