Question

Use an addition or subtraction formula to find the exact value of sin(19π/12)=−\sqrt{A}(\sqrt{B}+1)/4.
A,B=?

Answer 1

Answer:

$B = {(\frac{4A \times sin( \frac{19π}{12})}{\sqrt{A}})}^{2} + 1$

Step-by-step explanation:

$sin(19π/12)=−\sqrt{A}(\sqrt{B}+1)/4A \to \\ \\ sin( \frac{19π}{12})=− \frac{\sqrt{A}(\sqrt{B}+1)}{4A}

\\ \\ \sqrt{A}(\sqrt{B}+1) = 4A \times sin( \frac{19π}{12}) \\ \\ (\sqrt{B}+1) = \frac{4A \times sin( \frac{19π}{12})}{\sqrt{A}} \\ \\ B = {(\frac{4A \times sin( \frac{19π}{12})}{\sqrt{A}})}^{2} + 1$

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