Question

sin(θ + 45◦
) = 2 cos(θ − 30◦
), solve the equation

Answer 1

\[\sin \left( A+B \right)=\sin A \cos B+\cos A \sin B\]
\[\cos \left( A-B \right)=\cos A \cos B+\sin A \sin B\]

Answer 2

\[\sin(θ + 45^0 ) = 2 \cos(θ − 30^0) \]
\[\sinθ \cos45^0+\cosθ \sin45^0 = 2 [\cosθ \cos30^0+ \sinθ \sin30^0]\]
\[\sinθ \frac{1}{\sqrt2}+\cosθ \frac{1}{\sqrt2} = 2 [\cosθ \frac{\sqrt 3}{2}+ \sinθ \frac{1}{2}]\]
\[\frac{1}{\sqrt2}[\sinθ +\cosθ] =2 \times \frac{1}{2} [{\sqrt 3}\cosθ + \sinθ ]\]
\[[\sinθ +\cosθ] = \sqrt2 [{\sqrt 3}\cosθ + \sinθ ]\]
\[\sinθ +\cosθ = {\sqrt 6}\cosθ + \sqrt2 \sinθ \]
\[\sinθ -\sqrt2 \sinθ = \sqrt 6\cosθ -\cosθ \]
\[(1-\sqrt2) \sinθ = (\sqrt 6-1)\cosθ \rightarrow \frac {\sin θ } {\cos θ } =\frac {(\sqrt 6-1) } {(1-\sqrt2)}\]
i.e. \[\tan θ =\frac {(\sqrt 6-1) } {(1-\sqrt2)}\]

Answer 3

\[\sin 45=\frac{ 1 }{ \sqrt{2} },\cos 45=\frac{ 1 }{\sqrt{2} }\]
\[\sin 30=\frac{ 1 }{2 },\cos 30=\frac{ \sqrt{3} }{ 2 }\]