anyone with experience in trigonometric identities please help.?

Question

I take pride in figuring out problems by myself, but these two darned questions have me stuck.

I know that sin 2x = 2sinxcosx and the basics, but I can't seem to reduce these problems in the right way:

sin2x/sinx - cos2x/cosx = secx

sin (x+y) cos (x-y) + cos (x+y) sin (x-y) = sin2x

please show how you got these, in case I run into similar ones. thanks tons!

Answer 1

sin2x/sinx - cos2x/cosx
=2sinxcosx/sinx - [cos^2x-sin^2x]/cosx
=2cosx - cosx + sin^2x/cosx
=cosx + sin^2x/cosx
=cosx + (1-cos^2x)/cosx
=cosx + 1/cosx - cos^2x/cosx
=cosx + secx - cosx
=sec x

sin(x+y)cos(x-y) + cos(x+y)sin(x-y)
=(sinxcosy+cosxsiny)*(cosxcosy + sinxsiny) + (cosxcosy - sinxsiny)*(sinxcosy - cosxsiny)
=sinxcosxcos^2y + sin^2xsinycosy + cos^2xsinycosy + cosxsinxsin^2y + cosxsinxcos^2y - cos^2xcosysiny - sin^2xsinycosy - sinxcosxsin^2y
= sinxcosx(cos^2y + sin^2y) + sinycosy(sin^2x + cos^2x) + sinxcosx(sin^2y + cos^2y) - sinycosy(sin^2x + cos^2x)
= sinxcosx + sinycosy + sinxcosx - sinycosy
= sinxcosx + sinxcosx
= 2sinxcosx
= sin2x

Answer 2

For the first problem, you also need to know that cos(2x)= 2cos^2(x)-1. so then just plug this into the cos(2x) that you had.
here's how to solve the first:
(2sinxcosx/sinx)-((2cos^2(x)-1)/cosx)) -the sinx on the first cancels leaving you with:
(2cosx)-((2cos^2(x)-1)/cosx)- next, make the 1 into cosx/cosx. so you get
(2cosx)-((2cos^2(x)-(cosx/cosx)/cosx).
, then factor out the cosx from the second part.

(2cosx) - [cosx(2cosx-(1/cosx))]/cosx. Now the cosx that you factored out cancels with the bottom and you get

2cosx - (2cosx-(1/cosx)) and then just add/sub. (don't forget to distribute that negative sign.)

and get 1/cosx which is equal to secx.

for the second part you shouldn't have any trouble with that b/c its pretty much algebra. It will look very overwhelming but take the time to do it. Use these identities which can be found at this webpage
http://www.math.com/tables/trig/identities.htm

Answer 3

sin2x/sinx -cos2x/cosx =
(sin2x * cosx - cos2x * sinx)/(sinx * cosx) =

sinACosB - CosASinB = Sin (A-B)

therefore = Sin(2x-x)/(sinx * cosx) = Sinx / (sinx * cosx)

sinx gets cancelled and only = 1/cosx will remain

therefore = 1/cosx = secx


sin(x+y)cos(x-y) + cos(x+y)sin(x-y) = Sin ((x+y)+(x-y))
= sin (x+y+x-y) = sin2x

Answer 4

sec(x) = 1/cos(x) (definition of secant)
sin(2x) = 2 Sin(x)Cos(x)
Cos(2x) = Cos^2(x) - Sin^2(x) = Cos(x)Cos(x) - Sin(x)Sin(x)

sin2x/sinx - cos2x/cosx = secx
2Sin(x)Cos(x)/sin(x) - (Cos(x)Cos(x) - Sin(x)Sin(x))/cos(x) = 1/Cos(x)
2Cos(x) - Cos(x) + Sin(x)Sin(x)/cos(x) = 1/Cos(x)
Cos(x) + Sin(x)Sin(x)/cos(x) = 1/Cos(c)
Cos(x)Cos(x)/Cos(x) + Sin(x)Sin(x)/cos(x) = 1/Cos(c)
We now have Cos(x) as the denominator on both sides, we can get rid of it (by multiplying both sides by Cos(x))

Cos(x)Cos(x) + Sin(x)Sin(x) = 1
cos^2 + sin^2 = 1 (basic identity)

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The reason you had sin(2x) = 2Sin(x)Cos(x) is the general identity:

Sin(x+y) = Sin(x)Cos(y) + Cos(x)Sin(y)
Sin(x-y) = Sin(x)Cos(y) - Cos(x)Sin(y)
from which we can get:
Sin(x+x) = Sin(x)Cos(x) + Cos(x)Sin(x) = 2Sin(x)Cos(x)

Cos(x+y) = Cos(x)Cos(y) - Sin(x)SIn(y)
Cos(x-y) = Cos(x)Cos(y) + Sin(x)SIn(y)
(note the reversal of signs)

So, the question
sin (x+y) cos (x-y) + cos (x+y) sin (x-y) = sin2x
becomes

[Sin(x)Cos(y) + Cos(x)Sin(y)][Cos(x)Cos(y) + Sin(x)SIn(y)] + [Cos(x)Cos(y) - Sin(x)SIn(y)][ Sin(x)Cos(y) - Cos(x)Sin(y)] = Sin(2x)

Take it from there
(it is long but if you are careful, you'll get there)

Answer 5

sin(2x) = 2*sin(x)*cos(x)
cos(2x) = [cos(x)]^2 - [sin(x)]^2 = 2*[cos(x)]^2 - 1

sin(2x)/sin(x) - cos(2x)/cos(x)
=2*sin(x)*cos(x)/sin(x) - (2*[cos(x)]^2 - 1)/cos(x)
=2*cos(x) - 2*cos(x) +1/cos(x)
=1/cos(x)
=sec(x)
Q.E.D.


sin(A+B) = sin(A)*cos(B) + cos(A)*sin(B)
A = x+y, B = x-y

sin(x+y)*cos(x-y) + cos(x+y)*sin(x-y)
=sin[(x+y)+(x-y)]
=sin(2x)
Q.E.D.