Question

If $\cos \alpha =\frac{3}{5}$ and $\cos \beta =\frac{5}{13}$, then
(Note : $\alpha ,\beta$ are in the first quadrant)

• 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 $\cos (\alpha -\beta )=\frac{63}{65}$.
D ${\sin }^2\frac{\alpha -\beta }{2}=\frac{1}{65}$

$\cos \alpha =\frac{3}{5}\Rightarrow \sin \alpha =\frac{4}{5}$
And $\cos \beta =\frac{5}{13}\Rightarrow \sin \beta =\frac{12}{13}$
$\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta =\left(\frac{3}{5}\right)\left(\frac{5}{13}\right)-\left(\frac{4}{5}\right)\left(\frac{12}{13}\right)\therefore \cos (\alpha +\beta )=\frac{-33}{65}$
$\cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta =\left(\frac{3}{5}\right)\left(\frac{5}{13}\right)+\left(\frac{4}{5}\right)\left(\frac{12}{13}\right)\therefore \cos (\alpha -\beta )=\frac{63}{65}$
$\Rightarrow 1-2{\sin }^2\left(\frac{\alpha -\beta }{2}\right)=\frac{63}{65}\therefore {\sin }^2\left(\frac{\alpha -\beta }{2}\right)=\frac{1}{65}$
$\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta =\left(\frac{4}{5}\right)\left(\frac{5}{13}\right)+\left(\frac{3}{5}\right)\left(\frac{12}{13}\right)\therefore \sin (\alpha +\beta )=\frac{56}{65}$
Ans: B,C,D