Prove that type question of Trigonometric Identities

Question

If $3\sin A + 5\cos A = 5$, then prove that: $$5\sin A + 3\cos A = ±3.$$

Answer 1

The question as stated seems to be incorrect. Most likely it is asking for $$\color{blue}{5 \sin A} \color{red}{-} \color{blue}{3 \cos A}.$$ If such is the case then from the first equation if we divide throughout by $\sqrt{34}$, we get $$\frac{3}{\sqrt{34}} \sin A + \frac{5}{\sqrt{34}} \cos A=\frac{5}{\sqrt{34}}.$$ Let $\sin \alpha=\frac{3}{\sqrt{34}}$, then we get $$\cos(A-\alpha)=\frac{5}{\sqrt{34}}.$$ But then we also get $$\cos(A-\alpha)=\frac{5}{\sqrt{34}}=\cos \alpha.$$ This implies that $$A-\alpha = 2n \pi \pm \alpha, \qquad \text{ for } n \in \mathbb{Z}.$$ Thus $$A=2n \pi \qquad \text{ or } \qquad A=2n\pi+2\alpha.$$

Let $\color{blue}{5 \sin A} \color{red}{-} \color{blue}{3 \cos A}=x$.

1. If $A=2n \pi$, then $x=-3$.
2. If $A=2n\pi+2\alpha$, then $5 \sin A \color{red}{-} 3 \cos A=5 \sin 2\alpha \color{red}{-} 3 \cos 2\alpha$.

Using the values of $\sin \alpha$ and $\cos \alpha$, we get $x=3$.

Thus $x = \pm 3$.

NOTE: If the actual problem (with $+$ sign) stands as is then $x=-3$ or $x=\frac{99}{17}$.

Answer 2

HINT. Try squaring the first identity and then use $\sin^2A+\cos^2A=1$