Solving Trig Problems: 1/32(2+cos2x-2cos4x-cos6x) & More

[Laceylb]

ok here is the first problem:

sin^2 xcos^4 x and you are to simplify it, the answer is 1/32(2+cos2x-2cos4x-cos6x)

i will skip a few steps just to save some time:

1/8(1+2cos2x + cos^2 2x- cos2x -2cos^2 2x -cos^3 2x)

1/8( 1+2cos2x + 1/2(1+cos4x) - cos2x- 1+cos4x - 1/2cosx(1+cos4x) )

1/16( 2+ 4cos2x + 1 +cos4x - 2 cos2x - 2 + 2cos4x - cosx(1+cos4x))

and the rest just gives me a headache, any ideas where to go from there



the next one:

(1-secx)/(-sinx - tanx) = -cscx

prove the identity


would you multiply by the conjugate or use an identity.. I am not sure really how to start it. so far what i have tried it doesn't look like it leads anywhere




and the final one:


sqrt((1-cos^2 x)/(1+cos^2 x)) = sqrt(2)|sin x/2|

prove the identity

on the right side i have

sqrt(2)|sqrt((1-cosx)/2)|

1-cosx

im not sure how to simplify the left, i thought about using the conjugate, but it didnt look like it led anywhere



so if you can help with any of these, please do thanks

 

[pizzasky]

For the first question, you may want to start working on the 2nd expression \frac{1}{32}(2+cos2x-2cos4x-cos6x) first. After all, this question is about simplifying an expression and this definitely looks less simple than the other!

For the second question, are you supposed to prove \frac{1-secx}{-sinx-tanx}\ =\ -cosecx? If I am not mistaken, this is not an identity. Did you copy the question correctly?

Same goes for the 3rd question. Are you supposed to prove \sqrt{\frac{1-cos^{2}x}{1+cos^{2}x}}\ = \sqrt{2}\ \mid sin\frac{x}{2}\mid? Why not take another look at the question?

 

[Curious3141]

For the first one, you have the right idea. For further simplification, use Factor Formulae.

 

[arunbg]

If I am not mistaken, the second question should look like
\frac{1+secx}{-sinx-tanx}\ =\ -cosecx

 

[Laceylb]

If I am not mistaken, the second question should look like
\frac{1+secx}{-sinx-tanx}\ =\ -cosecx

the original for that was (1 + sex(-x))/sin-x) + tan(-x) = -cscx

 

[Laceylb]

For the first question, you may want to start working on the 2nd expression \frac{1}{32}(2+cos2x-2cos4x-cos6x) first. After all, this question is about simplifying an expression and this definitely looks less simple than the other!

For the second question, are you supposed to prove \frac{1-secx}{-sinx-tanx}\ =\ -cosecx? If I am not mistaken, this is not an identity. Did you copy the question correctly?

Same goes for the 3rd question. Are you supposed to prove \sqrt{\frac{1-cos^{2}x}{1+cos^{2}x}}\ = \sqrt{2}\ \mid sin\frac{x}{2}\mid? Why not take another look at the question?

yes for last two you are to prove the identity


im not really sure what you mean about the first one? a good part of it my teacher said was right and he said from there on it was just an algebra mess

 

[VietDao29]

Can you recheck your second, and your third problem. It does not look right to me.
I think the second problem should look like:
\frac{1 + \sec x}{- \sin x - \tan x} = -csc x (as arunbg has already pointed out)
And the third problem should read:
\sqrt{\frac{1 - \cos ^ 2 x}{1 + \cos x}} = \sqrt{2} \left| \sin \left( \frac{x}{2} \right) \right|
-------------------
Have you given any of the 2 problems above a try?
I'll give you a hint to start off the second problem:
You can either multiply boths sides of the equation by the LHS's denominator (i.e -sin x - tan x), or you can try to convert every tan, csc, and sec functions to just sin, and cos functions, and try factoring to get the RHS.
Give the third problem a try, it won't do you any harm. :)
Do you know the Double-Angle formulae?
Can you go from here? :)

 

[Laceylb]

