I have two questions how do you solve cos x=sqrt2-cosx on the interval of [0,2pi) and how do you verify sin^2(theta)/1-cos(theta)=1+cos(theta) without negating it to equal 1?
1.
cosx = √2 - cosx
2cosx = √2
cosx = √2/2 which is the same as 1/√2
if you are familiar with the 1,1,√2 right-angled triangle you should know that x = 45° or π/4 radians
(you can use your calculator to verify this)
but the cosine is positve in quadrants I and IV
so x = π/4 or x = 2π - π/4 = 7π/4
2.
For this one you should know that
sin^2 Ø + cos^2 Ø = 1 or
sin^2 Ø = 1 - cos^2 Ø
LS = sin^2Ø/(1- cosØ)
= (1 - cos^2 Ø)/(1-cosØ)
= (1+cosØ)(1-cos‚/(1-cosØ)
= 1 + cosØ
= RS
Thank you so much!
To solve the equation cos(x) = √2 - cos(x) on the interval [0, 2π), we can follow these steps:
Step 1: Move all the terms involving cos(x) to one side of the equation:
cos(x) + cos(x) = √2
Step 2: Combine the terms:
2cos(x) = √2
Step 3: Divide both sides of the equation by 2:
cos(x) = √2 / 2
Step 4: Determine the reference angle:
Since cos(x) = √2 / 2 is positive, the reference angle is π/4.
Step 5: Find the solutions in the given interval:
In the interval [0, 2π), the solutions for cos(x) = √2 / 2 are x = π/4 and x = 7π/4.
Now, regarding the second question, to verify sin^2(theta) / (1 - cos(theta)) = 1 + cos(theta) without negating it to equal 1, we will need to manipulate the equation using trigonometric identities:
Step 1: Start with the left-hand side of the equation:
sin^2(theta) / (1 - cos(theta))
Step 2: Apply the identity sin^2(theta) = 1 - cos^2(theta):
(1 - cos^2(theta)) / (1 - cos(theta))
Step 3: Factor out a common term from the numerator:
[(1 - cos(theta)) ∙ (1 + cos(theta))] / (1 - cos(theta))
Step 4: Cancel out the common term (1 - cos(theta)):
1 + cos(theta)
Therefore, the left-hand side of the equation simplifies to 1 + cos(theta), which is the same as the right-hand side. Hence, the equation sin^2(theta) / (1 - cos(theta)) = 1 + cos(theta) is verified.