Thank You for taking the time to look at this.
If x = 2cos^2(theta) and y = sin2(theta), show that (x -1)^2 + y^2=1
I tried my best but I couldn't solve this! Where did you get this question from?
substitute:
(2cos^2(theta)-1)^2 + (sin2(theta))^2 = 1
then recall some identities:
*2cos^2(theta)-1=cos2(theta)
*cos2(theta)=cos^2(theta)-sin^2(theta)
*sin^2(theta)=1-cos^2(theta)
therefore, it will become:
(cos2(theta))^2 + (sin2(theta))^2=1
>>actually, here you can conclude they are equal because of the identity:
cos^2(theta) + sin^2(theta) = 1
anyway, continuing and substituting the 3rd identity to the equation,,
(cos2(theta))^2 + 1-(cos2(theta))^2 = 1
therefore, the cosines will cancel each other, leaving 1,,
i hope i was able to help.. =)
To show that (x - 1)^2 + y^2 = 1, we will start by substituting the given expressions for x and y into the equation and simplifying.
Given: x = 2cos^2(theta) and y = sin2(theta)
Substitute the values of x and y into (x - 1)^2 + y^2:
(2cos^2(theta) - 1)^2 + (sin2(theta))^2
Now, let's simplify each term at a time. Starting with (2cos^2(theta) - 1)^2:
(2cos^2(theta) - 1)^2
= (2cos^2(theta))^2 - 2(2cos^2(theta)) + 1
= 4cos^4(theta) - 4cos^2(theta) + 1
Next, we will simplify (sin2(theta))^2:
(sin2(theta))^2
= sin^2(2(theta))
Now, we can substitute these simplified terms back into the original equation:
(4cos^4(theta) - 4cos^2(theta) + 1) + sin^2(2(theta))
To make it easier to manipulate, let's express sin^2(2(theta)) using the double angle identity for sine:
sin^2(2(theta)) = (1 - cos(4(theta)))/2
Now, substitute sin^2(2(theta)) back into the equation:
(4cos^4(theta) - 4cos^2(theta) + 1) + (1 - cos(4(theta)))/2
To simplify further, let's combine like terms:
= 4cos^4(theta) - 4cos^2(theta) + 1 + 1/2 - cos(4(theta))/2
= 4cos^4(theta) - 4cos^2(theta) + 3/2 - cos(4(theta))/2
Finally, if we factor out a 2 from the first two terms and a 2/2 = 1 from the last two terms, we can rewrite it as:
2(2cos^4(theta) - 2cos^2(theta)) + 3/2 - cos(4(theta))/2
Now, notice that 2cos^4(theta) - 2cos^2(theta) can be expressed as cos^2(2(theta)). Using this, we have:
2(cos^2(2(theta))) + 3/2 - cos(4(theta))/2
We now use another trigonometric identity, cos(4(theta)) = 2cos^2(2(theta)) - 1, to further simplify:
2(cos^2(2(theta))) + 3/2 - (2cos^2(2(theta)) - 1)/2
Distribute the 2 to both terms:
2cos^2(2(theta)) + 3/2 - (4cos^2(2(theta)) - 2)/2
Combine like terms:
2cos^2(2(theta)) + 3/2 - 4cos^2(2(theta)) + 2)/2
Simplify further:
(-2cos^2(2(theta)) + 7/2)/2
Now, distribute the 1/2 to both terms:
-2cos^2(2(theta))/2 + 7/4
Simplify:
-cos^2(2(theta)) + 7/4
Finally, we need to express cos^2(2(theta)) in terms of x using the equation given:
cos^2(2(theta)) = (x - 1)/2
Substitute cos^2(2(theta)) back into the equation:
-(x - 1)/2 + 7/4
Now, simplify by combining the terms with a common denominator:
-(x - 1 + 7)/4
Combine like terms:
-(x + 6)/4
To further simplify, we can multiply the numerator by -1:
(-x - 6)/4
Finally, the expression (-x - 6)/4 is equal to 1 when they have the same value. Therefore, we have proven that:
(x - 1)^2 + y^2 = 1