Question

if 2sinA + 3cosA=2, then find 3sinA - 2cosA ?

Answer 1

Answer:

3sinA - 2cosA = ±3.

Step-by-step explanation:

Hi,

Let 3sinA - 2cosA = x,

Given that 2sinA + 3cosA = 2,

Now, Consider

(3sinA - 2cosA)² + (2sinA + 3cosA)²

= x² + 2²

= x² + 4

= 9sin²A - 12sinAcosA + 4cos²A

+ 4sin²A + 9cos²A + 12sinAcosA

which can be simplified as

= 13sin²A  + 13cos²A

= 13(sin²A  + cos²A),

But we know that from trigonometric identities

sin²A  + cos²A = 1,

Hence,

x² + 4 = 13

x² = 9

x = ±3

Hence, 3sinA - 2cosA = ±3.

Hope, it helps !

Answer 2

Answer:

3sinA - 2cosA = ±3

Step-by-step explanation:

if 2sinA + 3cosA=2, then find 3sinA - 2cosA ?

2sinA + 3cosA=2

squaring both side

(2sinA + 3cosA)² = 2²

=> 4Sin²A + 9Cos²A + 12SinACosA = 4

=> 12SinACosA = 4 - 4 Sin²A - 9Cos²A

=> 12SinACosA = 4(1 -  Sin²A) - 9Cos²A

as Cos²A + Sin²A = 1

=> 12SinAcosA = 4Cos²A - 9Cos²A

=> 12SinACosA = -5Cos²A - eq 1

=> 5Cos²A + 12SinACosA = 0

=> CosA(5CosA + 12SinA) = 0

CosA = 0 or CosA = -12SinA/5

when CosA = 0

2sinA + 0 = 2 => SinA = 1

so 3sinA - 2cosA = 3 - 0 = 3

3sinA - 2cosA = 3

when

CosA = -12SinA/5

2SinA + 3(-12SinA/5) = 2

=> 10SinA - 36SinA = 10

=> -26SinA = 10

=> -13SinA = 5

=> SinA = -5/13

CosA = -12SinA/5

=> CosA = -12 * (-5/13) /5 = 12/13

=> CosA = 12/13

3sinA - 2cosA

= 3(-5/13) -2(12/13)

= -15/13 -24/13

= -39/13

= -3

3sinA - 2cosA = -3

3sinA - 2cosA = ±3

Another way to solve without finding value of SinA & CosA

let say

3sinA - 2cosA = x

Squaring both sides

(3sinA - 2cosA)² = x²

=> x² = 9Sin²A + 4Cos²A - 12SinACosA

putting value of 12SinACosA from Eq 1

=> x² = 9Sin²A + 4Cos²A - (-5Cos²A)

=> x² = 9Sin²A + 4Cos²A + 5Cos²A

=> x² = 9 (Sin²A + Cos²A)

=> x² = 94

=> x = ±3

=> 3sinA - 2cosA = ±3