blue powder
Registered User regular

Hey, guys, I have an exam tomorrow (well two exams) and essentially I'm freaking out. I've always had a lot of difficulty understanding this and I've put off trigonometric identities and using them to solve expressions and I've just now realized this.

so if I have

cos4x/((cos^2x-sin^2x)(cos2xsin2x))

whenever I ask my professor or someone to help me out with it he'll tell me to use double angle here and there but I think he's rearranging the actual expressions in a weird way because I sort of see how it applies, but I can't do it myself and make it work.

I have all the rules written out, so if you could just tell me how on earth you apply them that would be really fantastic. I'm not looking for just an answer here, I'm doing a practice exam so it's not assessed, I just need to know how to do this!

Thanks a lot in advance, I know from experience that there are a whole bunch of you out there who are great at math and explaining it (which is why I cam here).

(This is from an introductory calc and algebra unit at college if that's of any help.)

so if I have

cos4x/((cos^2x-sin^2x)(cos2xsin2x))

whenever I ask my professor or someone to help me out with it he'll tell me to use double angle here and there but I think he's rearranging the actual expressions in a weird way because I sort of see how it applies, but I can't do it myself and make it work.

I have all the rules written out, so if you could just tell me how on earth you apply them that would be really fantastic. I'm not looking for just an answer here, I'm doing a practice exam so it's not assessed, I just need to know how to do this!

Thanks a lot in advance, I know from experience that there are a whole bunch of you out there who are great at math and explaining it (which is why I cam here).

(This is from an introductory calc and algebra unit at college if that's of any help.)

0

## Posts

When I see [cos(2x)*sin(2x)] in the quotient, I recognize that as looking similar to the result of the sine double angle formula, so it'd be equivalent to [0.5 * sin(4x)].

Just say the word if if this hint doesn't poke you far enough in the right direction and you need another.

Edit: Nevermind, sorry. I just tried to work out the problem fully and I can't do it, even though there are lots of rules I can apply.

To be fair, I haven't done this stuff in over 4 years.

Raiden333oncos^2x - (1 - cos^2x) = 2cos^2x - 1

Now on the top you have cos4x. Remember that cos2x = 2cos^2x - 1, so

cos4x = 2cos^2(2x) - 1 = 2(2cos^2x - 1) - 1 = 4cos^2x - 3

Add in the hint above and you get the whole thing to be

(4cos^2x - 3) / (2cos^2x - 1) * sin(4x)

Although I'm not really sure if that improves anything...

Smug DucklingonThe best I was able to work it out to was 2[cot(4x)] / [cos(2x)sin(4x)], which is probably just as helpful.

Raiden333onuse double angle formula for cos

(cos(4*x)/((cos(2*x)*(cos(2*x)*sin(2*x))))

again

(cos(2*x)^2-sin(2*x)^2)/((cos(2*x)*(cos(2*x)*sin(2*x))));

substitute sin/cos = tan

(1 - tan(2*x)^2)/sin(2*x);

subs sec^2 = tan^2 +1

(2 - sec(2*x)^2)/sin(2*x);

Thats what i can get it too.

What should we be aiming for?

romanqwertyonblue powderon