Use the power-reducing formulas as many times as possible to rewrite the expression in terms of the first power of the cosine. sin^2 3x cos^2 3x
@johnweldon1993 Do you know how to do this?
Well first, how is this written? \[\large sin^2(3x)cos^2(3x)\] Like that?
@johnweldon1993 Yes! But without the parenthesis
Okay that's fine, the addition of the parenthesis doesn't change anything here So, what are the "power reduction formulas?" Moreso, what are the ones for sin and cos since that's all we'll need here...
Are they also considered the sum and differences formulas?
Wait never mind
I found them
Awesome, so list those out for me and we'll get to work! :)
Sin^2= 1-cos(2x)/2 Cos^2=1+cos(2x)/2
Great! So from the original question \[\large sin^2(3x)cos^2(3x)\] Replace sin^2 with the power reduction formula above and we get \[\large (\frac{1-cos(2\times 3x)}{2})cos^2(3x)\] Now replace that cos^2 with it's power reduction formula...we now get \[\large (\frac{1-cos(2\times 3x)}{2})(\frac{1+cos(2\times 3x)}{2})\] With me so far?
Also, this particular problem CAN be done a quicker way, but I'm gonna make you work first :D
Please tell me the quicker way!
So you just replaced the (x) with what was in the parenthesis?
That wouldn't be a good idea here because the question states to do the process AS MANY TIMES AS POSSIBLE The quicker way is just for reference for later :) And yes...our 'x' was '3x'
Why are both of them cos when substituted?
So comparing \[\large sin^2(x) = \frac{1 - cos(2x)}{2}\] \[\large sin^2(3x) = \frac{1 - cos(2(3x))}{2}\]
That's just how the power reduction formulas work out, cos in both cases Take a look at the equations you listed out for me, they are both in terms of cosine
So how would you solve it from this step? I am a little confused
First, do you see why i subbed in what I did and how we got to this point?
On sin2(3x)=1−cos(2(3x))/2 I am confused why the (2x) just became a 2 and the x went on the 3 to make it 3x
Ohhh, so basically it was just substituted
Sorry, sometimes I have to stare at it for a second
Okay :) *Glad you asked and didn't just saw yeah So...the ... Oh maybe you get it :O haha yeah we just substituted If your question had just sin^2(x) it would be the exact power reduction formula if it was sin^2(2x) it would be 2 times whatever your 'x' was which would be 2x here And here we have sin^2(3x) so its 2 times whatever the 'x' was...which was 3x So that's how we got to this point :)
Thank you!!)
No problem!!! Okay so now we're at the same point, we subbed in the 2 formulas and we have \[\large (\frac{1-cos(2\times 3x)}{2})(\frac{1+cos(2\times 3x)}{2})\] So let's just do that little multiplication quick \[\large (\frac{1-cos(6x)}{2})(\frac{1+cos(6x)}{2})\] Good?
Not done yet...but good so far?
Yes!
That is one for cos and one for sin and then multiplied? Right
Correct indeed!
\[\large (\color \red{\frac{1-cos(6x)}{2}})(\color \green{\frac{1+cos(6x)}{2}})\] Red is from the sin^2(3x) part and green is from the cos^2(3x) part and we're multiplying them just like the original problem was
\[\large (\frac{1-cos(6x)}{2})(\frac{1+cos(6x)}{2})\] So since this is just multiplication...how would we multiply these two?
Will it be 1−cos(6x)/4 ?
But negative
Oh so close... try it out 1 more time :) \[\large (1-cos(6x))(1+cos(6x))=?\]
Or -1−cos(6x)^2/2 ?
Oh Foil?
Mmhmm, I think you were doing that but messing up somewhere!
When foiling I got 1 6x 6x 12x
Am I doing it wrong?
1 cos(6x) -cos(6x) 0?
\[\large 1 \times 1=?\] \[\large 1 \times cos(6x)=?\] \[\large 1 \times -cos(6x)=?\] \[\large -cos(6x)\times cos(6x)=?\] There we go...just to clarify what is negative and what not YOUR 3 answers are good, good, good, BUT not good cos times cos is not zero :)
1 cos(6x) -cos(6x) -cos(6x)^2
PERFECT!
So then do I add all those answers together or multiply them?
So lets write everything again \[\large (\frac{1-cos(6x)}{2})(\frac{1+cos(6x)}{2}) = \frac{1 + cos(6x) - cos(6x) - cos^2(6x)}{4}\]
Yes!
