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Math  /  Trigonometry

QuestionUnit 4: Trigonometric Functions Activity 8: Trigonometric Identities
Assignment Prove each of the following trigonometric identities.
1. sin⁡xsin⁡2x+cos⁡xcos⁡2x=cos⁡x\sin x \sin 2 x+\cos x \cos 2 x=\cos x
2. cot⁡x=sin⁡xsin⁡(π2−x)+cos⁡2xcot⁡x\cot x=\sin x \sin \left(\frac{\pi}{2}-x\right)+\cos ^{2} x \cot x
3. 2csc⁡2x=sec⁡xcsc⁡x2 \csc 2 x=\sec x \csc x

Studdy Solution

STEP 1

1. We are given three trigonometric identities to prove.
2. We will use known trigonometric identities and algebraic manipulation to prove each identity.
3. We assume all trigonometric functions are defined for the given expressions.

STEP 2

1. Prove identity 1: sin⁡xsin⁡2x+cos⁡xcos⁡2x=cos⁡x\sin x \sin 2x + \cos x \cos 2x = \cos x.
2. Prove identity 2: cot⁡x=sin⁡xsin⁡(π2−x)+cos⁡2xcot⁡x\cot x = \sin x \sin \left(\frac{\pi}{2} - x\right) + \cos^2 x \cot x.
3. Prove identity 3: 2csc⁡2x=sec⁡xcsc⁡x2 \csc 2x = \sec x \csc x.

STEP 3

Use the product-to-sum identities for sin⁡xsin⁡2x\sin x \sin 2x and cos⁡xcos⁡2x\cos x \cos 2x.
sin⁡xsin⁡2x=12[cos⁡(x−2x)−cos⁡(x+2x)]=12[cos⁡(−x)−cos⁡(3x)]\sin x \sin 2x = \frac{1}{2} [\cos(x - 2x) - \cos(x + 2x)] = \frac{1}{2} [\cos(-x) - \cos(3x)]
cos⁡xcos⁡2x=12[cos⁡(x−2x)+cos⁡(x+2x)]=12[cos⁡(−x)+cos⁡(3x)]\cos x \cos 2x = \frac{1}{2} [\cos(x - 2x) + \cos(x + 2x)] = \frac{1}{2} [\cos(-x) + \cos(3x)]

STEP 4

Combine the expressions from STEP_1:
sin⁡xsin⁡2x+cos⁡xcos⁡2x=12[cos⁡(−x)−cos⁡(3x)]+12[cos⁡(−x)+cos⁡(3x)]\sin x \sin 2x + \cos x \cos 2x = \frac{1}{2} [\cos(-x) - \cos(3x)] + \frac{1}{2} [\cos(-x) + \cos(3x)]
Simplify:
=12[2cos⁡(−x)]=cos⁡x= \frac{1}{2} [2\cos(-x)] = \cos x

STEP 5

Use the co-function identity sin⁡(π2−x)=cos⁡x\sin\left(\frac{\pi}{2} - x\right) = \cos x and simplify the right side of the equation:
sin⁡xsin⁡(π2−x)=sin⁡xcos⁡x\sin x \sin\left(\frac{\pi}{2} - x\right) = \sin x \cos x

STEP 6

Substitute and simplify:
cot⁡x=sin⁡xcos⁡x+cos⁡2xcot⁡x\cot x = \sin x \cos x + \cos^2 x \cot x
Rearrange terms:
cot⁡x−cos⁡2xcot⁡x=sin⁡xcos⁡x\cot x - \cos^2 x \cot x = \sin x \cos x
Factor out cot⁡x\cot x:
cot⁡x(1−cos⁡2x)=sin⁡xcos⁡x\cot x (1 - \cos^2 x) = \sin x \cos x
Since 1−cos⁡2x=sin⁡2x1 - \cos^2 x = \sin^2 x, we have:
cot⁡xsin⁡2x=sin⁡xcos⁡x\cot x \sin^2 x = \sin x \cos x
Divide both sides by sin⁡x\sin x:
cot⁡xsin⁡x=cos⁡x\cot x \sin x = \cos x
Since cot⁡x=cos⁡xsin⁡x\cot x = \frac{\cos x}{\sin x}, the identity holds true.

STEP 7

Use the double angle identity for cosecant and rewrite the left side:
2csc⁡2x=2×1sin⁡2x=22sin⁡xcos⁡x=1sin⁡xcos⁡x2 \csc 2x = 2 \times \frac{1}{\sin 2x} = \frac{2}{2 \sin x \cos x} = \frac{1}{\sin x \cos x}

STEP 8

Rewrite the right side using the definitions of secant and cosecant:
sec⁡xcsc⁡x=1cos⁡x×1sin⁡x=1sin⁡xcos⁡x\sec x \csc x = \frac{1}{\cos x} \times \frac{1}{\sin x} = \frac{1}{\sin x \cos x}
Both sides are equal, thus proving the identity.

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