Prove the identity
sin(3 pi /2 + x) + sin(3pi/2 -x) = -2cosx
L.S = (sin 3pi/2)(cos x) + cos(3pi/2)(sin x) + sin(3pi/2)cos x) + cos(3pi/2)(sin x)
= -cos x + (-cos x), because sin(3pi/2)=-1 and cos(3pi/2) = 0
= -2cos x
= R.S.
To prove the given identity, we need to simplify the left side of the equation and show that it is equal to -2cos(x).
Let's start by using the sum and difference formulas for sine:
sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
sin(A - B) = sin(A)cos(B) - cos(A)sin(B)
Now, let's apply these formulas to the left side of the equation:
sin(3π/2 + x) + sin(3π/2 - x)
= sin(3π/2)cos(x) + cos(3π/2)sin(x) + sin(3π/2)cos(x) - cos(3π/2)sin(x)
Since sin(3π/2) = -1 and cos(3π/2) = 0, we can simplify further:
= -1*cos(x) + 0*sin(x) + -1*cos(x) - 0*sin(x)
= -cos(x) - cos(x)
= -2cos(x)
Therefore, the left side of the equation simplifies to -2cos(x), which confirms the given identity is true.
To prove the given identity, we can manipulate the trigonometric expressions using the sum-to-product formula for sines and the cosine of the difference formula. Here are the steps:
Step 1: Write the left side of the equation using the sum-to-product formula for sines:
sin(3π/2 + x) + sin(3π/2 - x)
According to the sum-to-product formula for sines, it states that:
sin(A + B) + sin(A - B) = 2sin(A)cos(B)
By applying this formula to the left side of the equation, we get:
2sin(3π/2)cos(x)
However, sin(3π/2) equals -1:
2(-1)cos(x)
-2cos(x)
Step 2: Simplify the expression obtained from step 1:
-2cos(x)
As you can see, the left side of the equation is equal to the right side of the equation (-2cos(x)). Therefore, we have proven the given identity.