So real quick ...what do you see about that numerator? what can go away?
Wait why is it −cos^2(6x) and not −cos(6x)^2
1−cos^2(6x) /4
I would say same difference but I need to know that you're not thinking that 6x gets squared as well \[\large cos(x)\times cos(x) = cos^2(x)\] ^Generally how it's written
Oh okay! Got it!
Great! And yes so we have \[\large \frac{1-cos^2(6x)}{4}\] But damn, we have to use the power reduction formula AGAIN because we need cosine to be first power!
So what would we get if we did that?
So we replace cos^2 with the formula?
Correct :)
cos^2 (6x) *
Haha I knew what ya meant :)
1- ( 1-cos(2)(6x) /2 ) ____________________ ? 4
Beautiful! Simplify it down a bit
Oh wait...not beautiful...you used the COS formula right?
Oh it should be a +
There we go...so NOWWWW simplify a bit
1- ( 1-cos(12x) /2 ) 1- ( 1-cos(12x) _____________________ -----> ________________ Is this right? or is it not the right way to 4 8 get rid of the /2 fraction?
Medals 0 1- ( 1-cos(12x) /2 ) 1- ( 1-cos(12x) _____________________ -----> ________________ Is this right? or is it not the right way 4 8
1- ( 1-cos(12x) /2 ) 1- ( 1-cos(12x) _____________________ -----> ________________ 4 8
Sorry!
Not quite! I noticed you didn't replace that '-' with the '+' thought! *Lol don't worry about it!* So we have from the start: \[\large \frac{1 - (\frac{1 + cos(12x)}{2})}{4}\] Right?
Oh yes! Sorry I was copy and pasting!
No worries :) Now from here...look at that top part...\(\large 1 - (\frac{1 + cos(12x)}{2})\) What do we know about adding/subtracting fractions? We need a ________denominator ...?
Woah why is that 1- in front of the whole thing
I thought it was how you wrote it before that one
Yes I thought it was written like that
We ONLY focus on the cos^2 part....we don't even touch that 1 - out in front \[\large \frac{1-cos^2(6x)}{4}\] Now replace cos^2(6x) with that power reduction formula \[\large \frac{1-(\frac{1 + cos(12x)}{2})}{4}\]
Yes!
Okay good :) Back on the same page! But how would you simplify that top part?
Honestly I have no idea! Simplifying is hard for me sometimes
Calculus is the hardest thing thats ever happened to me
Don't worry about it! Hint* Completely unrelated but what is \(\large 1 + \frac{1}{2}\) ?
1.5
Lol damn not the answer I wanted you to say...what is the answer in fraction form?
3/2
Perfect, and how did you get that? We needed a common denominator right? Which was 2 \[\large 1 + \frac{1}{2} = \frac{2 + 1}{2} = \frac{3}{2}\] Right?
Yes!
Awesome...back to your problem...focus on the top part \[\large 1 - (\frac{1 + cos(12x)}{2})\] What is the answer in fraction form?
-3/2 ?
Or 1/2
Not quite :) We need the common denominator...which again is 2 So we have: \[\large \frac{2 - (1 + cos(12x))}{2}\] Do you see how I got to that step? Just did the common denominator
Why isn't it 2/2 − .... Since 2/2 is the same as 1 and it makes the common denominator
EXACTLY RIGHT! Let me write it out that way quick \[\large 1 - (\frac{1 + cos(12x)}{2})\] Let's do the common denominator like you said \[\large \frac{2}{2} - (\frac{1 + cos(12x)}{2})\] Right?
Yes!!!
The whole point of finding the common denominator...is so you can combine both fractions OVER the common denominator :) Think back to \[\large 1 + \frac{1}{2}\] That would be \[\large \frac{2}{2} + \frac{1}{2}\] Which we then combine OVER that common 2...and we get \[\large \frac{2+1}{2}\] Right?
1+cos(12x) ____________ ? 2
Oh yes!!!
Okay good, back on track :D So that's why I wrote \[\large \frac{2 - (1 + cos(12x))}{2}\] NOWWWWWWW...I see that you did 2 - 1.....but what do we know about having something outside parenthesis? We distribute it into them right? Into every term in the parenthesis... What should we do with that minus sign? \[\large \frac{2 \color \red{-} (1 + cos(12x))}{2}\]
Ohhhh okay so -2-cos(12x)) _____________ 2
OHHHH so close....just look at it 1 more time and tell me why it shouldn't be -2 :)
Join our real-time social learning platform and learn together with your friends!