Question

Sin B = 1/2 . Show that
3CosB-4CosB3(cube)=0

Answer 1

Given:

$\sin B$ = $\dfrac{1}{2}$

We have to show that $3\cos B-4\cos^3 B=0.$

Solution:

∴ $\sin B$ = $\dfrac{1}{2}=\dfrac{p}{h}$

Here, p = perpendicular = 1 and h = hypotaneous = 2

By Pythagoras Theorem,

Base, b = $\sqrt{h^{2}-p^{2}}$

= $\sqrt{2^{2}-1^{2}}$

= $\sqrt{4-1}$

= $\sqrt{3}$

$\cos B$ = $\dfrac{b}{h}=\dfrac{\sqrt{3}}{2}$

L.H.S. = $3\cos B-4\cos^3 B$

= $3(\dfrac{\sqrt{3}}{2})-4(\dfrac{\sqrt{3}}{2})^3$

= $(\dfrac{3\sqrt{3}}{2})-4(\dfrac{3\sqrt{3}}{8})$

= $(\dfrac{3\sqrt{3}}{2})-(\dfrac{3\sqrt{3}}{2})$

= 0

= R.H.S., proved.

Thus, if $\sin B$ = $\dfrac{1}{2}$, then $3\cos B-4\cos^3 B=0$, shown.