Question

If 3sin (theta)+4cos(theta)=5 then find the value of 4sin(theta)-3cos(theta)

Answer 1

$3 \sin \theta + 4 \cos\theta = 5 ..... < 1 > \\ \\ let \\ \\ 4 \sin\theta - 3 \cos\theta = x...... < 2 > \\ \\ squaring \: and \: adding \: 1 \: and \: 2 \\ \\ 9 { \sin}^{2} \theta + 16 { \cos}^{2} \theta + 24 \sin\theta \cos\theta = 25 \\ \\ 16 { \sin}^{2} \theta + 9 { \cos}^{2} \theta - 24 \sin\theta \cos\theta = {x}^{2} \\ \\ 9 {( { \cos}^{2}\theta + { \sin}^{2} \theta) } + 16{( { \cos}^{2}\theta + { \sin}^{2} \theta) } = 25 + {x}^{2} \\ \\ x = 0 \\ \\ 4 \sin\theta - 3 \cos\theta =0$

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