Question

no spamming.... please help ASAP​

Answer 1

Answer:

refer to the attachment

Explanation:

hope this helps

peacy^-^

Answer 2

$\underline\mathfrak{Answer}$

$Given, \\ sin \theta = \frac{1}{2}sinθ=21 \\ </p><p></p><p>Since, we \: know \: that, \\ </p><p></p><p>cos^2\theta = 1 - sin^2\theta \: cos2θ=1−sin2θ \\ </p><p></p><p>=1-(\frac{1}{2})^2 =1-\frac{1}{4} \\ =\frac{3}{4} =1−(21)2 \\ =1−41 \\ =43</p><p> \\ \\ </p><p>\implies cos \theta = \sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2} \\ ⟹cosθ=43=23 \\ </p><p></p><p>\implies cos^3 \theta = (\frac{\sqrt{3}}{2})^3=\frac{3\sqrt{3}}{3} \\ ⟹cos3θ=(23)3=833 \\ \\ </p><p></p><p>Thus, \\ </p><p></p><p>3 cos \theta - 4 cos^3 \theta3cosθ−4cos3θ \\ </p><p></p><p>=3\times \frac{\sqrt{3}}{2} - 4 \times \frac{3\sqrt{3}}{8} \\ =3×23−4×833 \\ </p><p></p><p>=\frac{3\sqrt{3}}{2}-\frac{3\sqrt{3}}{2} \\ =233−233 \\ </p><p></p><p>=0=0 \\ </p><p></p><p>Hence, proved.</p><p></p><p>$

His account got deleted yesterday_! so, he's on his new account named potato95. check my following list for that. Btw, he's offline now.

well,why you asked me about this?? ಠಿ_ಠ