Question

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.

Answer 1

See below.

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. )

Answer 2

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