Guys please help me with this Trigonometry question based on De Moivre's theorem.
Q: Find ¦È such that 0¡Ü¦È¡Ü360.
cos7¦È + cos3¦È = cos5¦È
I attempted to solve it with limited knowledge but I kind of doubt my answer. Anyway, this is how I did it :
1) [cos7¦È + cos3¦È] - cos5¦È = 0
2) (cos5¦È x cos3¦È) - cos5¦È = 0
[Application of cos (C) and cos (D)
= cos5(C+D) x cos5(C-D)]
3) cos5¦È[cos3¦È - 1] = 0
4) => either cos5¦È = 0 or cos3¦È = 1
5) So, when cos5¦È = 0, ¦È = (2n ¡À 1)(¦Ð/4) wher n is an integer and when cos¦È = 1
¦È=2n¦Ð where n is an integer.
I'm stuck here. Please show me step by step how to solve this. :/
I am going to rephrase your question, without fancy symbols
cos(7E) + cos(3E) = cos(5E) , where 0 ≤ E ≤ 360°
Correction on the formula you are using:
should say:
cosA + cosB = 2cos((A+B)/2)cos((A-B)/2)
so cos(7E) + cos(3E)
= 2cos(5E)cos(2E)
but cos(7E) + cos(2E) = cos(5E) <----- given
thus:
2cos(5E)cos(2E) = cos(5E)
2cos(5E)cos(2E) - cos(5E) = 0
cos 5E (2cos 2E - 1) = 0
cos 5E = 0 or cos 2E = 1/2
case1,
if cos 5E = 0
5E = 90° or 5E = 270°
E = 18° or E = 54°
The period of cos 5E is 360/5° = 72°
so adding multiples of 72 to each answer yields a new answer.
E = 18°, 90° 162°, 234°, 306°, 54°, 126°, 198°, 270°, 342°
case 2
cos 2E = 1/2
2E = 60° or 2E = 300°
E = 30° or E = 150°
to get the other angles, proceed in the same way I showed you for case1
To solve the equation cos(7¦È) + cos(3¦È) = cos(5¦È), you can use De Moivre's theorem along with some trigonometric identities. Here's a step-by-step guide to solving this equation:
1) Start with the equation cos(7¦È) + cos(3¦È) = cos(5¦È).
2) Apply the trigonometric identity: cos(A) + cos(B) = 2cos((A+B)/2)cos((A-B)/2). This simplifies the equation to:
2cos((7¦È + 3¦È)/2)cos((7¦È - 3¦È)/2) = cos(5¦È).
3) Simplify further. Combine like terms within the cosines:
2cos(5¦È)cos(2¦È) = cos(5¦È).
4) Divide both sides by cos(5¦È) to isolate the product of the two cosines:
2cos(2¦È) = 1.
5) Divide both sides by 2:
cos(2¦È) = 1/2.
6) Solve for the principal values of ¦È. Take the inverse cosine (also known as arccos) of both sides to isolate ¦È:
2¦È = arccos(1/2).
7) Use the inverse cosine function to find the principal angle ¦È:
¦È = arccos(1/2) / 2.
Note: Since the question specifies that 0 ≤ ¦È ≤ 360, you should consider the principal values in the first quadrant and the corresponding angles in the other quadrants.
8) Use a calculator or reference table to find the value of arccos(1/2) (which is π/3 or 60°):
¦È = (π/3) / 2.
9) Divide π/3 by 2:
¦È = π/6.
So, the solution to the equation cos(7¦È) + cos(3¦È) = cos(5¦È) with 0 ≤ ¦È ≤ 360 is ¦È = π/6 or 30° (since radians and degrees can both be used in trigonometry).
Keep in mind that there may be additional solutions beyond the principal values, such as when adding multiples of 2π or 360° to the angle ¦È.