Get the solution to "Find the value(s) of θ such that: sin⁡(θ+30∘)=cos⁡50∘"

Ashleigh Mitchell

Answered question

2022-04-05

Find the value(s) of

θ

such that:

\sin (θ + 30^{\circ}) = {\cos 50}^{\circ}

prangijahnot

Beginner2022-04-06Added 17 answers

Method 1:

\sin (θ + 30^{\circ}) = {\cos 50}^{\circ}

\Rightarrow \cos (90^{\circ} - (θ + 30^{\circ})) = {\cos 50}^{\circ}

\Rightarrow \cos (60^{\circ} - θ) = {\cos 50}^{\circ}

\Rightarrow \cos (\frac{π}{3} - θ) = \cos \frac{5 π}{18}

Writing the general solution as follows

\frac{π}{3} - θ = 2 n π \pm \frac{5 π}{18}

\Rightarrow θ = \frac{π}{3} - (2 n π \pm \frac{5 π}{18})

titanspokey20urvn

Beginner2022-04-07Added 4 answers

Method 2: ,

\sin (θ + 30^{\circ}) = {\cos 50}^{\circ}

\Rightarrow \sin (θ + 30^{\circ}) = \sin (90^{\circ} - 50^{\circ})

\Rightarrow \sin (θ + 30^{\circ}) = {\sin 40}^{\circ}

\Rightarrow \sin (θ + \frac{π}{6}) = \sin \frac{2 π}{9}

Writing the general solution as follows

θ + \frac{π}{6} = 2 n π + \frac{2 π}{9}

\Rightarrow θ = 2 n π + \frac{2 π}{9} - \frac{π}{6}

\Rightarrow θ = 2 n π + \frac{π}{18}

θ + \frac{π}{6} = (2 n + 1) π - \frac{2 π}{9}

\Rightarrow θ = 2 n π + π - \frac{2 π}{9} - \frac{π}{6}

\Rightarrow θ = 2 n π + \frac{11 π}{18}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?