Question : If $\sin (\theta +18^{\circ})=\cos 60^{\circ}(0< \theta < 90^{\circ})$, then the value of $\cos 5\theta$ is:
Option 1: $\frac{1}{2}$
Option 2: $0$
Option 3: $\frac{1}{\sqrt{2}}$
Option 4: $1$
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Correct Answer: $\frac{1}{2}$
Solution : $\sin (\theta +18^{\circ})=\cos 60^{\circ}$ We know that $\cos (90^{\circ}-\theta)=\sin\theta$ $⇒\sin (\theta +18^{\circ})=\cos(90^{\circ}-30^{\circ})$ $⇒\sin (\theta +18^{\circ})=\sin30^{\circ}$ $⇒\theta +18^{\circ}=30^{\circ}$ $⇒\theta=12^{\circ}$ $\therefore \cos 5\theta=\cos 5×12^{\circ}=\cos 60^{\circ}=\frac{1}{2}$ Hence, the correct answer is $\frac{1}{2}$.
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Question : If $\sin 2\theta=\frac{\sqrt{3}}{2}$, then the value of $\sin 3\theta$ is equal to $(0^{\circ}\leq \theta\leq 90^{\circ})$:
Option 2: $1$
Option 3: $0$
Option 4: $\frac{\sqrt{3}}{2}$
Question : If $0^{\circ} < \theta < 90^{\circ}$ and $2 \sin^{2}\theta +3\cos\theta =3$, then the value of $\theta$ is:
Option 1: 30°
Option 2: 60°
Option 3: 45°
Option 4: 75°
Question : If $(4 \sin \theta+5 \cos \theta)=3$, then the value of $(4 \cos \theta-5 \sin \theta)$ is:
Option 1: $3 \sqrt{2}$
Option 2: $4 \sqrt{2}$
Option 3: $2 \sqrt{3}$
Option 4: $2 \sqrt{5}$
Question : If $\cos A+\sin A=\sqrt{2}\cos A$, then $\cos A-\sin A$ is equal to: (where $0^{\circ}< A< 90^{\circ}$)
Option 1: $\sqrt{2}\sin A$
Option 2: $2\sin A$
Option 3: $2\sqrt{\sin A}$
Option 4: $\sqrt{2\sin A}$
Question : If $\frac{\cos \theta}{(1+\sin \theta)}+\frac{\cos \theta}{(1-\sin \theta)}=4$ and $\theta$ is acute, then the value of $\theta$ is:
Option 1: $60^{\circ}$
Option 2: $15^{\circ}$
Option 3: $45^{\circ}$
Option 4: $30^{\circ}$
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