# 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)

By expanding (1) using algebraic identity, we get

(a cos θ−b sin θ)=  (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 + b- (a cos θ − b sin θ)

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

(a sin θ − b cos θ)2    a2 + b2  - c

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

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

=  (cosθ + sinθ)3 - 3 sinθcosθ(cosθ + sin2 θ)

=  (1)3 - 3 sinθ cosθ(1)

=  1 - 3 sinθ cosθ  -------(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 ((m- 1)/2)2

=  1 - [3 (m- 1)2/4]

=  [4 - 3 (m- 1)2]/4

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