Trigonometric functions and their properties

In the second quadrant, (\cos x) is negative. From (\sin^2 x+\cos^2 x=1), (\cos x=-\frac{12}{13}).

In the third quadrant, both (\sin x) and (\cos x) are negative, while (\tan x) is positive. Hence (\sin x=-\frac{4}{5}).

The standard triple-angle identity is \(\sin 3x=3\sin x-4\sin^3 x\), so option B is correct. Option C is the negative of option B and hence equals \(-\sin 3x\). Option D is related to the triple-angle formula for cosine, not sine. Exam tip: remember that the \(\sin^3 x\) term in \(\sin 3x\) has coefficient 4 and a minus sign.

The triple angle formula is (\cos 3x=4\cos^3 x-3\cos x). In exams, do not confuse it with (\sin 3x).

(\sin x) is positive in the first and second quadrants. Hence (x=\frac{\pi}{6},\frac{5\pi}{6}).

(\cos x) is negative in the second and third quadrants. Hence (x=\frac{2\pi}{3},\frac{4\pi}{3}).

The maximum value of (a\sin x+b\cos x) is (\sqrt{a^2+b^2}). So (\sqrt{5^2+(-12)^2}=13).

The period of (\sin 2x) is (\pi) and the period of (\sin 4x) is (\frac{\pi}{2}). Their common fundamental period is (\pi).

Since (\sin x(1+\sin x)=0), (\sin x=0) or (\sin x=-1). In the given interval, (x=0,\pi,\frac{3\pi}{2}).

In the fourth quadrant, (\sin \theta) is negative and (\cos \theta) is positive. Hence (\sin \theta=-\frac{15}{17}) and (\tan \theta=-\frac{15}{8}).