if cosA-4sinA=1, then sinA+4cosA=??

Question

if cosA-4sinA=1, then sinA+4cosA=??

Answer 1

(cosA -4sinA)² = cos²A - 8cosAsinA + 16sin²A = 1

(sinA +4cosA)² = sin²A + 8cosAsinA +16cos²A = X

 

we add the two equations to get

 

17cos²A + 17sin²A = x + 1

17 (cos²A + sin²A) = x + 1

17 = x + 1

x = 16

 

Therefore sinA + 4cosA = √16 = ±4

Answer 2

Squaring cosA - 4sinA =1, we get

 

cos²A - 8cosAsinA + 16sin²A = 1.

 

Using cos²A = 1 - sin²A in the above equation, we get

 

15sin²A - 8cosAsinA = 0                                             (1)

 

Now,

 

(sinA + 4cosA)² = sin²A + 8cosAsinA + 16cos²A

  = sin²A + 8cosAsinA + 16(1 - sin²A)

  = 16 - 15sin²A + 8cosAsinA

  = 16 - ( 15sin²A - 8cosAsinA )

  = 16                                                  Using (1)

 

So, we have 

(sinA + 4cosA)² = 16

which implies

sinA + 4cosA = ±4