1. Which is a solution to the equation below? sin2xsecx=1
2. What is the exact value of cos15 degrees?
3. Which expression is equivalent to the expression below? cos^2 xsin^2?
4. What is the exact value of cos2theta? one side is 8 and the other is 15
5. The angle x lies in quadrant ll. If sinx=3/5, then what is the exact value of tan2x?
PLEASE HELP ME OUTTTTTTTTTTTTTTTTTTT
in #2 I had my formula backwards
should say
cos 2A = 2cos^2 A - 1
cos 30 = 2cos^2 15 - 1
1. To solve the equation sin^2(x)sec(x) = 1, we can rewrite the equation using the trigonometric identities. The identity sec(x) = 1/cos(x) can be substituted into the equation.
So, sin^2(x) * (1/cos(x)) = 1
Simplifying this equation, we get sin^2(x) / cos(x) = 1.
Now, let's further simplify the equation using the identity sin^2(x) = 1 - cos^2(x).
Substituting this into the equation, we have (1 - cos^2(x)) / cos(x) = 1.
Expanding the numerator, we have (1 / cos(x)) - (cos^2(x) / cos(x)) = 1.
Simplifying the equation, we get 1 / cos(x) - cos(x) = 1.
To solve for cos(x), we can multiply the entire equation by cos(x) to eliminate the denominator.
This gives us 1 - cos^2(x) = cos(x).
Rearranging the equation, we have cos^2(x) + cos(x) - 1 = 0.
This is a quadratic equation in terms of cos(x). We can solve this by factoring or using the quadratic formula to find the solutions for cos(x).
2. To find the exact value of cos(15 degrees), we can use the trigonometric identity cos(2θ) = 2cos^2(θ) - 1. We can rewrite cos(15 degrees) as cos(2 * 7.5 degrees).
Using the double angle formula, we have cos(2 * 7.5 degrees) = 2cos^2(7.5 degrees) - 1.
Since cos(15 degrees) lies in the first quadrant, we can find the exact value of cos(15 degrees) by knowing the value of cos(7.5 degrees).
To find cos(7.5 degrees), we can use the half-angle formula cos(θ/2) = sqrt((1 + cos(θ))/2).
Substituting cos(θ) with cos(15 degrees) in the equation, we have cos(7.5 degrees) = sqrt((1 + cos(15 degrees))/2).
Now, we can use the equation cos(15 degrees) = 2cos^2(7.5 degrees) - 1 and the equation for cos(7.5 degrees) to ultimately find the exact value of cos(15 degrees).
3. The expression cos^2(x)sin^2(x) can be simplified using the trigonometric identity sin^2(x) = 1 - cos^2(x).
Substituting this identity into the expression, we have cos^2(x)(1 - cos^2(x)).
Expanding this expression, we get cos^2(x) - cos^4(x).
Therefore, the expression cos^2(x)sin^2(x) is equivalent to cos^2(x) - cos^4(x).
4. To find the exact value of cos(2θ), we need to use the Pythagorean identity and the given information.
Using the Pythagorean identity sin^2(θ) + cos^2(θ) = 1, we have sin^2(θ) = 1 - cos^2(θ).
Substituting this identity into the equation cos(2θ) = 8/15, we get cos^2(θ) - (1 - cos^2(θ)) = 8/15.
Simplifying the equation, we have 2cos^2(θ) - 1 = 8/15.
Now, we can solve for cos^2(θ) by isolating the variable.
Adding 1 to both sides of the equation, we have 2cos^2(θ) = 23/15.
Dividing both sides by 2, we get cos^2(θ) = 23/30.
Finally, we can take the square root of both sides of the equation to find the exact value of cos(θ).
5. Since the angle x lies in the second quadrant and sin(x) = 3/5, we can find the value of cos(x) using the Pythagorean identity.
The Pythagorean identity sin^2(x) + cos^2(x) = 1 can be used to find cos(x) when sin(x) is known.
Substituting sin(x) = 3/5 into the equation, we have (3/5)^2 + cos^2(x) = 1.
Simplifying the equation, we get 9/25 + cos^2(x) = 1.
Subtracting 9/25 from both sides gives us cos^2(x) = 16/25.
Taking the square root of both sides of the equation, we find cos(x) = ±4/5.
Since x lies in the second quadrant, the value of cos(x) is negative.
Thus, cos(x) = -4/5.
To find the exact value of tan(2x), we can use the formula tan(2x) = (2tan(x)) / (1 - tan^2(x)).
Substituting tan(x) = sin(x) / cos(x) and the values of sin(x) and cos(x) that we found, we get
tan(2x) = (2 * (3/5)) / (1 - (3/5)^2).
Simplifying the expression, we have tan(2x) = 6/(-4/25).
This results in tan(2x) = -75/4.
This assignment is obviously meant to study the double angle formulas
1.
sin 2x secx = 1
2sinxcosx(1/cosx) = 1
sinx = 1/2
so x is in I or II
x = 30 or 150 degrees or pi/6, 5pi/6 radians
2. use cos2A = 1 - 2cos^2 A
cos 30 = 1 - 2cos^2 A
you should know cos 30, sub it in and simplify
how about you trying some of these.