Circular functions

[d0minicz]

solve each of the following for

a)

b)

c)

thank you !!

[TrueTears]

a) cosx(cosx - sinx) = 0

therefore cosx = 0 and cosx - sinx = 0

first one cosx = 0 you can solve

but the other one cosx - sinx = 0 needs further simplifying

times both sides by yields

so (because = or )

using the cos(x+y) compound angle formula yields

which i think you can solve XD

[d0minicz]

lol thanks truetears
and how do you know to times both sides?
i get the rest though =]

[TrueTears]

sorry check that post again i accidently stuffed up my LaTeX EDITED now

doing the other 2 now

[TrueTears]

b) sin2x = sinx

2sinxcosx = sinx

2sinxcosx - sinx = 0 (notice here we don't divide both sides by sinx to get rid of it because we will be losing possible solutions. Consider , you wouldn't divide both sides by x here because it gets rid of possible solutions, you would do this though: then factor and solve. Same principle applies here)

so factor out a sinx leaves

sinx(2cosx -1 ) = 0

now using the null factor law

sinx = 0, i think you can solve

and 2cosx - 1 = 0, rearranging leaves which you can solve as well

c) This is the same principle to b) , do not divide both sides by cosx because it gets rid of possible solutions. Instead bring it over to the other side of the equation and factor it out.

basically these questions just always try to get it to =0 and then use the null factor law.

[TrueTears]

lol thanks truetears
and how do you know to times both sides?
i get the rest though =]

Basically whenever you have a situation like

cosx + sinx = 1

or cosx - sinx = 1

or sinx - cosx = 1

People will often try to square both sides and then solve. Obviously you can, but remember squaring leaves us with REDUNDANT roots. (meaning answers which will not actually work for the equation if you sub it back in)

so in this case, always try to aim to get it into the compound angle formula form, we know sinx and cosx share a same value, that is . Hence if we multiply both sides by , we can change one of them into and another into , that way we can change the whole thing into a compound angle formula, which in this case is cos(x+y)

Hope that explains it, but if you don't understand, feel free to ask.

[d0minicz]

i get it, thanks =]
ill let you know if i have any other troubles !

[d0minicz]

Solve a)  for

b) ,
thanks =]

[TrueTears]

sin8x = cos4x



so (notice you don't divide both sides by cos4x as it gets rid of possible solutions)

now factor out a cos4x yields

cos4x(2sin4x -1 ) =0

so cos4x = 0  You can solve this

and which you can solve too XD

[TrueTears]

b) cos2x  = cosx

so

so



let a = cosx



solve this quadratic yields a = , 1

but a = cosx

cosx = or cosx = 1 (which you can solve XD)

[d0minicz]

Given than the domain of sinx and cosx are restricted to and defined the implied domain and range :

a)

b)

c)

having trouble restricting etc..
thanks

[TrueTears]

a) let and

now

So for f o g to exist



and

Now, ran g is not a subset of dom f, so we must restrict ran g to AT LEAST

and in order to do we must restrict the domain of g to [0,1], (another way of thinking about it is just solving )

so we know dom f o g = dom g = [0,1]

EDIT: to find the range we know that the output of will give the end points of the range for .

so since is restricted to

now sub and 0 into to find end points.

and

therefore range of

[pHysiX]

for this one:
\ cos^2 x - cosxsinx = 0, factor out cos(x) and u'll get cos(x)(cos(x)-sin(x))=0.

so cos(x)=0 and cos(x)-sin(x)=0. Now, for me, i prefer the brute force way, so instead of thinking about pi/4, i'd go:

cos(x)-sin(x)=0
cos(x)=sin(x)
therefore, tan(x)=1

just my 2 cents =]

[d0minicz]

thanks guys

Truetears can you pls show me ur working out for b) and c)
thanks a bunch

and i need help with this question

sinx + cosx = 1

thanks

[TrueTears]

thanks guys

Truetears can you pls show me ur working out for b) and c)
thanks a bunch

and i need help with this question

sinx + cosx = 1

thanks

for sinx + cosx = 1

look at the 2nd post on page 3 its a minus in the middle, but it's the same principle, in the meanwhile I shall answer b) and c) for you.