I am stuck on a problem that has to do with the double angle formulas. I have to solve this: 4sinxcosx = 1. I tried using the formula 2sinu = 2sinucosu by substituting it in like this: 2(2sinxcosx)= 1 but then I am stuck
remember that sin 2x = 2sinxcosx
so 4sinxcosx = 1
2(2sinxcosx) = 1
2(sin2x) = 1
sin2x = 1/2
I will assume you are working in degrees, we can always switch later on
so 2x = 30° or 2x = 150° because of the CAST rule
x = 15° or x = 75°
the period of sin 2x is 180°, so adding 180 to any answer will produce new answers
x = 15°, 195°, ...
or
x = 75°, 255°, ...
In Radians:
15° = π/12 radians
x = π/12, 13π/12....
x = 5π/12, 17π/12, ...
To solve the equation 4sinxcosx = 1, you are on the right track by using the double angle formula. However, your substitution of the formula isn't correct. The correct double angle formula for sin2u is sin2u = 2sinucosu.
Let's start solving the equation correctly using the double angle formula. We have 4sinxcosx = 1, and we can rewrite it as 2(2sinxcosx) = 1 to use the double angle formula for sin2u.
Using the double angle formula sin2u = 2sinucosu, we can rewrite the left-hand side (LHS) of the equation as sin(2x). So, the equation becomes:
sin(2x) = 1
Now, to solve for x, we need to find the values of 2x that have a sine function equal to 1. Since the sine function has a maximum value of 1, we know that the angle whose sine is 1 is 90 degrees or π/2 radians.
So, we set 2x equal to π/2 radians (or 90 degrees):
2x = π/2
Now, let's solve for x by dividing both sides of the equation by 2:
x = π/4
Therefore, the solution to the equation 4sinxcosx = 1 is x = π/4.