SOLUTION: for what value of theta the expression is real and imaginary. 3+2*sin(theta)/1-2*sin(theta)

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Question 17687: for what value of theta the expression is real and imaginary.
3+2*sin(theta)/1-2*sin(theta)
Answer by xcentaur(357) About Me  (Show Source):
You can put this solution on YOUR website!
+%283%2B2%2Asin%28theta%29%29%2F%281-2%2Asin%28theta%29%29+
Now sin(theta) is 1 for 90 degrees
So if theta is 90 we get
+%283%2B2%2Asin%2890%29%29%2F%281-2%2Asin%2890%29%29+
+%283%2B2%29%2F%281-2%29+
+5%2F-1+
+-5+
Hence it is an imaginary (non-real) value.


Again,for theta=0,sin becomes 0
+%283%2B2%2Asin%280%29%29%2F%281-2%2Asin%280%29%29+
+%283%2B0%29%2F%281-0%29+
+3%2F1+
+3+
Which is real.


Hence for given expression we get,
Real values,theta=0 degreees
Imaginary values,theta=90 degrees


Hope this helps,
Prabhat