F4 Trigo Question

(a)

P(n): sin@ + sin3@ + sin5@ + ... + sin(2n-1)@ = sin2n@/sin@

for all positive integers n, with sin@ ≠ 0


When n = 1:

L.S. = sin@

R.S. = sin2@/sin@ = sin@

L.S = R.S.

P(1) is true.


Assume that P(k) is true, i.e.

sin@ + sin3@ + sin5@ + ... + sin(2k-1)@ = sin2k@/sin@


When n = k + 1:

To prove that: sin@ + sin3@ + sin5@ + ... + sin(2k+1)@ = sin2(k+1)@/sin@

L.S.

= sin@ + sin3@ + sin5@ + ... + sin(2k-1)@ + sin(2k+1)@

= [sin@ + sin3@ + sin5@ + ... + sin(2k-1)@] + sin(2k+1)@

= sin2k@/sin@ + sin(2k+1)@sin@/sin@

= [sin2k@ + sin(2k+1)@sin@]/sin@

= [sin2k@ + sin(2k@ + @)sin@]/sin@

= [sin2k@ + (sin2k@cos@ + cos2k@sin@)sin@]/sin@

= [sin2k@ + sin2k@cos@sin@ + cos2k@sin2@]/sin@

= [sin2k@ + 2sink@cosk@)sin@cos@ + (cos2k@-sin2k@)(1 - cos2@)]/sin@

= [sin2k@ + 2sink@cosk@)sin@cos@ + cos2k@- sin2k@ - cos2k@cos2@ + sin2k@cos2@]/sin@

= [2sink@cosk@)sin@cos@ + cos2k@ - cos2k@cos2@ + sin2k@cos2@]/sin@

= [2sink@cosk@)sin@cos@ + cos2k@(1 - cos2@) + sin2k@cos2@]/sin@

= [2sink@cosk@)sin@cos@ + cos2k@sin2@ + sin2k@cos2@]/sin@

= [(sink@cos@)2 + 2(sink@cos@)(cosk@sin@) + (cosk@sin@)]/sin@

= [sink@cos@ + cosk@sin@]2/sin@

= [sin(k@ +@)]2/sin@

= sin2(k+1)@/sin@

= R.S.

Hence, P(k+1) is true when P(k) is true.


By the principle of mathematical induction,

P(n) is true for all positive integers n, with sin@ ≠ 0

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(b)

sin@ + sin3@ + sin5@ + ... + sin(2n-1)@ = sin2n@/sin@


When @ = (π/2)-x, and n = 5

sin[(π/2)-x] + sin3[(π/2)-x] + sin5[(π/2)-x] + sin7[(π/2)-x] + sin9[(π/2)-x] = sin25[(π/2)-x]/sin[(π/2)-x]

sin[(π/2)-x] + sin[(3π/2)-3x] + sin[(5π/2)-5x] + sin[(7π/2)-7x] + sin[(9π/2)-x] = sin2[(5π/2)-5x]/cosx

sin[(π/2)-x] + sin[(3π/2)-3x] + sin[(π/2)-5x] + sin[(3π/2)-7x] + sin[(π/2)-x] = sin2[(π/2)-5x]/cosx

cosx - cos3x + cos5x - cos7x + cos9x = cos25x/cosx
=

a. Let P(n) be the proposal "sin@ + sin3@ + sin5@ + ... + sin(2n - 1)@ = sin^2 n@ / sin@".

When n = 1, Left = sin@
Right = sin^2 @ / sin@ = sin@, thus Left = Right, i.e. P(1) is true.

Here, assuming P(k) is true where k is any positive integer,
i.e. sin@ + sin3@ + sin5@ + ... + sin(2k - 1)@ = sin^2 k@ / sin@

For P(k+1), Left = sin@ + sin3@ + sin5@ + ... + sin(2k - 1)@+ sin(2k + 1)@
= (sin^2 k@ / sin@) + sin(2k + 1)@
= [sin^2 k@ + sin(2k + 1)@ sin@]/ sin@
= [2sin^2 k@ + cos2k@ - cos(2k+2)@]/ 2sin@
= [2sin^2 k@ + cos2k@ - cos(2k+2)@]/ 2sin@
= [2sin^2 k@ + 1 - 2sin^2 k@ - cos(2k+2)@]/ 2sin@
= [1 - cos(2k+2)@]/ 2sin@
= sin^2 (k+1)@ / sin@ (By half angle formula) = Right.
So P(k+1) is true.

By mathematic induction,P(n) is true for all positive integers n.

*********
b. From(a),
sin@ + sin3@ + sin5@ + sin7@ + sin9@ = sin^2 5@ / sin@

Substituting @ = [(pi)/2-x],

sin[(pi)/2-x] + sin[3(pi)/2-3x] + sin[5(pi)/2-5x] + sin[7(pi)/2-7x] + sin[9(pi)/2-9x] = sin^2 [5(pi)/2-5x] / sin[(pi)/2-x]

cosx + sin[3(pi)/2-3x] + sin[(pi)/2-5x] + sin[3(pi)/2-7x] + sin[(pi)/2-9x] = sin^2 [(pi)/2-5x] / cosx
(Because sinx and cosx are periodic, 2pi is counted as one period of them)

cosx + sin[3(pi)/2-3x] + cos5x + sin[3(pi)/2-7x] + cos9x = cos^2 5x / cosx

cosx - sin[(pi)/2-3x] + cos5x - sin[(pi)/2-7x] + cos9x = cos^2 5x / cosx

cosx - cos3x + cos5x - cos7x + cos9x = cos^2 5x / cosx

Q.E.D.