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Math  /  Trigonometry

Question6. sec⁡2θ4[cos⁡8θ−1cos⁡4θ−1]=cos⁡2θ\frac{\sec 2 \theta}{4}\left[\frac{\cos 8 \theta-1}{\cos 4 \theta-1}\right]=\cos 2 \theta எனக் காட்டுக.

Studdy Solution

STEP 1

1. The given equation involves trigonometric identities.
2. We will use trigonometric identities to simplify and prove the equation.
3. The goal is to manipulate the left-hand side to match the right-hand side, cos⁡2θ\cos 2\theta.

STEP 2

1. Simplify the fraction inside the brackets.
2. Simplify the overall expression.
3. Prove the equation by matching both sides.

STEP 3

Start by simplifying the fraction inside the brackets. Notice that cos⁡8θ\cos 8\theta and cos⁡4θ\cos 4\theta can be expressed using double angle identities. Use the identity for cos⁡2A=2cos⁡2A−1\cos 2A = 2\cos^2 A - 1.
For cos⁡8θ\cos 8\theta, use: cos⁡8θ=2cos⁡24θ−1 \cos 8\theta = 2\cos^2 4\theta - 1
Substitute this into the fraction: cos⁡8θ−1cos⁡4θ−1=(2cos⁡24θ−1)−1cos⁡4θ−1 \frac{\cos 8\theta - 1}{\cos 4\theta - 1} = \frac{(2\cos^2 4\theta - 1) - 1}{\cos 4\theta - 1} =2cos⁡24θ−2cos⁡4θ−1 = \frac{2\cos^2 4\theta - 2}{\cos 4\theta - 1} =2(cos⁡24θ−1)cos⁡4θ−1 = \frac{2(\cos^2 4\theta - 1)}{\cos 4\theta - 1}

STEP 4

Notice that cos⁡24θ−1\cos^2 4\theta - 1 can be factored using the identity cos⁡2A−1=(cos⁡A−1)(cos⁡A+1)\cos^2 A - 1 = (\cos A - 1)(\cos A + 1):
cos⁡24θ−1=(cos⁡4θ−1)(cos⁡4θ+1) \cos^2 4\theta - 1 = (\cos 4\theta - 1)(\cos 4\theta + 1)
Substitute back into the fraction: 2(cos⁡4θ−1)(cos⁡4θ+1)cos⁡4θ−1 \frac{2(\cos 4\theta - 1)(\cos 4\theta + 1)}{\cos 4\theta - 1}
Cancel out (cos⁡4θ−1)(\cos 4\theta - 1) from the numerator and denominator: =2(cos⁡4θ+1) = 2(\cos 4\theta + 1)

STEP 5

Substitute the simplified expression back into the original equation:
sec⁡2θ4×2(cos⁡4θ+1)=cos⁡2θ \frac{\sec 2\theta}{4} \times 2(\cos 4\theta + 1) = \cos 2\theta
Simplify the expression: sec⁡2θ4×2cos⁡4θ+sec⁡2θ4×2=cos⁡2θ \frac{\sec 2\theta}{4} \times 2\cos 4\theta + \frac{\sec 2\theta}{4} \times 2 = \cos 2\theta sec⁡2θcos⁡4θ2+sec⁡2θ2=cos⁡2θ \frac{\sec 2\theta \cos 4\theta}{2} + \frac{\sec 2\theta}{2} = \cos 2\theta

STEP 6

Recall that sec⁡2θ=1cos⁡2θ\sec 2\theta = \frac{1}{\cos 2\theta}. Substitute this into the equation:
cos⁡4θ2cos⁡2θ+12cos⁡2θ=cos⁡2θ \frac{\cos 4\theta}{2\cos 2\theta} + \frac{1}{2\cos 2\theta} = \cos 2\theta
Combine the terms: cos⁡4θ+12cos⁡2θ=cos⁡2θ \frac{\cos 4\theta + 1}{2\cos 2\theta} = \cos 2\theta

STEP 7

Multiply both sides by 2cos⁡2θ2\cos 2\theta to eliminate the fraction:
cos⁡4θ+1=2cos⁡22θ \cos 4\theta + 1 = 2\cos^2 2\theta
Use the identity cos⁡4θ=2cos⁡22θ−1\cos 4\theta = 2\cos^2 2\theta - 1:
2cos⁡22θ−1+1=2cos⁡22θ 2\cos^2 2\theta - 1 + 1 = 2\cos^2 2\theta
Both sides are equal, proving the equation.
The equation is verified as true.

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