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Need a trig explanation?
I came across questions that might be asked on my Trig Final that I have no idea how to do. If you could tell me how to do these? An example problem:
Given that cos x + i sin y = sin x + i, find all correct x and y values between 0 and 2pi.
@Maurice
That response doesn't really tell me how to do these problems at all...
2 Answers
- DragonLv 67 years agoFavorite Answer
Here's what I did:
a) Rewrite equation to put one unknown (and i) on one side to get
cos(x) - sin(x) = i - i[sin(y)]
b) square both sides to get
cos^2(x) - 2(cos(x))(sin(x)) + sin^2(x) = i^2 - 2(i^2)(sin(y)) + i^2(sin^2(y))
c) substitute 1 for cos^2 + sin^2 on the left side; and i^2 = [sqrt(-1)]^2 or -1 on the right side to get
1 - 2(cos(x))(sin(x)) = -1 + 2(sin(y)) - 1(sin^2(y) = - (1 - sin(y))^2
d) multiply both sides of the equation by -1 and reverse left and right sides of equation to get
(1 - sin(y))^2 = 2(cos(x))(sin(x))
e) take sqrt of both sides to get 1 - sin(y) = sqrt[2(cos(x))(sin(x)]
f) subtract 1 from each side to get - sin(y) = sqrt[2(cos(x))(sin(x)] -1
g) divide both sides by -1 to get sin(y) = 1 - sqrt[2(cos(x))(sin(x)]
h) y = sin^-1 [1 - sqrt[2(cos(x))(sin(x)]
i) input above equation into graphing calculator and use trace function to input x values to get desired y values or manually input x values to get corresponding y values
Source(s): Good luck, Dragon - mauriceLv 68 years ago
equate real and imaginary parts
cosx = sinx; x = pi/4, pi+pi/4
siny = 1; y =pi/2