Expert: Scott A Wilson - 3/1/2009

Question
QUESTION: i need to verify each of the following identities.
1.sin x cos^3 x - cos x sin^3x = 1/4 sin4x
2. cos 3x = 4cos^3 x -3cos x
3. [sin (x + y + z) + sin (x + y + z)] over [cos (x + y + z) - cos (x + y + z)] = (tan y tan z -1) over (tan y + tan z)
4. (cos x over 1-tan x)+(sin x over 1-cot x)= cos x + sin x
5. (cos x + cos 4x + cos 7x) over (sin x + sin 4x + sin 7x)  
6. (1 + sin 2x)over(cos x + sin x)=(cos 2x) over (cos x-sin x)
7. (csc x)over(1 + csc x) - (csc x)over(1 - csc x) = 2 sec^2 x

I have just had a hard time figuring out these few. I  really appreciate your helping.
Thanks, Keah

ANSWER: The trig identities to use are
sin²(x) + cos²(x) = 1
1 + ctn²(x) = csc²(x)
1 + tan²(x) = sec²(x)
sin(2x) = 2sin(x)cos(x)
cos(2x) = 1 - 2sin²(x)
tan(x) = sin(x)/cos(x)
ctn(x) = cos(x)/sin(x)

Is there an equality on (5) that was missed?

By the way, I convert everything to sin(x) and cos(x).

There's two I forgot to give you -
sec(x) = 1/cos(x) and csc(x) = 1/sin(x).

There is also the sums of angles:
sin(x+y) = sin(x)cos(y) + sin(y)cos(x) and
cos(x+y) = cos(x)cos(y) - sin(x)sin(y).

Work that over and if you need further help, write tomorrow.
I'm almost out of time today and doing these would make me
think for too long, but tomorrow morning I can try to do them.


---------- FOLLOW-UP ----------

QUESTION: ok, so I have worked with your advice, but my problem is i will get down towards the end of the problem, but i can't get the answer. I was hoping if you could work through some of them, I also applied the things you gave me through the other 50 problems i had and it worked well, these are the few that i can not get through. The end of is = cot 4x. As I said i will get down to the last couple of lines and get stuck and spend hours trying to get through these last 7 problems. If you could find the time to work them out i would really really appreciate it.
Thank you,
Keah

Answer
It uses the basic identities sin²x+cos²x=1, sin2x=2sinxcosx,
sin(x+y) = sinx cosy siny cosx, and
cos(x+y) = cos²x - sin²x.

1.sin x cos^3 x - cos x sin^3x = 1/4 sin4x

 Rearrange right side

 = sin2x cos2x/2, since sin4x = 2sin2x cos2x

 = sinx cosx(cos²x – sin²x), by the identities for sin2x, cos2x

 = left side by multiplying it out


2. cos 3x = 4cos³x -3cos x

 Rearrange the left side cos3x to

 cos3x = cos2xcos2 – sin2xsinx, since it's the cos(2x+x)

 = (2cos²x – 1)cosx – 2sin^2xcosx

 = 2cos³x – cosx – 2(1 - cos²)cosx

 = 4cos³x – 3cosx, which is what we were looking for


3. [sin (x + y + z) + sin (x + y + z)] over [cos (x + y + z) - cos (x + y + z)] = (tan y tan z -1) over (tan y + tan z)

 Looks like the left side is divided by zero with
 cos(x+y+z) – cos(x+y+z) in the denominator


4. (cos x over 1-tan x)+(sin x over 1-cot x)= cos x + sin x

redo left side, giving = cosx/(1-sinx/cosx) + sinx/(1–cosx/sinx),
and then redo the denominator as one fraction

= cosx((cosx-sinx)/cosx) + sinx/((sinx – cosx)/sinx)

Invert the bottom and not the top is still over 1, then multiply

= cos²(x)/(cosx-sinx) + sin²x/(sinx-cos)

= (cos²x – sin²x)/(cosx – sinx),

since the fractions are backwards, so a minus sign was added

= cosx + sinx, by cancelling a cosx - sinx on the top and bottom
this is what we we’re looking for


5. (cos x + cos 4x + cos 7x) over (sin x + sin 4x + sin 7x)

 where is the equality symbol?


6. (1 + sin 2x)/(cos x + sin x)=(cos 2x)/(cos x-sin x)

 Reworking the right side gives

 = (cos²x-sin²x)/(cosx – sinx) by the double angle cos formula

 = cosx + sinx by

    cancelling the cosx-sinx in the top and bottom of the fraction

 = (cos²x + 2cosxsinx + sin²x)/(cosx + sinx),

    by multiplying the top and bottom by cosx + sinx

 = (1 + sin2x)/(sinx + cosx),

    which is what we were look for due to the fact that

sin²x+cos²=1 and 2cosxsinx = sin2x.


7. (csc x)/(1 + csc x) - (csc x)/(1 - csc x) = 2 sec^2 x

 Let’s rework the left side by combining fractions, giving

 = (cscx – csc^2x – cscx – csc^2x)/(1-csc^2x)

 = -2csc^2x/(1-csc^2x), and divide by cscx,
         noting that 1/cscx = sinx

 = -2/(sin^2x-1)
    and apply the sum of squares for sin and cos, giving

 = -2/(-cos^2x)

 = 2sec^2x