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27 July, 09:45

1. Derive this identity from the sum and difference formulas for cosine:

sin a sin b = (1 / 2) [cos (a - b) - cos (a + b) ]

Calculation:

1.

2.

3.

Reason:

1.

2.

3.

2. Use the trigonometric subtraction formula for sine to verify this identity:

sin ((π / 2) - x) = cos x

Calculation:

1.

2.

3.

Reason:

1.

2.

3.

+1
Answers (2)
  1. 27 July, 11:14
    0
    1. sin a sin b = (1 / 2) [cos (a - b) - cos (a + b) ]

    Calculation:

    Taking L. H. S. of above equation

    (1 / 2) [cos (a - b) - cos (a + b) ]

    = (1 / 2) [ (cos a cos b + sin a sin b) - (cos a cos b - sin a sin b) ]

    {∵ cos (a - b) = cos a cos b + sin a sin b & cos (a + b) = cos a cos b - sin a sin b}

    = (1 / 2) [ cos a cos b + sin a sin b - cos a cos b + sin a sin b]

    = (1 / 2) [2 sin a sin b]

    = sin a sin b

    2. sin ((π / 2) - x) = cos x

    Calculation:

    sin ((π / 2) - x) = sin (π / 2) cos x - cos (π / 2) sin x

    {∵sin (a - b) = sin a cos b - cos b sin a

    & sin (π / 2) = 1 & cos (π / 2) = 0}

    = 1 * cos x - 0 * sin x

    = cos x
  2. 27 July, 13:21
    0
    See below.

    Step-by-step explanation:

    1. (1 / 2) [cos (a - b) - cos (a + b) ]

    = 1/2 (cosa cosb + sina sinb - (cosa cosb - sina sinb)

    = 1/2 (cosa cosb - cosa cos b + sina sinb + sina sinb)

    = 1/2 (2 sina sinb)

    = sina sinb.

    (I used the 2 identities cos (a - b) = cosa cosb + sina sinb) and

    cos (a + b) = cosa cosb - sina sinb.)

    2. sin (π/2 - x) = sin (π/2) cos x - cos (π/2) sin x

    = 1 * cos x - 0 * sinx

    = cosx - 0

    = cos x.

    (I used the identity sin (a - b) = sina cosb - cosa sinb

    and the fact that sin (π/2) = 1 and cos (π/2) = 0.)
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