**How to Express Products of Trig Functions as Sum or Difference :**

Here we are going to see some example problems to show expressing the product of trig functions as sum or difference.

sin(A + B) = sin A cos B + cos A sin B -------(1)

sin(A − B) = sin A cos B − cos A sin B -------(2)

cos(A + B) = cos A cos B − sin A sin B ------(3)

cos(A − B) = cos A cos B + sin A sin B -------(4)

(1) + (2)

sin (A + B) + sin (A - B) = 2 sin A cos B

(1) - (2)

sin (A + B) - sin (A - B) = 2 cos A sin B

(3) + (4)

cos (A + B) + cos (A - B) = 2 cos A cos B

(3) - (4)

cos (A + B) - cos (A - B) = -2 sin A sin B

**Question 1 :**

Express each of the following as a sum or difference

(i) sin 35°cos 28°

**Solution :**

= sin 35°cos 28°

Multiply and divide the given trigonometric ratio by 2.

= (2/2) sin 35°cos 28°

= (1/2) (2 sin 35°cos 28°)

It exactly matches the formula 2 sin A cos B

2 sin A cos B = sin (A + B) + sin (A - B)

= (1/2) [sin (35°+28°) + sin (35°-28°)]

= (1/2) [sin 63°+ sin 7°]

(ii) sin 4x cos 2x

**Solution :**

= sin 4x cos 2x

Multiply and divide the given trigonometric ratio by 2.

= (2/2) sin 4x cos 2x

= (1/2) (2 sin 4x cos 2x)

It exactly matches the formula 2 sin A cos B

2 sin A cos B = sin (A + B) + sin (A - B)

= (1/2) [sin (4x+2x) + sin (4x-2x)]

= (1/2) [sin 6x + sin 2x]

(iii) 2 sin 10θ cos 2θ

**Solution :**

= 2 sin 10θ cos 2θ

It exactly matches the formula 2 sin A cos B

2 sin A cos B = sin (A + B) + sin (A - B)

= (1/2) [sin (10θ+2θ) + sin (10θ+2θ)]

= (1/2) [sin 12θ + sin 8θ]

(iv) cos 5θ cos 2θ

**Solution :**

= cos 5θ cos 2θ

Multiply and divide the given trigonometric ratio by 2.

= (2/2) cos 5θ cos 2θ

= (1/2) (2 cos 5θ cos 2θ)

It exactly matches the formula 2 cos A cos B

2 cos A cos B = cos (A + B) + cos (A - B)

= (1/2) [cos (5θ + 2θ) + cos (5θ - 2θ)]

= (1/2) [cos 7θ + cos 3θ]

(v) sin 5θ sin 4θ.

**Solution :**

= sin 5θ sin 4θ

Multiply and divide the given trigonometric ratio by 2.

= (-2/-2) sin 5θ sin 4θ

= (-1/2) (-2 sin 5θ sin 4θ)

It exactly matches the formula -2 sin 5θ sin 4θ

-2 sin A sin B = cos (A + B) - cos (A - B)

= (-1/2) [cos (5θ + 4θ) - cos (5θ - 4θ)]

= (-1/2) [cos 9θ - cos θ]

= (1/2)[cos θ - cos 9θ]

After having gone through the stuff given above, we hope that the students would have understood, "How to Express Products of Trig Functions as Sum or Difference"

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