Expert: Josh - 3/23/2007

Question
Can you please help me in solving these trigonometric problems :

a)Verify the following identities :

ì) sin2x = (2tanx )( 1+ tan^2 x )

ìì) [( sin2theta/ sin theta )-( cos theta /cos theta )] =
   sec theta

ììì) sin(x+y)cos(x-y)+cos(x+y)sin(x-y) = sin 2x

ìv)  cos2x = (cot^2 x-1)/cot^2 x+1

v) Derive the identity for sin 3x in terms of sin x


Answer
Hi Robin,

Here are some pointers.

For part a (i):
You can use the t-substitution technique, where t=tan(x). From simple geometry, it can be shown that sin(2x)=2t/(1+t^2)=2tan(x)/(1+tan^2(x)).
Work through the tutorial at http://www.geocities.com/joshcameron_ae/Archive/T0001.html You will find further explanation there.

ii) To understand what I am about to show you, you need to first complete the tutorial at the said website.
The derivation produces the following identities:
sin(2x)=2t/(1+t^2), cos(2x)=(1-t^2)/(1+t^2),
sin(x)=t/sqrt(1+t^2), cos(x)=1/sqrt(1+t^2).

To keep things simple, I rewrite "theta" as "x". Using t-substitution,
sin(2x)/sin(x)-cos(2x)/cos(x)
=[2t/(1+t^2)]/[t/sqrt(1+t^2)]-[(1-t^2)/(1+t^2)]/[1/sqrt(1+t^2)]
=[2t/(1+t^2)]/[tsqrt(1+t^2)/(1+t^2)]-[(1-t^2)/(1+t^2)]/[sqrt(1+t^2)/(1+t^2)] ...<note 1>
=(2t)/[t*sqrt(1+t^2)]-(1-t^2)/sqrt(1+t^2)
=(2t-t+t^3)/[t*sqrt(1+t^2)]...<note2>
=t(1+t^2)/[t*sqrt(1+t^2)]...<note3>
=sqrt(1+t^2)...<note4>
=sec(x) as required.

<note 1> In the previous line, for fractions in the denominator, I multiplied top and bottom by sqrt(1+t^2). The whole expression then takes the form (a/b)/(c/b)-(d/b)/(e/b) which is equivalent to (a/c)-(d/e).
<note 2> for the second term, I multiplied both the numerator and denominator by t, so that the two fractions have a common denominator.
<note 3> factorizing here
<note 4> this is the inverse of cos(x)=1/sqrt(1+t^2)

iii) This uses two basic trig identities:
sin(x+y)=sin(x)cos(y)+cos(x)sin(y)
cos(x+y)=cos(x)cos(y)-sin(x)sin(y)
From these, it follows that
sin(x-y)=sin(x)cos(y)-cos(x)sin(y)
cos(x-y)=cos(x)cos(y)+sin(x)sin(y)
LHS = [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)]
=Sx*Cx*[Cy]^2+[Sx]^2*Sy*Cy+[Cx]^2*Sy*Cy+Sx*Cx*[Sy]^2 +Cx*Sx*[Cy]^2-[Cx]^2*Sy*Cy-[Sx]^2*Sy*Cy+Sx*Cx*[Sy]^2 ...<note 1>
=Sx*Cx*[Cy]^2+Sx*Cx*[Sy]^2+Cx*Sx*[Cy]^2+Sx*Cx*[Sy]^2 ...<note 2>
=2*Sx*Cx*(cos^2(y)+sin^2(y))
=2sin(x)cos(x) since cos^2(y)+sin^2(y)=1
=sin(2x)

<note 1> Abbreviation: Sx=sin(x),Cx=cos(x) etc.
<note 2> Observe that 2nd term cancels with 7th term, while 3rd term cancels with 6th term

iv) You can use t-substitution, applying it to the RHS. Let t=tan(x). Then, cot(x)=1/tan(x)=1/t.
RHS=((1/t^2)-1)/((1/t^2)+1)
  =(1-t^2)/(1+t^2)

v) Recognize that sin(3x)=sin(2x+x)
Using sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
sin(3x)=sin(2x)cos(x)+cos(2x)sin(x)
...
I'll let you finish this yourself.

Hints: Substitute sin(2x)=2sin(x)cos(x). Also, cos(2x)=cos^2(x)-sin^2(x)=1-2sin^2(x). Don't forget that cos^2(x)=1-sin^2(x)