Trigonometric Proof:

Question

Question:

If $m\cos\alpha-n\sin\alpha=p$ then prove that $m\sin\alpha+n\cos\alpha=\pm \sqrt{m^2+n^2-p^2}$

My Efforts:

$(m\cos\alpha-n\sin\alpha)^{2}=p^2$

$m^2\cos^2\alpha+n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=p^2$

Now i think we need to add something on both side but i can't figure out what to add

Answer 1

$$m^2\cos^2\alpha+n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=p^2$$

multiply by $-1$ and add $m^2+n^2$ to both sides :

$$m^2+n^2-m^2\cos^2\alpha-n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=m^2+n^2-p^2$$ $$m^2(1-\cos^2\alpha)+n^2(1-\sin^2\alpha)-2mn\cos\alpha\ \sin\alpha=m^2+n^2-p^2$$ $$m^2(\sin^2\alpha)+n^2(\cos^2\alpha)-2mn\cos\alpha\ \sin\alpha=m^2+n^2-p^2$$ $$(m\sin\alpha+n\cos\alpha)^2=m^2+n^2-p^2$$ $$m\sin\alpha+n\cos\alpha=\pm \sqrt{m^2+n^2-p^2}$$

Answer 2

HINT:

$$(m\cos\alpha-n\sin\alpha)^2+(m\sin\alpha+n\cos\alpha)^2=?$$

See also : Brahmagupta–Fibonacci identity

Answer 3

From $$m^2\cos^2\alpha+n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=p^2$$

we can add $m^2\sin^2\alpha+n^2\cos^2\alpha$ to both sides (to make use of the $\sin^2x+\cos^2x=1$ identity) to achieve

$$m^2(\sin^2\alpha+\cos^2\alpha)+n^2(\sin^2\alpha+\cos^2\alpha)-2mn\cos\alpha\ \sin\alpha=p^2+m^2\sin^2\alpha+n^2\cos^2\alpha \\\iff m^2+n^2-p^2=m^2\sin^2\alpha+n^2\cos^2\alpha+2mn\cos\alpha\ \sin\alpha=(m\sin\alpha+n\cos\alpha)^2$$ The result follows.