Question

If $\cos \alpha =\frac{3}{5}$ and $\cos \beta =\frac{5}{13}$ , then

• A. $\cos (\alpha +\beta )=\frac{33}{65}$
• B. $\sin (\alpha +\beta )=\frac{56}{65}$
• C. ${\sin }^2\frac{\alpha -\beta }{2}=\frac{1}{65}$
• D. $\cos (\alpha -\beta )=\frac{63}{65}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $\sin (\alpha +\beta )=\frac{56}{65}$
C ${\sin }^2\frac{\alpha -\beta }{2}=\frac{1}{65}$
D $\cos (\alpha -\beta )=\frac{63}{65}$

$cos(\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta$
$=\frac{3}{5}\times \frac{5}{13}-\sqrt{1-(\frac{3}{5}{)}^2}\sqrt{1-(\frac{5}{13}{)}^2}$
$=\frac{15}{65}-\frac{48}{65}$
$cos(\alpha +\beta )=-\frac{33}{65}$
$\sin (\alpha +\beta )=\sqrt{1-\left(\cos \right(\alpha +\beta ){)}^2}=\sqrt{1-(\frac{-33}{65}{)}^2}$
$\sin (\alpha +\beta )=\frac{56}{65}$ --------------B
$\cos (\alpha -\beta )=\frac{15}{65}+\frac{48}{65}=\frac{63}{65}$ - ---------------------D
$\therefore {\sin }^2\left(\frac{(\alpha -\beta )}{2}\right)=\frac{1-\cos (\alpha -\beta )}{2}=\frac{1-\frac{63}{65}}{2}=\frac{1}{65}$ ------------------- C