Question

Please help! (100 points!)

Answer 1

`y=(\pi)/(3)`

Explanation:

Given equation:

`\sin (x+y) =(1)/(2) \sin(x) +(√(3))/(2) \cos(x)`

`\boxed{\begin{minipage}{6 cm}\underline{Trigonometric Identity}\\\\$\sin (A \pm B) \equiv \sin A \cos B \pm \cos A \sin B$\\\end{minipage}}`

Use the sine trigonometric identity to rewrite sin(x+y) in terms of sin and cos:

`\implies \sin(x+y)= \sin (x) \cos (y) + \cos (x) \sin (y)`

Substitute this into the given equation:

`\begin{aligned}\sin (x+y) & =(1)/(2) \sin(x) +(√(3))/(2) \cos(x)\\\\\implies \sin (x) \cos (y) + \cos (x) \sin (y) & =(1)/(2) \sin(x) +(√(3))/(2) \cos(x)\end{aligned}`

Compare the left side of the equation with the right side:

`\implies \cos (y) = (1)/(2)`

`\implies \sin (y) = (√(3))/(2)`

Therefore, solving for y:

`\begin{aligned}\implies \cos (y) & = (1)/(2)\\y & = \cos^(-1)\left((1)/(2)\right)\\y & = (\pi)/(3)\end{aligned}`

`\begin{aligned}\implies \sin (y) & = (√(3))/(2)\\y & = \sin^(-1)\left((√(3))/(2)\right)\\y & = (\pi)/(3)\end{aligned}`

Answer 2

`\\ \rm\Rrightarrow sin(x+y)=(1)/(2)sinx+(√(3))/(2)cosx`

`\\ \rm\Rrightarrow sin(x+y)=sinxcos\left((\pi)/(3)\right)+cosxsin\left((\pi)/(3)\right)`

• sin(a+b)=sinacosb+cosasinb

`\\ \rm\Rrightarrow sin(x+y)=sin\left(x+(\pi)/(3)\right)`

`\\ \rm\Rrightarrow x+y=x+(\pi)/(3)`

`\\ \rm\Rrightarrow y=(\pi)/(3)`

Option B is correct