Can you recheck your second, and your third problem. It does not look right to me.
I think the second problem should look like:
\frac{1 + \sec x}{- \sin x - \tan x} = -csc x (as arunbg has already pointed out)
And the third problem should read:
\sqrt{\frac{1 - \cos ^ 2 x}{1 + \cos x}} = \sqrt{2} \left| \sin \left( \frac{x}{2} \right) \right|
-------------------
Have you given any of the 2 problems above a try?
I'll give you a hint to start off the second problem:
You can either multiply boths sides of the equation by the LHS's denominator (i.e -sin x - tan x), or you can try to convert every tan, csc, and sec functions to just sin, and cos functions, and try factoring to get the RHS.
Give the third problem a try, it won't do you any harm. :)
Do you know the Double-Angle formulae?
Can you go from here? :)

the first problem is correctly written the way i had it, but i did mess up it should be cos raised to the first on the bottom


i have tried these problems forever, a few days at least, we had two assignments 1-71 every other odd and 2-116 every other even... these are the only three i just can't seem to get...

 

[arunbg]

the original for that was (1 + sec(-x))/sin(-x) + tan(-x) = -cscx

sec(-x)=sec(x) and not -sec(x).
Third question, Viet Dao's right about the correction.

 

[VietDao29]

i have tried these problems forever, a few days at least, we had two assignments 1-71 every other odd and 2-116 every other even... these are the only three i just can't seem to get...

Ack, have you looked at the hints I've previously posted, and others' as well?
Why don't you give it another try using my hints? It's just for your own sake that we don't give out complete solutions here, at PF. We, however can guide you through the problem, and help you whenever you get stuck. If you'd like, you can skim through the rules https://www.physicsforums.com/showthread.php?t=28.
Now, where do you get stuck? :)

 

[Curious3141]

To reiterate, for the first one, go as far as you can with the double and quadruple angle identities, then when you get something like cos(mx)cos(nx), just apply Factor Formula. The problem will crumble like old cheese.

http://library.thinkquest.org/C0110248/trigonometry/formfactor.htm

 

[Laceylb]

i have all but the first and third done... for the third i have sqrt(cosx-1) but i don't know how to get rid of the sqrt, am i allowed to square it ?

 

[Curious3141]

i have all but the first and third done... for the third i have sqrt(cosx-1) but i don't know how to get rid of the sqrt, am i allowed to square it ?

I would suggest for the first one (if you haven't solved it yet), start afresh and apply Factor Formulae as needed. I did it and it's solvable within five steps.

 

[Laceylb]

I would suggest for the first one (if you haven't solved it yet), start afresh and apply Factor Formulae as needed. I did it and it's solvable within five steps.

what is factor formulae?

 

[Curious3141]

what is factor formulae?

I linked it above.

 

[VietDao29]

i have all but the first and third done... for the third i have sqrt(cosx-1) but i don't know how to get rid of the sqrt, am i allowed to square it ?

Uhmm, as I've pointed out before, for #3, you should use the Double Angle formulae, i.e:
\cos (x) = \cos ^ 2 \left( \frac{x}{2} \right) - \sin ^ 2 \left( \frac{x}{2} \right) = 1 - 2 \sin ^ 2 \left( \frac{x}{2} \right) = 2 \cos ^ 2 \left( \frac{x}{2} \right) - 1.
Can you go from here? :)
-------------------
For problem 1, there are many ways to tackle it. Hence, it's very hard to check your answer if you just give us the second half of your work.
So if you can, just show your complete work, and we may check it for you.
Or you can try this way.
Since there are even power of sin, and cos function in this problem, one should consider the Power Reduction Formulae. But first, you can apply the Double Angle Formulae to somplify the expression a bit.
So I'll start off the problem for you:
\sin ^ 2 x \cos ^ 4 x = \sin ^ 2 x \cos ^ 2 x \cos ^ 2 x = \frac{\sin ^ 2 (2x)}{4} \cos ^ 2 x = ...
Can you go from here? :)
Don't forget to use the Product To Sum Formulae. :)

 

[Curious3141]

Don't forget to use the Product To Sum Formulae. :)

For the OP, that's another name for the Factor Formulae.

