How do you use the double angle or half angle formulas to solve #2 sinx cosx=cos 2x#?

Question

Anna Andrews · Accepted Answer

Solve 2sin x.cos x = cos 2x

Ans: #x = pi/8# Apply the 2 trig identities: 2sin x.cos x = sin 2x. #cos x = sin (pi/2 - x)# We have: #sin 2x = cos 2x = sin (pi/2 - 2x)# #2x = (pi/2 - 2x)# #4x = pi/2# --> #x = pi/8# Check by calculator. #x = pi/8 = 22^@5# --> 2x = 45 --> --> sin 2x = 0.71 --> cos 2x = 0.71 .OK

Carter Anderson · Answer

undefined To solve the equation $ 2 \sin(x) \cos(x) = \cos(2x) $ using double angle or half-angle formulas, we can start by using the double angle identity for cosine:

$$ \cos(2x) = \cos^2(x) - \sin^2(x) $$

Now, we substitute this expression into the given equation:

$$ 2 \sin(x) \cos(x) = \cos^2(x) - \sin^2(x) $$

Next, we can use the identity $ \sin^2(x) = 1 - \cos^2(x) $ to replace $ \sin^2(x) $ in the equation:

$$ 2 \sin(x) \cos(x) = \cos^2(x) - (1 - \cos^2(x)) $$

$$ 2 \sin(x) \cos(x) = \cos^2(x) - 1 + \cos^2(x) $$

Now, we simplify the equation:

$$ 2 \sin(x) \cos(x) = 2 \cos^2(x) - 1 $$

$$ 2 \sin(x) \cos(x) - 2 \cos^2(x) + 1 = 0 $$

Now, let's factor out $ \cos(x) $ from the first two terms:

$$ 2 \cos(x)(\sin(x) - \cos(x)) + 1 = 0 $$

Now, we have a quadratic equation in terms of $ \cos(x) $. We can solve it using the quadratic formula or factoring techniques.

After finding the values of $ \cos(x) $, we can then use the inverse cosine function to find the values of $ x $ within the desired range (typically $ 0 \leq x \leq 2\pi $ or $ 0^\circ \leq x \leq 360^\circ $).