VCE Specialist 3/4 Question Thread!

[polar]

[Homer]

Thanks polar, however I am stuck on another one of these questions :/

if and then equals?

[Stick]

Here's a hint - use the double angle formula using cos(B) rather than cos(2B).

[TrueTears]

cos(2B) = 2cos^2(B)-1

cos(B) = 2cos^2(B/2)-1

[Homer]

smaartt thanks truetears and stick

[Homer]

how would i find the implied domain of

[brightsky]

-1 =< sin2x =< 1
which is true for all x
so implied domain is R

[Homer]

the answer is :/

[Homer]

however, if I solve for x from -1 =< sin2x =< 1 i would get the answer. Thanks Brightsky

[Planck's constant]

For the inverse of a function to be defined, the function has to be one-to-one (else the inverse will fail the vertical line test and it will not be a function)
For this reason we restrict the domain of functions so that the inverse may be defined.

EDIT (got confused at this point),
For the sine function we (conventionally) restrict the domain to [-pi/2 , pi/2]

With this in mind, if we apply brightsky's method, we obtain jai's result

[brightsky]

however, if I solve for x from -1 =< sin2x =< 1 i would get the answer. Thanks Brightsky

For the inverse of a function to be defined, the function has to be one-to-one (else the inverse will fail the vertical line test and it will not be a function)
For this reason we restrict the domain of functions so that the inverse may be defined.

EDIT (got confused at this point),
For the sine function we (conventionally) restrict the domain to [-pi/2 , pi/2]

With this in mind, if we apply brightsky's method, we obtain jai's result

hmm...calculator seems to agree with my answer. i don't see a need to restrict the sine function....

[Homer]

EDIT (got confused at this point),
For the sine function we (conventionally) restrict the domain to [-pi/2 , pi/2]

With this in mind, if we apply brightsky's method, we obtain jai's result

 

I think theoretically  [-pi/4 , pi/4] is right, however when you sketch it on cas its R :/

[Jenny_2108]

EDIT (got confused at this point),
For the sine function we (conventionallyl) restrict the domain to [-pi/2, pi/2]

Why do we have to restrict domain of sine function?

Its a composite function so it'll be defined when range of sin(2x) E domain of inverse cos (which is what brightsky did above)

-1 =< sin(2x) =< 1
-pi/2 +2npi =< 2x =< pi/2 + 2npi (n is integer)
-pi/4 + npi =< x =< pi/4 + npi

I think the ans should be R

[Planck's constant]

Yes, I am now also convinced that the implied domain is R
Sorry for the confusion.

[Homer]

solve 4cos^3(t)-3cos(t) = cos(t) x belongs to [0,2pi]
I keep on getting 0,pi,2pi, but the answer has more to it :/