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Calculus/Trigonometric reduction

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Expert: - 12/5/2008


Question
I am trying to figure out how to use the Reduction Identity to write the following in terms of sine only:

f(x) = -√2 sin2x + √2 cos2x

only the beginning "2's" have the square root, not the whole term.

Answer
These are the equations that will be used
[A] sinē(y) + cosē(y) = 1,
[B] sin(2x) = 2sin(x)cos(x), and
[C] cos(2x) = 1 - 2sinē(x).

These are derived in most trigonometry courses
and are known to be true.

For the problem, factor the √2 out front.

Case 1 (uses [A] only): Reduce the equation to be in terms of sin(2x)

 Put in 2x for y in [A] and solve for the value of the cos(2x).
 The result should be cos(2y) = √[1-sinē(x)].

 Substitue this back into f(x) and this will leave
 the f(x) with only sin(2x) in it and no cos(2x).

Case 2 (uses [A], [B] and [C]):
Reduce the equation to be in terms of sin(x)

 Note that from [A], we have [D] cos(x) = [1 - sinē(x)]^0.5.

 Using [D], we can rewrite [B] so that we now
 have [B] in terms of sin(x) only.

 Using what we just got for [B] and also [C],
 f(x) can be rewritten to be in terms of sin{x} only.

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