Question

If 3 sin theta + 5cos theta =5 ,prove that 5 sin theta - 3 cos theta =+or - 3

Answer 1

Given:

(3 sinθ+5cosθ)²= 5²

Squaring on both sides.

(3sinθ)²+(5cosθ)²+2× 3sinθ 5cosθ= 25

[a+b= a²+b²+2ab]

9sin²θ+ 25cos²θ+30sinθcosθ= 25

9 (1-cos²θ) + 25(1-sin²θ)+30sinθcosθ=25

[sin²θ + cos²θ =1]

9-9cos²θ + 25-25sin²θ +30sinθcosθ=25

9+25 -(9cos²θ +25sin²θ -30sinθcosθ) =25

34 - (9cos²θ +25sin²θ -30sinθcosθ) =25

- (25sin²θ +9cos²θ-30sinθcosθ) =25-34

(25sin²θ+9cos²θ -30sinθcosθ) =9

(5sinθ - 3cosθ)²= 9

(5sinθ - 3cosθ)= √9

(5sinθ - 3cosθ)= ±3

L.H.S = R.H.S

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Hope this will help you...

Answer 2

Given :
3 sinθ + 5cosθ =5

To Find :
5Sinθ -3cosθ =±3

Solution :
3sinθ + 5cosθ =5

Squaring on both the sides we get,

(3sinθ + 5cosθ)² = 5²

9sin²θ + 25 cos²θ +30 sinθcos θ =25   [ ∵(a+b)² = a² + b² +2ab]

9(1-cos²θ) + 25 (1 -sin²θ) + 30sinθcosθ  =25

∵ sin²θ + cos²θ =1

9-9cos²θ + 25 -25sin²θ +30 sinθcosθ =25

-9cos²θ  -25sin²θ +30 sinθcosθ =25-34

-9cos²θ  -25sin²θ +30 sinθcosθ =-9

9cos²θ + 25sin²θ -30 sinθcosθ =9

25sin²θ + 9cos²θ -30 sinθcosθ=9

(5sinθ-3cosθ)² = 9

∵[(a-b)²= a²+b²-2ab] 

(5sinθ-3cosθ) = √9

(5sinθ-3cosθ) = ±3


Hence proved