Question

5sinx+4cosx=3 then 4sinx-5cosx= ?

Answer 1

I hope this answer is correct

Answer 2

Answer:

4sinx-5cosx=0

Step-by-step explanation:

Taking, $5sinx+4cosx=3$

⇒$\frac{5}{3}sinx+\frac{4}{3}cosx=1$

Let $cosA=\frac{5}{3}$ and $sinA=\frac{4}{3}$

⇒$cosAsinx+sinAcosx=1$

⇒$sin(x+a)=1$                                                                                 (1)

Also, $4sinx-5cosx$

Multiplying and dividing by 3, we have

$3(\frac{4}{3}sinx-\frac{5}{3}cosx)$

=$3(sinAsinx-cosAcosx)$

=$-3(cos(x+A))$

=$-3(\sqrt{1-sin^{2}(x+A)})$

using equation (1),

=$-3(\sqrt{1-1})$

=$0$

therefore,  $4sinx-5cosx$=$0$