Prove:
sin2x*sinx/2cosx + cos^2 0 = 1
I corrected your typo
sin(2x)*sinx/2cosx + cos^2 x = 1
LS = 2sinxcosx(sinx)/(2cosx) + cos^2x
= sinx(sinx) + cos^2 x
= sin^2 x + cos^2 x
= 1
= RS
Thank you! :)
To prove the given equation, we need to simplify the left-hand side (LHS) until it equals the right-hand side (RHS).
Let's start by simplifying the LHS of the equation step by step:
sin 2x * sin(x/2) / (2cosx) + cos^2 0
First, we can use the identities sin 2x = 2sin x * cos x and sin(x/2) = √[(1 - cos x) / 2] to rewrite the equation as:
(2sin x * cos x) * √[(1 - cos x) / 2] / (2cos x) + cos^2 0
The 2s in the numerator and denominator cancel out, and we can simplify further:
sin x * cos x * √[(1 - cos x) / 2cos x] + cos^2 0
Next, the cos^2 0 can be simplified to 1 since cos 0 equals 1:
sin x * cos x * √[(1 - cos x) / 2cos x] + 1
Now, let's focus on the term √[(1 - cos x) / 2cos x]. We can simplify it by rationalizing the denominator. Multiply the numerator and denominator by √(2cos x) to eliminate the radical from the denominator:
sin x * cos x * [√(2cos x) * √(1 - cos x)] / [2cos x]
This simplifies to:
sin x * cos x * √(2cos x * (1 - cos x)) / [2cos x]
Again, we can cancel out cos x in the numerator and denominator:
sin x * √(2cos x * (1 - cos x)) / 2
Now, we can simplify further by using the identity 2sin x * cos x = sin 2x:
(1/2) * √[2cos x * (1 - cos x)] * sin 2x
Finally, we can use the identity 2cos x * (1 - cos x) = sin^2 x to rewrite the equation:
(1/2) * √[sin^2 x] * sin 2x
The square root of sin^2 x is simply sin x:
(1/2) * sin x * sin 2x
Now, let's simplify the expression (1/2) * sin x * sin 2x:
(1/2) * sin x * sin 2x = (1/2) * sin x * 2sin x * cos x
= sin^2 x * cos x
Finally, remembering the identity sin^2 x + cos^2 x = 1, we can replace sin^2 x with 1 - cos^2 x:
= (1 - cos^2 x) * cos x
= cos x - cos^3 x
After simplifying, we see that the LHS expression cos x - cos^3 x is indeed equal to the RHS expression 1.
Therefore, we have proven that sin^2 x * sin(x/2) / (2cos x) + cos^2 0 simplifies to 1.