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C3 - Trig

Not letting me upload the image for some reason.

But the question is:

7i) Prove that for cos x cannot equal 0,

sin2x-tan x = tanx cos2x

With the identity sign instead of the equal sign.




Need some help with question 7. I tried starting with the LHS and saying
sin2x= 2sinxcosx
Which gives 2sinxcos-tanx
Changed the tanx into sinx/cosx. Then had no idea what to do.
Right I have tried combing them.
So I think it is: (2sinxcos^2x-sinx)/cosx
Took a factor of sinx from the top: sinx(2cos^2x-1)/(cosx)
Thanks :smile:
(edited 9 years ago)
Original post by Super199
Not letting me upload the image for some reason.

But the question is:

7i) Prove that for cos x cannot equal 0,

sin2x-tan x = tanx cos2x

With the identity sign instead of the equal sign.




Need some help with question 7. I tried starting with the LHS and saying
sin2x= 2sinxcosx
Which gives 2sinxcos-tanx
Changed the tanx into sinx/cosx. Then had no idea what to do.

Thanks :smile:


Factorise sin(x) out of the expression.
Reply 2
Avatar for Super199
Super199
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Original post by ThatPerson
Factorise sin(x) out of the expression.

Right so:

2sinxcosx-sinx/cosx

sinx[2cosx-(1/cosx)]
sinx [ (2cos^2x-1)/(cosx)]

sinx[(cos2x)/(cosx)]?

No idea if that is right?
Original post by Super199
Right so:

2sinxcosx-sinx/cosx

sinx[2cosx-(1/cosx)]
sinx [ (2cos^2x-1)/(cosx)]

sinx[(cos2x)/(cosx)]?

No idea if that is right?


Yeah that's right. There's just one step left to prove the identity.
Reply 4
Avatar for Super199
Super199
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20
Original post by ThatPerson
Yeah that's right. There's just one step left to prove the identity.

Which is.... ?
Original post by Super199
Which is.... ?


Compare what you're trying to prove with what you have. What identity can you use to change your expression into tan(x)cos(2x)?
Reply 6
Avatar for Super199
Super199
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20
Original post by ThatPerson
Compare what you're trying to prove with what you have. What identity can you use to change your expression into tan(x)cos(2x)?

That is the same as saying:

Sinx/cosx (cos2x)
which gives tanxcos2x
Reply 7
Avatar for Super199
Super199
OP
20
I am stuck on a modulus question:

The functions g is defined by g(x)= |2x-1|+4, x<0
Find an expression for g^-1(x)

Swapped the y and x around

x=|2y-1|+4
x-4=|2y-1| .

No idea how you do this question. How do you 'undo' a mod sign?
Reply 8
Avatar for ztibor
ztibor
10
Original post by Super199
I am stuck on a modulus question:

The functions g is defined by g(x)= |2x-1|+4, x<0
Find an expression for g^-1(x)

Swapped the y and x around

x=|2y-1|+4
x-4=|2y-1| .

No idea how you do this question. How do you 'undo' a mod sign?


THe f(x) is not strictly monotone on all interval but monotone decreasing from
minus infinity to 1/2 and monotone insreasing from 1/2 to infinity.
Find separate g^-1(x) for both intervals.
Reply 9
Avatar for Super199
Super199
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20
Original post by ztibor
THe f(x) is not strictly monotone on all interval but monotone decreasing from
minus infinity to 1/2 and monotone insreasing from 1/2 to infinity.
Find separate g^-1(x) for both intervals.

Sorry I don't understand?
Reply 10
Avatar for ztibor
ztibor
10
Original post by Super199
Sorry I don't understand?


Sorry, I misred the question

the function is g(x)=∣2x−1∣+4\displaystyle g(x)=|2x-1|+4 for
Unparseable latex formula:

\displaystyel x<0



First you have to eliminate the modulus, From the definition of |x|

Unparseable latex formula:

\displaystyle |2x-1|=\begin {Bmatrix} 2x-1 & \text{when} & 2x-1>=0 & (x>=\frac{1}{2} )\\ 1-2x & \text{when} &2x-1<0 & (x<\frac{1}{2}) \end



Substituting the proper form in g(x) you can invert the g(x) on the given intervals separately.
i.e
x<0 means x<1/2 that is in this interval [2x-1|=1-2x

So the

g(x)=∣2x−1∣+4=1−2x+4→y=5−2xg(x)=|2x-1|+4=1-2x+4 \rightarrow y=5-2x

Now you can invert this.
(edited 9 years ago)

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