well, you are correct, in that
2θ = 231.655° or 308.345°
That makes θ = 115.83° or 154.17°
However, since we're dealing with 2θ, that gives us answers in the 0<θ<180 range. To get all the values up to 360, we need to add 360 to 2θ (or, add 180 to θ) to our answers to get all values for 0<2θ<720, giving us the other two answers.
Solve the equation on 0° ≤ θ < 360° and express in degrees to two decimal places.
sin(2θ) = -0.7843
I've gotten 308.345° (QIV) and 231.655° (QIII). I'm unsure how to get the answer though? The final answer is: 115.83°, 295.83°, 154.17°, 334.17°. Thank you to anyone who can help me!
2 answers
treat the 2Ø as one angle , let's say 2Ø = A
then sin A = -.7843
so we know that A is in III or IV , like you had
I then find the "angle in standard position" by finding
sin^-1 (+.7843) which is 51.67°
so in III, A = 180 + 51.66 = 231.66°
in IV , A = 360 - 51.66 = 308.34°
But A = 2Ø
2Ø = 231.66 -----> Ø = 115.83°
2Ø = 308.34 -----> Ø = 154.17°
Now we know that the period of sin 2Ø = 360/2 = 180
So by adding or subtracting 180 to any existing answer will yield a new answer
115.83 + 180 = 295.83°
154.17 + 180 = 334.17°
And that is how they got those 4 answers.
then sin A = -.7843
so we know that A is in III or IV , like you had
I then find the "angle in standard position" by finding
sin^-1 (+.7843) which is 51.67°
so in III, A = 180 + 51.66 = 231.66°
in IV , A = 360 - 51.66 = 308.34°
But A = 2Ø
2Ø = 231.66 -----> Ø = 115.83°
2Ø = 308.34 -----> Ø = 154.17°
Now we know that the period of sin 2Ø = 360/2 = 180
So by adding or subtracting 180 to any existing answer will yield a new answer
115.83 + 180 = 295.83°
154.17 + 180 = 334.17°
And that is how they got those 4 answers.
1 answer