If sin (alpha - beta ) = frac{1}{2} and cos ( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $\sin (\alpha - \beta ) = \frac{1}{2}$. Since $\sin 30^\circ = \frac{1}{2}$,we have $\alpha - \beta = 30^\circ$ $(i)$.Given $\cos (\alpha +

Vedclass · Accepted Answer

$\alpha = 45^\circ, \beta = 15^\circ$

Vedclass · Answer

(A) Given $\sin (\alpha - \beta ) = \frac{1}{2}$. Since $\sin 30^\circ = \frac{1}{2}$,we have $\alpha - \beta = 30^\circ$ $(i)$.Given $\cos (\alpha + \beta ) = \frac{1}{2}$. Since $\cos 60^\circ = \frac{1}{2}$,we have $\alpha + \beta = 60^\circ$ $(ii)$.Adding equations $(i)$ and $(ii)$ gives $2\alpha = 90^\circ$,so $\alpha = 45^\circ$.Substituting $\alpha = 45^\circ$ into $(ii)$,we get $45^\circ + \beta = 60^\circ$,so $\beta = 15^\circ$.Thus,$\alpha = 45^\circ$ and $\beta = 15^\circ$.

If $\sin (\alpha - \beta ) = \frac{1}{2}$ and $\cos (\alpha + \beta ) = \frac{1}{2},$ where $\alpha$ and $\beta$ are positive acute angles,then:

Similar Questions

If $\left[1-\cos \left(\frac{\pi}{2}+\alpha\right)+\sin \left(\frac{3 \pi}{2}+\alpha\right)\right]^2+\left[1-\sin \left(\frac{3 \pi}{2}-\alpha\right)-\cos \left(\frac{3 \pi}{2}+\alpha\right)\right]^2=a+b \sin ^2\left(\frac{\pi}{4}+\alpha\right)$,then $a^2+b^2=$

If $x = \tan 15^{\circ}$,$y = \operatorname{cosec} 75^{\circ}$ and $z = 4 \sin 18^{\circ}$,then :

If $a = \sin 175^{\circ} + \cos 175^{\circ}$,then

In a right-angled triangle $ABC$ where $\angle C = 90^{\circ}$,$AC = BC$. If $D$ is the midpoint of $AC$,then the cotangent of $\angle DBC$ is equal to:

Find the value of the trigonometric function $\sin \left(-\frac{11 \pi}{3}\right)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module