how do you solve: sin(2arctan(-24/7) - arccos (-20/29)) how do you solve: sin(2a
ID: 2838082 • Letter: H
Question
how do you solve: sin(2arctan(-24/7) - arccos (-20/29))
how do you solve: sin(2arctan(-24/7) - arccos (-20/29)) I have to find the exact value and I have to show all my work. I know that this problem uses the sum and difference formula of Sin(u-v) = sin(u)cos(v)-cos(u)sin(v). For the most part I am knowledgable on how to solve this problem, however the ''2'' infront of the arctan is what is throwing me off. I have started with 2tan Theta = -24/7 then divide by 2 to get rid of the 2 and come with tan Theta = -24/14 and then I proceed to solve from there but I do not know if this is the right approach or not. If someone could please help I would greatly appreciate it! Thank you in advance!Explanation / Answer
sin(2arctan(-24/7) - arccos (-20/29))
=
sin(2arctan(-24/7))cos(arccos (-20/29)) - cos(2arctan(-24/7))sin(arccos (-20/29))
=
BY triangle equality, we find the values
sin(2arcsin(24/25)) * (-20/29) - cos(2arccos(-7/25))sin(arcsin(21/29))
=
2sin(arcsin(24/25)) * cos(arcsin(24/25)) * (-20/29) - (2 * 49/625 - 1) * 21/29
=
2 * 24/25 * cos(arccos(7/25)) * (-20/29) - (98/625 - 1) * 21/29
=
2 * 24/25 * 7/25 * (-20/29) - (-527/625) * 21/29
=
-6720/18125 + 11067/18125
=
4347/18125
I HOPE IT IS CORRECT....PLZ CHK CALCULATIONS
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