Question : If $\sin (p+q)=1$ and $\cos (p-q)=\frac{\sqrt{3}}{2}$, then find $p$.
Option 1: 90°
Option 2: 80°
Option 3: 60°
Option 4: 120°
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Correct Answer: 60°
Solution : Given: $\sin (p+q)=1$ and $\cos (p-q)=\frac{\sqrt{3}}{2}$ $\sin (p+q)=\sin 90°$ and $\cos (p-q)=\cos 30°$ ⇒ $(p+q)=90°$ ------------(1) And $(p-q)=30°$ --------- (2) Adding equations (1) and (2), we have, ⇒ $2p=120°$ ⇒ $p=60°$ Hence, the correct answer is 60°.
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Question : If $\sin A-\cos A=\frac{\sqrt{3}-1}{2}$, then the value of $\sin A\cdot \cos A$ is:
Option 1: $\frac{\sqrt{3}}{2}$
Option 2: $\frac{3}{2}$
Option 3: $\frac{\sqrt{3}}{4}$
Option 4: $\frac{1}{\sqrt{3}}$
Question : If $\sin(A+B)=\sin A\cos B+\cos A \sin B$, then the value of $\sin75°$ is:
Option 1: $\frac{\sqrt{3}+1}{\sqrt{2}}$
Option 2:
$\frac{\sqrt{2}+1}{2\sqrt{2}}$
Option 3:
$\frac{\sqrt{3}+1}{2\sqrt{2}}$
Option 4:
$\frac{\sqrt{3}+1}{2}$
Question : If $\sin m+\sin n=p, \cos m+\cos n=q$, then find the value of $\sin m \times \sin n+\cos m \times \cos n$.
Option 1: $p^2+q^2-2$
Option 2: $\frac{p^2+q^2-2}{2}$
Option 3: $p+q-p q$
Option 4: $p+q+p q$
Question : If $\theta$ is an acute angle and $\sin \theta \cos \theta=2 \cos ^3 \theta-\frac{1}{4} \cos \theta$, then the value of $\sin \theta$ is:
Option 1: $\frac{\sqrt{15}-1}{8}$
Option 2: $\frac{\sqrt{15}-1}{4}$
Option 3: $\frac{\sqrt{15}+1}{4}$
Option 4: $\frac{\sqrt{15}-1}{2}$
Question : If $\sin \alpha=\frac12$ and $\sin \beta=\frac12$, then what is the value of $\cos (\alpha+\beta)$? $(0°<\alpha, \beta<90° )$
Option 2: $\frac{1}{ 4}$
Option 3: $\frac12$
Option 4: $1$
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