Question

If $\theta ={18}^o$ , then what is the value of $4si{n}^2\theta +2sin\theta$ ?

• A. -1
• B. 1
• C. 0
• D. 2

Answer 1

Answer: (B)

1

To find the value of $sin{18}^o$

Let $A={18}^o$

Therefore, $5A={90}^o$

$2A={90}^o-3A$

Taking $\sin$ on both sides, we get

$\sin 2A=\sin ({90}^o-3A)=\cos 3A$

$2\sin A\cos A=4{\cos }^{3}A-3\cos A$

$2\sin A\cos A-4{\cos }^{3}A+3\cos A=0$

$\cos A(2\sin A-4{\cos }^{2}A+3)=0$

Dividing both sides by $\cos A$, we get

$2\sin A-4(1-{\sin }^{2}A)+3=0$

$4{\sin }^{2}A+2\sin A-1=0,$

$\Rightarrow 4{\sin }^{2}A+2\sin A=1$

Where $\sin A=\sin {18}^0$

Hence, $4{\sin }^2\theta +2\sin \theta =1$