HOW TO EXPRESS PRODUCTS OF TRIG FUNCTIONS AS SUM OR DIFFERENCE

About "How to Express Products of Trig Functions as Sum or Difference"

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

Expressing Products of Trig Functions as Sum or Difference

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 + ) + cos (5θ - )]

  =  (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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