Question

If sin a = 1/2and a is an acute angle, find the value of
(3 cos a - 4cos cube a).​

Answer 1

Solution:-

GIVEN:-

$\rm \implies \sin a \: = \dfrac{1}{2}$

TO FIND THE VALUE OF

$\rm \implies \: 3 \cos a - {4} \cos {}^{3} a$

NOW TAKE

$\rm \implies \sin a \: = \dfrac{1}{2} = \dfrac{p}{h}$

WE HAVE

$\rm \implies \: p = 1,h = 2 \: and \: b = x$

USING PYTHAGORAS THEOREM

$\rm \implies \: {h}^{2} = {b}^{2} + {p}^{2}$

$\rm \implies \: {2}^{2} = {b}^{2} + {1}^{2}$

$\rm \implies \: 4 = {b}^{2} + 1$

$\rm \implies \: {b}^{2} = 3$

$\rm \implies \: b = \sqrt{3}$

NOW WE GET

$\rm \implies \: p = 1,h = 2 \: and \: b = \sqrt{3}$

SO

$\rm \implies \: \cos a = \dfrac{ \sqrt{3} }{2} = \dfrac{b}{h}$

WE HAVE TO FIND THE VALUE OF

$\rm \implies \: 3 \cos a - {4} \cos {}^{3} a$

$\rm \implies \: 3 \times \dfrac{ \sqrt{3} }{2} - 4 \times \bigg( \dfrac{ \sqrt{3} }{2} \bigg)^{3}$

$\rm \implies \: \dfrac{3 \sqrt{3} }{2} - 4 \times \dfrac{3 \sqrt{3} }{8}$

$\rm \implies \: \dfrac{3 \sqrt{3} }{2} - \cancel4 \times \dfrac{3 \sqrt{3} }{ \cancel8}$

$\rm \implies \: \dfrac{3 \sqrt{3} }{2} - \dfrac{3 \sqrt{3} }{2} = 0$

ANSWER IS 0