PROVING SUMS IN TRIGONOMETRY FOR GRADE 11

Example 1 :

If a cos θ − b sin θ = c, show that

a sin θ + b cos θ  =  ± √a² + b² − c².

Solution :

Let (a cos θ − b sin θ)²   ----(1)

(a sin θ − b cos θ)²   ----(2)

By expanding (1) using algebraic identity, we get

(a cos θ−b sin θ)²  =  (a cos θ)² + (b sin θ)² + 2ab sin θcos θ

 =  a² cos² θ + b² sin² θ + 2ab sin θcos θ  ---(1)

By expanding (2) using algebraic identity, we get

(a sin θ−b cos θ)²  =  (a sin θ)² + (b cos θ)² - 2ab sin θcos θ

 =  a² sin² θ + b² cos² θ - 2ab sin θcos θ ---(2)

(1) + (2)  

  =  a² cos² θ + a² sin² θ + b² sin² θ + b² cos² θ 

  =  a² (cos² θ + sin² θ) + b² (cos² θ + sin² θ)

(a cos θ − b sin θ)² + (a sin θ − b cos θ)²   =  a² + b²

(a sin θ − b cos θ)²   =  a² + b²  - (a cos θ − b sin θ)² 

Here the value of a cos θ − b sin θ is c.

(a sin θ − b cos θ)²   =  a² + b²  - c² 

(a sin θ − b cos θ)   =  ± √a² + b²  - c² 

Example 2 :

If sin θ + cos θ = m, show that

cos⁶ θ + sin⁶ θ = 4 − 3 (m² − 1)²/4, where m² ≤ 2.

Solution :

L.H.S  =  cos⁶ θ + sin⁶ θ  =  (cos² θ)³ + (sin² θ)³

a³ + b³  =  (a + b)³ - 3ab (a + b) 

(cos²θ)³+(sin²θ)³

  =  (cos² θ + sin² θ)³ - 3 sin² θcos² θ(cos² θ + sin² θ)

  =  (1)³ - 3 sin² θ cos² θ(1)

  =  1 - 3 sin² θ cos² θ  -------(1)

  =  1 - 3 (sin θ cos θ)²  -------(1)

sin θ + cos θ = m

Taking squares on both sides, we get

(sin θ + cos θ)²  =  m²

sin² θ + cos² θ + 2sin θ cos θ  =  m²

1 + 2sin θ cos θ  =  m²

sin θ cos θ  =  (m² - 1)/2

Applying the value of sin θ cos θ in (1)

  =  1 - 3 ((m² - 1)/2)²

  =  1 - [3 (m² - 1)²/4]

  =  [4 - 3 (m² - 1)²]/4

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