 

[Laceylb]

I linked it above.

oh, i have them written down already

 

[Laceylb]

Uhmm, as I've pointed out before, for #3, you should use the Double Angle formulae, i.e:
\cos (x) = \cos ^ 2 \left( \frac{x}{2} \right) - \sin ^ 2 \left( \frac{x}{2} \right) = 1 - 2 \sin ^ 2 \left( \frac{x}{2} \right) = 2 \cos ^ 2 \left( \frac{x}{2} \right) - 1.
Can you go from here? :)
-------------------
For problem 1, there are many ways to tackle it. Hence, it's very hard to check your answer if you just give us the second half of your work.
So if you can, just show your complete work, and we may check it for you.
Or you can try this way.
Since there are even power of sin, and cos function in this problem, one should consider the Power Reduction Formulae. But first, you can apply the Double Angle Formulae to somplify the expression a bit.
So I'll start off the problem for you:
\sin ^ 2 x \cos ^ 4 x = \sin ^ 2 x \cos ^ 2 x \cos ^ 2 x = \frac{\sin ^ 2 (2x)}{4} \cos ^ 2 x = ...
Can you go from here? :)
Don't forget to use the Product To Sum Formulae. :)

i actually can't understand the third problem... i never even knew you could divide the oringinal double angle identities... this is what i have on the left side = to 1-cosx:

sqrt((1-cosx)(1+cosx)/(1+cosx))

sqrt(1-cosx)... i tryed to put an identity in...but i don't know if i did it right because it didnt look like it led any where

 

[Laceylb]

Uhmm, as I've pointed out before, for #3, you should use the Double Angle formulae, i.e:
\cos (x) = \cos ^ 2 \left( \frac{x}{2} \right) - \sin ^ 2 \left( \frac{x}{2} \right) = 1 - 2 \sin ^ 2 \left( \frac{x}{2} \right) = 2 \cos ^ 2 \left( \frac{x}{2} \right) - 1.
Can you go from here? :)
-------------------
For problem 1, there are many ways to tackle it. Hence, it's very hard to check your answer if you just give us the second half of your work.
So if you can, just show your complete work, and we may check it for you.
Or you can try this way.
Since there are even power of sin, and cos function in this problem, one should consider the Power Reduction Formulae. But first, you can apply the Double Angle Formulae to somplify the expression a bit.
So I'll start off the problem for you:
\sin ^ 2 x \cos ^ 4 x = \sin ^ 2 x \cos ^ 2 x \cos ^ 2 x = \frac{\sin ^ 2 (2x)}{4} \cos ^ 2 x = ...
Can you go from here? :)
Don't forget to use the Product To Sum Formulae. :)

\sin ^ 2 x \cos ^ 4 x = \sin ^ 2 x \cos ^ 2 x \cos ^ 2 x = \frac{\sin ^ 2 (2x)}{4} \cos ^ 2 x = ...

im not really sure how you got to the last part of it with sin^2 over 4

 

[Laceylb]

im seriously mildy retarded in trig, i will kill myself if i have to take it in college

 

[pizzasky]

\sin ^ 2 x \cos ^ 4 x = \sin ^ 2 x \cos ^ 2 x \cos ^ 2 x = (sinx cosx)^2 \cos ^ 2 x =\ (\frac{1}{2} \sin(2x))^2 \cos ^ 2 x = \frac{\sin ^ 2 (2x)}{4} \cos ^ 2 x = ...

And the third problem is \sqrt{\frac{1 - \cos ^ 2 x}{1 + \cos x}} = \sqrt{2} \left| \sin \left( \frac{x}{2} \right) \right| (as corrected by VietDao29). Why not try to find

an alternate expression for 1 + \cos x using this hint given by VietDao29? \cos (x) = \cos ^ 2 \left( \frac{x}{2} \right) - \sin ^ 2 \left( \frac{x}{2} \right) = 1 - 2 \sin ^ 2 \left( \frac{x}{2} \right) = 2 \cos ^ 2 \left( \frac{x}{2} \right) - 1