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b) Find the general solution of 
cos theta+Cos 3theta+cos 5theta=0
OR
Prove that: 
tan^(-1)x-tan^(-1)y=tan^(-1)((x-y)/(1+xy))

b) Find the general solution of cosθ+Cos3θ+cos5θ=0 \cos \theta+\operatorname{Cos} 3 \theta+\cos 5 \theta=0 \newlineOR\newlineProve that: tan1xtan1y=tan1xy1+xy \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} \frac{x-y}{1+x y}

Full solution

Q. b) Find the general solution of cosθ+Cos3θ+cos5θ=0 \cos \theta+\operatorname{Cos} 3 \theta+\cos 5 \theta=0 \newlineOR\newlineProve that: tan1xtan1y=tan1xy1+xy \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} \frac{x-y}{1+x y}
  1. Apply sum-to-product identities: Use the sum-to-product identities to simplify cosθ+cos5θ\cos \theta + \cos 5\theta.cosθ+cos5θ=2cos(θ+5θ2)cos(θ5θ2)\cos \theta + \cos 5\theta = 2 \cdot \cos\left(\frac{\theta + 5\theta}{2}\right) \cdot \cos\left(\frac{\theta - 5\theta}{2}\right)= 2cos(3θ)cos(2θ)2 \cdot \cos(3\theta) \cdot \cos(-2\theta)= 2cos(3θ)cos(2θ)2 \cdot \cos(3\theta) \cdot \cos(2\theta)
  2. Factor out cos(3θ)\cos(3\theta): Now we have 2×cos(3θ)×cos(2θ)+cos(3θ)2 \times \cos(3\theta) \times \cos(2\theta) + \cos(3\theta). Factor out cos(3θ)\cos(3\theta). cos(3θ)×(2×cos(2θ)+1)\cos(3\theta) \times (2 \times \cos(2\theta) + 1)
  3. Find general solution: Set the equation equal to zero to find the general solution. cos(3θ)(2cos(2θ)+1)=0\cos(3\theta) \cdot (2 \cdot \cos(2\theta) + 1) = 0
  4. Solve for cos(3θ)\cos(3\theta): Solve for when each factor is equal to zero.\newlineFor cos(3θ)=0\cos(3\theta) = 0, the general solution is:\newline3θ=(2n+1)π23\theta = (2n + 1) \cdot \frac{\pi}{2}, where nn is an integer.\newlineθ=(2n+1)π6\theta = (2n + 1) \cdot \frac{\pi}{6}
  5. Solve for cos(2θ)\cos(2\theta): For 2cos(2θ)+1=02 \cdot \cos(2\theta) + 1 = 0, solve for cos(2θ)\cos(2\theta).\newlinecos(2θ)=12\cos(2\theta) = -\frac{1}{2}\newline2θ=±(2n+1)π32\theta = \pm (2n + 1) \cdot \frac{\pi}{3}, where nn is an integer.\newlineθ=±(2n+1)π6\theta = \pm (2n + 1) \cdot \frac{\pi}{6}
  6. Combine solutions: Combine the solutions for θ\theta.θ=(2n+1)×π6\theta = (2n + 1) \times \frac{\pi}{6} and θ=±(2n+1)×π6\theta = \pm (2n + 1) \times \frac{\pi}{6}

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