Question

The value of ${\cos }^3{40}^{\circ }-3\sin {50}^{\circ }$ is equal to

Answer 1

$\Rightarrow {\cos }^3{40}^o-3\sin {50}^o$

$\Rightarrow {\cos }^3{40}^o-3\sin (90-40)$

$\Rightarrow {\cos }^{3}40-3\cos 40$

Let $\cos 40=\cos x$

by complementing trigonometric function

We can write $\cos (n\frac{\pi }{2}\pm \theta )=\cos \theta$

Now if $\cos x=\cos 40$

then $\cos 3x=\cos 120$

Now $\Rightarrow \cos 3x=cos(2x+x)=\cos 2x\cos x-\sin 2x\sin x$

$\Rightarrow \cos 120=({\cos }^{2}x-{\sin }^{2}x)\cos x-2{\sin }^{2}x\cos x$

$\Rightarrow \frac{-1}{2}={\cos }^{3}x-(1-{\cos }^{2}x)\cos x-2{\sin }^{2}x\cos x$

$\Rightarrow \frac{-1}{2}={\cos }^{3}x-\cos x+{\cos }^{3}x-2(1-{\cos }^{2}x)\cos x$

$\Rightarrow \frac{-1}{2}=2{\cos }^{3}x-\cos x-2\cos x+2{\cos }^{3}x$

$\Rightarrow 8{\cos }^{3}x-6\cos x+1=0$

Solving this we get

$\cos x=-0.94,\mathrm{0.174.0.766}$

as $\cos x$ cannot be negative between $[0,\frac{\pi }{2}]$ and always decreasing i.e $\cos 45<\cos 40$ and $\cos 45=0.707$

$\therefore \cos x=\cos 40=0.766$

$\therefore {\cos }^{3}40-3\cos 40=-1.848$