To prove the given equation, we will utilize the product-to-sum identities for cosine and sine. These identities are given as:
Given Equation: cos(3x)cos(2x)−sin(x)sin(6x)=cos(7x)\cos(3x)\cos(2x) - \sin(x)\sin(6x) = \cos(7x)cos(3x)cos(2x)−sin(x)sin(6x)=cos(7x)
Expand the terms using the product-to-sum identities.
cos(3x)cos(2x)=12[cos(3x−2x)+cos(3x+2x)]\cos(3x)\cos(2x) = \dfrac{1}{2}[\cos(3x-2x) + \cos(3x+2x)]cos(3x)cos(2x)=21 [cos(3x−2x)+cos(3x+2x)] = 12[cos(x)+cos(5x)]\dfrac{1}{2}[\cos(x) + \cos(5x)]21 [cos(x)+cos(5x)]
sin(x)sin(6x)=12[cos(x−6x)−cos(x+6x)]\sin(x)\sin(6x) = \dfrac{1}{2}[\cos(x-6x) - \cos(x+6x)]sin(x)sin(6x)=21 [cos(x−6x)−cos(x+6x)] = 12[cos(−5x)−cos(7x)]\dfrac{1}{2}[\cos(-5x) - \cos(7x)]21 [cos(−5x)−cos(7x)] = −12[cos(5x)+cos(7x)]-\dfrac{1}{2}[\cos(5x) + \cos(7x)]−21 [cos(5x)+cos(7x)]
Therefore, cos(3x)cos(2x)−sin(x)sin(6x)=12[cos(x)+cos(5x)]+12[cos(5x)+cos(7x)]\cos(3x)\cos(2x) - \sin(x)\sin(6x) = \dfrac{1}{2}[\cos(x) + \cos(5x)] + \dfrac{1}{2}[\cos(5x) + \cos(7x)]cos(3x)cos(2x)−sin(x)sin(6x)=21 [cos(x)+cos(5x)]+21 [cos(5x)+cos(7x)] = 12[cos(x)+cos(5x)+cos(5x)+cos(7x)]\dfrac{1}{2}[\cos(x) + \cos(5x) + \cos(5x) + \cos(7x)]21 [cos(x)+cos(5x)+cos(5x)+cos(7x)] = 12[2cos(5x)+cos(x)+cos(7x)]\dfrac{1}{2}[2\cos(5x) + \cos(x) + \cos(7x)]21 [2cos(5x)+cos(x)+cos(7x)] = cos(5x)+12[cos(x)+cos(7x)]\cos(5x) + \dfrac{1}{2}[\cos(x) + \cos(7x)]cos(5x)+21 [cos(x)+cos(7x)] = cos(5x)+cos(6x)\cos(5x) + \cos(6x)cos(5x)+cos(6x)
Hence, cos(3x)cos(2x)−sin(x)sin(6x)=cos(7x)\cos(3x)\cos(2x) - \sin(x)\sin(6x) = \cos(7x)cos(3x)cos(2x)−sin(x)sin(6x)=cos(7x) has been proven correct.
To prove the given equation, we will utilize the product-to-sum identities for cosine and sine. These identities are given as:
cos(a)cos(b)=12[cos(a−b)+cos(a+b)]\cos(a)\cos(b) = \dfrac{1}{2}[\cos(a-b) + \cos(a+b)]cos(a)cos(b)=21 [cos(a−b)+cos(a+b)]sin(a)sin(b)=12[cos(a−b)−cos(a+b)]\sin(a)\sin(b) = \dfrac{1}{2}[\cos(a-b) - \cos(a+b)]sin(a)sin(b)=21 [cos(a−b)−cos(a+b)]Given Equation: cos(3x)cos(2x)−sin(x)sin(6x)=cos(7x)\cos(3x)\cos(2x) - \sin(x)\sin(6x) = \cos(7x)cos(3x)cos(2x)−sin(x)sin(6x)=cos(7x)
Expand the terms using the product-to-sum identities.
cos(3x)cos(2x)=12[cos(3x−2x)+cos(3x+2x)]\cos(3x)\cos(2x) = \dfrac{1}{2}[\cos(3x-2x) + \cos(3x+2x)]cos(3x)cos(2x)=21 [cos(3x−2x)+cos(3x+2x)] = 12[cos(x)+cos(5x)]\dfrac{1}{2}[\cos(x) + \cos(5x)]21 [cos(x)+cos(5x)]
sin(x)sin(6x)=12[cos(x−6x)−cos(x+6x)]\sin(x)\sin(6x) = \dfrac{1}{2}[\cos(x-6x) - \cos(x+6x)]sin(x)sin(6x)=21 [cos(x−6x)−cos(x+6x)] = 12[cos(−5x)−cos(7x)]\dfrac{1}{2}[\cos(-5x) - \cos(7x)]21 [cos(−5x)−cos(7x)] = −12[cos(5x)+cos(7x)]-\dfrac{1}{2}[\cos(5x) + \cos(7x)]−21 [cos(5x)+cos(7x)]
Therefore, cos(3x)cos(2x)−sin(x)sin(6x)=12[cos(x)+cos(5x)]+12[cos(5x)+cos(7x)]\cos(3x)\cos(2x) - \sin(x)\sin(6x) = \dfrac{1}{2}[\cos(x) + \cos(5x)] + \dfrac{1}{2}[\cos(5x) + \cos(7x)]cos(3x)cos(2x)−sin(x)sin(6x)=21 [cos(x)+cos(5x)]+21 [cos(5x)+cos(7x)] = 12[cos(x)+cos(5x)+cos(5x)+cos(7x)]\dfrac{1}{2}[\cos(x) + \cos(5x) + \cos(5x) + \cos(7x)]21 [cos(x)+cos(5x)+cos(5x)+cos(7x)] = 12[2cos(5x)+cos(x)+cos(7x)]\dfrac{1}{2}[2\cos(5x) + \cos(x) + \cos(7x)]21 [2cos(5x)+cos(x)+cos(7x)] = cos(5x)+12[cos(x)+cos(7x)]\cos(5x) + \dfrac{1}{2}[\cos(x) + \cos(7x)]cos(5x)+21 [cos(x)+cos(7x)] = cos(5x)+cos(6x)\cos(5x) + \cos(6x)cos(5x)+cos(6x)
Hence, cos(3x)cos(2x)−sin(x)sin(6x)=cos(7x)\cos(3x)\cos(2x) - \sin(x)\sin(6x) = \cos(7x)cos(3x)cos(2x)−sin(x)sin(6x)=cos(7x) has been proven correct.