SIN(α + β); SINα = 4/5, α IN QUADRANT II, TAN β = -√3/2, β IN QUADRANT IV.

Question

Can someone please show me how to find the exact value and the appropriate identity?

Answer 1

Hint: In all trigonometric problems we use the trigonometric unit property:

                                                             sin²α + cos² α = 1                                               (1)

To find sin of the sum of two angles we can use the formula:

                                             sin (α + β) = sinα∙cosβ + cosα∙sinβ                                 (2)

We know that sin α = 4/5 and the angle is in the II quadrant, hence cos α is also positive in this quadrant. To find its value we use trigonometric unit and can write:

                                                        cos α = √(1 - sin²α)                                                 (3)

Square the value of sin α, plug it in the equation (3) and you will get cos α = 3/5.

To find cos β, we can use the formula

                                                        1 + tan² β = 1/cos² β                                              (4)

Angle beta is in the IV quadrant, thus its value is positive. Putting the square of the tan β in equation (4), and solving it against cos β, you can get

                                                                      cos β = 2 / √7

Using trigonometric unit (like in equation 1), you can easy find, that

                                                                       sin β = - √(3/7).

Its value is negative because angle beta - in the IV quadrant.

Now if you put all values in equation (2), you will find:

                                       sin (α + β) = (4/5)∙(2/√7) + (3/5)∙(- √(3/7) = (8 - 3√3) / (5√7).

                                                                                    Answer: (8 - 3√3) / (5√7)