Question

if sin theta = 1/2 . Show that 3 cos theta minus 4 cos cube theta =0

Answer 1

Let theta be ₹.
sin ₹ = 1/2
sin ₹ = sin 30°
₹ = 30°
to prove:-
3cos₹ - 4cos^3₹ = 0
3cos30° - 4cos^3 30° = 0
$3( \sqrt{3} /2) - 4( \sqrt{3} /2)^{3} = 0$
0 = 0
Hence Proved

Answer 2

Answer:

Given, $sin \theta = \frac{1}{2}$

Since, we know that,

$cos^2\theta = 1 - sin^2\theta$

$=1-(\frac{1}{2})^2=1-\frac{1}{4}=\frac{3}{4}$

$\implies cos \theta = \sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$

$\implies cos^3 \theta = (\frac{\sqrt{3}}{2})^3=\frac{3\sqrt{3}}{8}$

Thus,

$3 cos \theta - 4 cos^3 \theta$

$=3\times \frac{\sqrt{3}}{2} - 4 \times \frac{3\sqrt{3}}{8}$

$=\frac{3\sqrt{3}}{2}-\frac{3\sqrt{3}}{2}$

$=0$

Hence, proved.