when solving for all solutions how do I work out this problem: cos4(alpha)=cos2(alpha)
and when verifying each identity how do I complete this problem: sin (alpha/2)=sec alpha -1/2sin (alpha/2)sec alpha
cos4x = cos2x
2cos^2(2x)-1 = cos2x
2cos^2(2x)-cos(2x)-1 = 1
(2cos2x+1)(cos2x-1) = 0
so, cos2x = -1/2 or 1
cos 2pi/3 = cos 4pi/3 = -1/2
cos 0 = 1
as for the identity, I get the feeling there's a typo, because secx appears twice on the right.
To solve the equation cos(4α) = cos(2α) and find all solutions, you can use trigonometric identities to simplify the equation and then solve for α. Here's a step-by-step approach:
Step 1: Apply the double-angle identity for cosine:
cos(4α) = 2cos^2(2α) - 1
Step 2: Substitute cos(2α) in the equation with u to simplify:
2u^2 - 1 = u
Step 3: Rearrange the equation to form a quadratic equation:
2u^2 - u - 1 = 0
Step 4: Solve the quadratic equation:
You can solve this equation by factoring or by using the quadratic formula.
If you factor the equation, you will get: (2u + 1)(u - 1) = 0.
This gives you two potential solutions:
2u + 1 = 0, which leads to u = -1/2
u - 1 = 0, which leads to u = 1
Step 5: Replace u with cos(2α):
Putting the values of u back into the equation, you get:
cos(2α) = -1/2 ... (1)
cos(2α) = 1 ... (2)
Step 6: Solve for α for each equation:
(1) Solving for cos(2α) = -1/2:
We use the inverse cosine function and find two solutions for 2α:
2α = ±2π/3 + 2πk, where k is an integer.
Dividing both sides by 2, we get:
α = ±π/3 + πk, where k is an integer.
(2) Solving for cos(2α) = 1:
We use the inverse cosine function and find the solution:
2α = 2πk, where k is an integer.
Dividing both sides by 2, we get:
α = πk, where k is an integer.
Therefore, the solutions to the given equation cos(4α) = cos(2α) are:
α = πk, or α = ±π/3 + πk, where k is an integer.
For verifying the identity sin(α/2) = sec(α) - 1/2sin(α/2)sec(α), you need to show that the equation holds for all valid values of α. Here's how to approach it:
Step 1: Simplify the right-hand side of the equation:
sec(α) - 1/2sin(α/2)sec(α)
Step 2: Use the definitions of secant and sine in terms of cosine:
1/cos(α) - 1/2sin(α/2)(1/cos(α))
Step 3: Combine the fractions over a common denominator:
(1 - sin(α/2)/2) / cos(α)
Step 4: Simplify the fraction:
(2 - sin(α/2)) / 2cos(α)
Step 5: Use the double-angle identity for sine:
(2 - 2sin^2(α/2)) / 2cos(α)
Step 6: Simplify further:
2(1 - sin^2(α/2)) / 2cos(α)
Step 7: Apply the pythagorean identity for sine:
2cos^2(α/2) / 2cos(α)
Step 8: Cancel out the common terms:
cos(α/2) / cos(α)
Step 9: Use the reciprocal identity for cosine:
sec(α/2) / sec(α)
Step 10: Since the left-hand side is sin(α/2), we can rewrite the equation as:
sin(α/2) = sec(α/2) / sec(α)
Step 11: The equation is now in simplest form, and you have verified the identity.
Note: It's important to keep in mind the restrictions on the domain for which the identity holds true. Trigonometric identities might have specific restrictions on the values of α for which they are valid.