Question

if cosα-4sinα=1 ,then sinα+4cosα=?

Answer 1

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Let the result be 'y'

$cos \alpha - 4 sin \alpha = 1$

$sin \alpha + 4 cos \alpha = y$

Square the equations

$(cos \alpha - 4 sin \alpha)^2 = 1$

$(sin \alpha + 4 cos \alpha)^2 = y^2$

Add the equations

$(cos \alpha - 4 sin \alpha)^2 + (sin \alpha + 4 cos \alpha)^2 = 1 + y^2$

$cos^2 \alpha - 8 sin \alpha cos \alpha + 16 sin^2 \alpha + sin^2 \alpha + 8 cos \alpha sin \alpha + 16 cos^2 \alpha = y^2 + 1$

$17 sin ^2 \alpha + 17 cos^2 \alpha = y^2 + 1$

$17 (sin^2 \alpha + cos^2 \alpha) = y^2 + 1$

$[ sin^2 \alpha + cos^2 \alpha = 1 ]$

$17(1) = y^2 + 1$

$16 = y^2$

$y = \sqrt {16}$

$y = 4$

Answer 2

Hey !!!

easy question xD ..

cos@ - 4sin@ = 1 ----1)

sin@ + 4 cos@ = x -----2) ✔let sin@ + 4cos@ = x

Adding both equation and squaring then .

(cos@ - 4sin@ )² + (sin@ + 4cos@ )²

cos²@ + 16sin²@ - 2*4sin@*cos@ + sin²@ + 16cos²@ + 2* 4sin@*cos@

cos²@ + 16sin²@ + sin²@ + 16cos²@

sin²@ + cos²@ + 16 cos²@ + 16sin²@

1 + 16(sin²@ + cos²@ )

1 + 16×1

17....... ------3)

again ,

squaring both above equation and adding them...

(cos@ - 4sin@ )² + ( cos@ + 4cos@ ) ² = 17
(from 3equation )

1² + (x) ² = 17 (from 1 and 2 equation )

x² = 17 - 1

x² = 16

x = √16

x = 4 .

hence , sin@ + 4cos@ = 4 Answer

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Hope it helps you !!!

@Rajukumar111