Highschool Trig....?

Question

Hi. I need some help with this question. Could you explain how its done? It might be easier if you write it down properly but this is the question.

"Solve 1-sin^2 (theta) =1/2 for theta between or equal to 0 and 360 degrees"

Thanks

The answers in the book are 45, 135, 225 or 315 degrees. I just need to know how you get it

Answer 1

x = theta
1- Sin(x)^2 = cos(x)^2 (trig property)
thus
cos(x)^2 = 1/2
sqrt(cos(x)^2) = sqrt (1/2)
thus
cos(x) = 1/sqrt(2)

x = 45, 135, 215, 315, all values work since you have to sq the cos anyways (so the negative values caused by 215 and 315 will become positive .5 after being squared)

or in radii values

x = pi/4, 3pi/4, 5pi/4, 7pi/4

Answer 2

theta = t for short
1-sin^2(t) = .5
1-.5 = sin^2(t)
(.5)^.5=(sin^2(t))^.5

so now you should be asking yourself when is
sin(t) = (+ or -)1/(2)^.5
or if you do not know that then you can use the inverse function of sin(t) which is is sin^-1(t) on your calculator
so
sin^-1(sin(t)) = sin^-1(1/2^.5)
t=45
now we need to check quadrants so in quad 1 we have t = 45
quad 2 -> t= 135 or 45 + 90
quad 3 -> t = 225 0r 45 + 180
quad 4 -> t = 315 or 45 + (3 * 90)
notice that 1/2^.5 = .707......
sin(45) = .707....
sin(135) = .707...
sin(225) = -.707...
sin(315) = -.707...

now remember we are asking when is sin(t) = .707 so only when t = 45 and 135 degrees does 1-sin^2(t)=.5

now if you find it interesting about the sign flip this is why notice that we are dealing with a 45 degree angle meaning if we make a right triangle then two sides will be equal and we can use a^2 + b^2 = c^2 -> c = (a^2 + b^2)^.5

now save that equation in your head for a min and realize that the sin(t) = opposite side facing the angle / hypotenuse when dealing with right triangles so when you are searching for sin(t) = + or - (1/2^.5) that implies opposite side = 1 and hypotenuse = 2^.5 now that you have your triangle and you know all the sides using the equation a^2 + b^2 = c^2. now place this triangle on a cartesian coordinate graph. you can move the triangle to each quadrant and realize from the picture where the negative signs are going to pop up i.e. the bottom half. whille moving the triangle the 45 degree angle needs to be based on the origin.

hope this helps

Answer 3

Applying trigonometric identities:
cos²(θ) = 1/2
©
cosθ = ±√2 / 2

cosθ = √2 / 2 ....... θ = 45° , 315°
cosθ = -√2 / 2 ...... θ = 135° , 225°

θ = 45°, 135°, 225°, 315°

Answer 4

Let x= sin^2(theta).
Then you have: 1-x=1/2.
This gives x=1/2.
Therefore, sin^2(theta)=1/2
Take the square root of 1/2, which is 1/2 * square root of 2.
Then, theta=arcsine(1/2 * square root of 2)
Therefore, theta = 45 degrees.

Answer 5

1 - sin²θ = 1/2

Using the identity, Cos²θ + sin²θ = 1, we get,

Cos²θ = 1/2
Cosθ = +/-(1/√2)

θ = 45 , 135 (Cosθ = 1/√2)
θ = 225, 315 (Cosθ = -1/√2)