Respuesta :
[tex]\bf \textit{Sum and Difference Identities}
\\ \quad \\
sin({{ \alpha}} + {{ \beta}})=sin({{ \alpha}})cos({{ \beta}}) + cos({{ \alpha}})sin({{ \beta}})
\\ \quad \\
\boxed{sin({{ \alpha}} - {{ \beta}})=sin({{ \alpha}})cos({{ \beta}})- cos({{ \alpha}})sin({{ \beta}})}[/tex]
[tex]\bf cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}}) \\ \quad \\ cos({{ \alpha}} - {{ \beta}})= cos({{ \alpha}})cos({{ \beta}}) + sin({{ \alpha}})sin({{ \beta}})\\\\ -------------------------------\\\\ sin(48^o)cos(15^o)-cos(48^o)sin(15^o)\implies sin(48^o+15^o) \\\\\\ sin(63^o)[/tex]
[tex]\bf cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}}) \\ \quad \\ cos({{ \alpha}} - {{ \beta}})= cos({{ \alpha}})cos({{ \beta}}) + sin({{ \alpha}})sin({{ \beta}})\\\\ -------------------------------\\\\ sin(48^o)cos(15^o)-cos(48^o)sin(15^o)\implies sin(48^o+15^o) \\\\\\ sin(63^o)[/tex]
Answer:
The expression in terms of the sine expression is given by:
[tex]\sin 48\cos 15-\cos 48\sin 15=\sin 33\degree[/tex]
Step-by-step explanation:
The expression is given as:
[tex]\sin 48\cos 15-\cos 48\sin 15[/tex]
Now we know that the formula is as follows:
[tex]\sin \alpha \cos \beta-\cos \alpha \sin \beta=\sin (\alpha-\beta)[/tex]
Here on comparing the given expression with the above formula we have:
[tex]\sin 48\cos 15-\cos 48\sin 15=\sin (48-15)[/tex]
i.e.
[tex]\sin 48\cos 15-\cos 48\sin 15=\sin 33[/tex]