Question

Please help me out with these questions. Its trigonometry.

Answer 1

Answer:  i) θ = 30°, 60°, 210°, & 240°

              ii) θ = 20° &  200°

Step-by-step explanation:

i) sin (2θ) = cos 30°

$\sin(2\theta)=\dfrac{\sqrt3}{2}\\\\.\quad 2\theta=\sin^{-1}\bigg(\dfrac{\sqrt3}{2}\bigg)\\\\.\quad 2\theta=60^o\qquad 2\theta=120^o\\\\.\quad \theta=30^o\qquad \theta=60^o$

To include all of the solutions for one rotation, add 360/2 = 180 to the solutions above.   θ = 30°, 60°, 210°, 240°

If you need ALL of the solutions (more than one rotation), add 180n to the solutions.   θ = 30° + 180n   &   60° + 180n

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ii) cos α = sin (50 + α)

Use the Identity: cos α = sin (90 - α)

Use Transitive Property to get:   sin (50° + α) = sin (90° - α)

50° + α = 90° - α

50° + 2α = 90°

        2α = 40°

          α = 20°

To find all solutions for one rotation, add 360/2 = 180 to the solution above.

α = 20°, 200°

If you need ALL of the solutions (more than one rotation), add 180n to the solution. α = 20° + 180n