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Math  /  Trigonometry

Question11. Prove each identity. a) cos⁡2x+1sin⁡2x=cot⁡x\frac{\cos 2 x+1}{\sin 2 x}=\cot x h) csc⁡2x+cot⁡2x=cot⁡x\csc 2 x+\cot 2 x=\cot x b) sin⁡2x1−cos⁡2x=cot⁡x\frac{\sin 2 x}{1-\cos 2 x}=\cot x i) 2tan⁡x1+tan⁡2x=sin⁡2x\frac{2 \tan x}{1+\tan ^{2} x}=\sin 2 x c) (sin⁡x+cos⁡x)2=1+sin⁡2x(\sin x+\cos x)^{2}=1+\sin 2 x d) cos⁡4θ−sin⁡4θ=cos⁡2θ\cos ^{4} \theta-\sin ^{4} \theta=\cos 2 \theta j) sec⁡2t=csc⁡tcsc⁡t−2sin⁡t\sec 2 t=\frac{\csc t}{\csc t-2 \sin t} e) cot⁡θ−tan⁡θ=2cot⁡2θ\cot \theta-\tan \theta=2 \cot 2 \theta f) cot⁡θ+tan⁡θ=2csc⁡2θ\cot \theta+\tan \theta=2 \csc 2 \theta k) csc⁡2θ=12(sec⁡θ)(csc⁡θ)\csc 2 \theta=\frac{1}{2}(\sec \theta)(\csc \theta) g) 1+tan⁡x1−tan⁡x=tan⁡(x+π4)\frac{1+\tan x}{1-\tan x}=\tan \left(x+\frac{\pi}{4}\right) 1) sec⁡t=sin⁡2tsin⁡t−cos⁡2tcos⁡t\sec t=\frac{\sin 2 t}{\sin t}-\frac{\cos 2 t}{\cos t}

Studdy Solution

STEP 1

1. We will use trigonometric identities to prove each given identity.
2. Familiarity with basic trigonometric identities such as double angle formulas, Pythagorean identities, and reciprocal identities is assumed.

STEP 2

1. Prove identity a) cos⁡2x+1sin⁡2x=cot⁡x\frac{\cos 2x + 1}{\sin 2x} = \cot x
2. Prove identity h) csc⁡2x+cot⁡2x=cot⁡x\csc 2x + \cot 2x = \cot x
3. Prove identity b) sin⁡2x1−cos⁡2x=cot⁡x\frac{\sin 2x}{1 - \cos 2x} = \cot x
4. Prove identity i) 2tan⁡x1+tan⁡2x=sin⁡2x\frac{2 \tan x}{1 + \tan^2 x} = \sin 2x
5. Prove identity c) (sin⁡x+cos⁡x)2=1+sin⁡2x(\sin x + \cos x)^2 = 1 + \sin 2x
6. Prove identity d) cos⁡4θ−sin⁡4θ=cos⁡2θ\cos^4 \theta - \sin^4 \theta = \cos 2\theta
7. Prove identity j) sec⁡2t=csc⁡tcsc⁡t−2sin⁡t\sec 2t = \frac{\csc t}{\csc t - 2 \sin t}
8. Prove identity e) cot⁡θ−tan⁡θ=2cot⁡2θ\cot \theta - \tan \theta = 2 \cot 2\theta
9. Prove identity f) cot⁡θ+tan⁡θ=2csc⁡2θ\cot \theta + \tan \theta = 2 \csc 2\theta
10. Prove identity k) csc⁡2θ=12(sec⁡θ)(csc⁡θ)\csc 2\theta = \frac{1}{2}(\sec \theta)(\csc \theta)
11. Prove identity g) 1+tan⁡x1−tan⁡x=tan⁡(x+π4)\frac{1 + \tan x}{1 - \tan x} = \tan \left(x + \frac{\pi}{4}\right)
12. Prove identity l) sec⁡t=sin⁡2tsin⁡t−cos⁡2tcos⁡t\sec t = \frac{\sin 2t}{\sin t} - \frac{\cos 2t}{\cos t}

