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add the sum and difference identities for sin x to derive 1/2 [sin(x+y)+sin(x-y)]=sin x cos y

Sagot :

Answer:

true

Step-by-step explanation:

1/2 (sin(x+y)+sin(x-y))

= 1/2 (sin x cos Y + cos x sin y + sin x cos y - cos x sin y)

= 1/2 * 2 sin x cos y

= sin x cos y

The expression is true.

The trigonometric identity 1/2 [sin (x + y)+sin (x - y)] = sin x cos y holds true, using the sum and difference identities for sin x.

What are the sum and difference identities for trigonometric ratios?

The sum and difference identities for trigonometric ratios are as follows:-

  • sin (x + y) = sin x cos y + cos x sin y
  • sin (x - y) = sin x cos y - cos x sin y
  • cos (x + y) = cos x cos y - sin x sin y
  • cos (x - y) = cos x cos y + sin x sin y
  • tan (x + y) = (tan x + tan y)/(1 - tan x tan y)
  • tan (x - y) = (tan x - tan y)/(1 + tan x tan y).

How to solve the question?

In the question, we are asked to add the sum and difference identities for sin x to derive 1/2 [sin (x + y)+sin (x - y)] = sin x cos y.

Going by the left-hand side of the identity:-

1/2 [sin (x + y) + sin (x - y)],

Putting sin (x + y) = sin x cos y + cos x sin y, and sin (x - y) = sin x cos y - cos x sin y, we get:

= 1/2 [(sin x cos y + cos x sin y) + (sin x cos y - cos x sin y)]

= 1/2 [sin x cos y + cos x sin y + sin x cos y - cos x sin y]

= 1/2 [ 2 sin x cos y] {Since, cos x sin y is cancelled}

= sin x cos y

= The right-hand side of the identity.

Hence, the identity is derived.

Learn more about trigonometric identities at

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