Which series of transformations will ALWAYS form an image that is congruent to its pre-image? Select all that apply.
1 answer:
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Answer:
Step-by-step explanation:
We are given the trigonometric equation of:
Let u = 4x then:
Find a measurement that makes sin(u) = √3/2 true within [0, π) which are u = 60° (π/3) and u = 120° (2π/3).
Convert u-term back to 4x:
Divide both sides by 4:
The interval is given to be 0 ≤ 4x < π therefore the new interval is 0 ≤ x < π/4 and these solutions are valid since they are still in the interval.
Therefore:
