Let $\alpha, \beta$ be such that $\pi<\alpha-\beta<3 \pi$. If $\sin \alpha+\sin \beta=-\frac{21}{65}$ and $\cos \alpha+\cos \beta=-\frac{27}{65}$, then the value of $\cos \frac{\alpha-\beta}{2}$ is

JEE Main · 2004 · 24 Apr 2004 · Offline · Trigonometric Ratios and Identities

Let

α, β

be such that

π < α - β < 3 π

. If

sin α + sin β = - \frac{21}{65}

and

cos α + cos β = - \frac{27}{65}

, then the value of

cos \frac{α - β}{2}

Answer: (A)

Apply the sum-to-product formulas:

sin α + sin β = 2 sin \frac{α + β}{2} cos \frac{α - β}{2} = - \frac{21}{65}

and

cos α + cos β = 2 cos \frac{α + β}{2} cos \frac{α - β}{2} = - \frac{27}{65}

. Squaring and adding these, the factor

sin^{2} \frac{α + β}{2} + cos^{2} \frac{α + β}{2} = 1

drops out and leaves

4 cos^{2} \frac{α - β}{2} = \frac{2 1 ^{2} + 2 7 ^{2}}{6 5 ^{2}} = \frac{1170}{4225}

, so

cos^{2} \frac{α - β}{2} = \frac{1170}{16900} = \frac{9}{130}

. Because

π < α - β < 3 π

, the half-angle lies in

(\frac{π}{2}, \frac{3 π}{2})

where cosine is negative, so

cos \frac{α - β}{2} = - \frac{3}{130}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option