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Help with Maths Homework! Compound and Double angle formula !!

Find the value of tanx and hence solve the equation for 0<x<360
a) sin(x-30)=cosx
For this one I got x=30, 210 but I'm not sure if that's correct!
b) cos(x+60) = 2cos(x+30)
I am COMPLETELY stuck ...
Any help would be appreciated ! Thanks!
expand sin( x - 30° ) using sin ( A - B ) ≡ sinAcosB - cosAsinB
Original post by the bear
expand sin( x - 30° ) using sin ( A - B ) ≡ sinAcosB - cosAsinB


This.

And I'll also add that you should always just draw out sine and cosine curves if you're in any doubt. For example, if x was 30 in the first question (sin(x-30) = cos(x)) you'd get sin(0) = cos(30) which can't ever be true.
Original post by the bear
expand sin( x - 30° ) using sin ( A - B ) ≡ sinAcosB - cosAsinB
Which i've done and got : sin(x)cos(-30) - cos(x)sin(-30) = cosx Then I divided it by (cosx)(cos(-30)) Should I have done that?!
here A is x

B is 30° not -30°
Original post by the bear
here A is x

B is 30° not -30°



Ok so I've worked it out and I got: x=60,240

Part b)??
Reply 6
Avatar for B_9710
B_9710
18
I wouldn't use addition formula unless you're specifically asked to here.
Note that cos⁡x≡sin⁡(90−x) \cos x\equiv \sin (90-x) . Then you get sin⁡(x−30)=sin⁡(90−x) \sin (x-30)=\sin (90-x) .
Original post by Sophie_Banwell
Ok so I've worked it out and I got: x=60,240

Part b)??


those look correct.

b) i would use the expansion for cos ( D + E )....
Original post by the bear
those look correct.

b) i would use the expansion for cos ( D + E )....



The cos(A+B)=cosAcosB+sinAsinB identity? So I would then get cosxcos60 + sinxsin60 = 2(cosxcos30 + sinxsin30) ??
Original post by Sophie_Banwell
The cos(A+B)=cosAcosB+sinAsinB identity? So I would then get cosxcos60 + sinxsin60 ??


there is a slight mistake in your version of the identity.
Original post by the bear
there is a slight mistake in your version of the identity.
Sorry a typo it should be : cos(A+B) = cosAcosB - sinAsinB So I would get: cosxcos60 - sinxsin60 = 2(cosxcos30 - sinxsin30) ??
Original post by Sophie_Banwell
Sorry a typo it should be : cos(A+B) = cosAcosB - sinAsinB So I would get: cosxcos60 - sinxsin60 = 2(cosxcos30 - sinxsin30) ??


yes
Original post by the bear
yes
Now what do I do? Do I divide it all by ((cosx)(cos30)) ?
Original post by Sophie_Banwell
Now what do I do? Do I divide it all by ((cosx)(cos30)) ?


you could divide each term by cosx ...
But if I divide by cos(x) I get :
tanx(2sin30-sin60) = 2cos30-cos60

Is this even right?
Original post by the bear
you could divide each term by cosx ...


But if I divide by cos(x) I get :
tanx(2sin30-sin60) = 2cos30-cos60

Is this even right?
Original post by Sophie_Banwell
But if I divide by cos(x) I get :
tanx(2sin30° -sin60° ) = 2cos30° -cos60°

Is this even right?


that looks fine...

now using your knowledge of special angles, replace the sin30° etc with fractions.

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