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. 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

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.)