1.

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

Answer»

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.


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