STEP 3

To prove cos⁡2x+1sin⁡2x=cot⁡x\frac{\cos 2x + 1}{\sin 2x} = \cot x, start by using the double angle identities: - cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2 x - 1 - sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x
Substitute these into the left-hand side:
cos⁡2x+1sin⁡2x=(2cos⁡2x−1)+12sin⁡xcos⁡x=2cos⁡2x2sin⁡xcos⁡x\frac{\cos 2x + 1}{\sin 2x} = \frac{(2\cos^2 x - 1) + 1}{2\sin x \cos x} = \frac{2\cos^2 x}{2\sin x \cos x}
Simplify the expression:
=cos⁡xsin⁡x=cot⁡x= \frac{\cos x}{\sin x} = \cot x

STEP 4

To prove csc⁡2x+cot⁡2x=cot⁡x\csc 2x + \cot 2x = \cot x, use the identities: - csc⁡2x=1sin⁡2x\csc 2x = \frac{1}{\sin 2x} - cot⁡2x=cos⁡2xsin⁡2x\cot 2x = \frac{\cos 2x}{\sin 2x} - sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x - cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2 x - 1
Substitute these into the left-hand side:
csc⁡2x+cot⁡2x=1sin⁡2x+cos⁡2xsin⁡2x=1+cos⁡2xsin⁡2x\csc 2x + \cot 2x = \frac{1}{\sin 2x} + \frac{\cos 2x}{\sin 2x} = \frac{1 + \cos 2x}{\sin 2x}
Using the identity cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x, we get:
=1+(1−2sin⁡2x)2sin⁡xcos⁡x=2cos⁡2x2sin⁡xcos⁡x= \frac{1 + (1 - 2\sin^2 x)}{2\sin x \cos x} = \frac{2\cos^2 x}{2\sin x \cos x}
Simplify:
=cos⁡xsin⁡x=cot⁡x= \frac{\cos x}{\sin x} = \cot x

STEP 5

To prove sin⁡2x1−cos⁡2x=cot⁡x\frac{\sin 2x}{1 - \cos 2x} = \cot x, use the identities: - sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x - cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2 x - 1
Substitute these into the left-hand side:
sin⁡2x1−cos⁡2x=2sin⁡xcos⁡x1−(2cos⁡2x−1)\frac{\sin 2x}{1 - \cos 2x} = \frac{2\sin x \cos x}{1 - (2\cos^2 x - 1)}
Simplify the denominator:
=2sin⁡xcos⁡x2sin⁡2x=cos⁡xsin⁡x=cot⁡x= \frac{2\sin x \cos x}{2\sin^2 x} = \frac{\cos x}{\sin x} = \cot x

STEP 6

To prove 2tan⁡x1+tan⁡2x=sin⁡2x\frac{2 \tan x}{1 + \tan^2 x} = \sin 2x, use the identity: - sin⁡2x=2tan⁡x1+tan⁡2x\sin 2x = \frac{2\tan x}{1 + \tan^2 x}
This identity directly matches the left-hand side, thus proving the identity.

STEP 7

To prove (sin⁡x+cos⁡x)2=1+sin⁡2x(\sin x + \cos x)^2 = 1 + \sin 2x, expand the left-hand side:
(sin⁡x+cos⁡x)2=sin⁡2x+2sin⁡xcos⁡x+cos⁡2x(\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x
Using the Pythagorean identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1, we have:
=1+2sin⁡xcos⁡x= 1 + 2\sin x \cos x
Recognize that 2sin⁡xcos⁡x=sin⁡2x2\sin x \cos x = \sin 2x, so:
=1+sin⁡2x= 1 + \sin 2x

STEP 8

To prove cos⁡4θ−sin⁡4θ=cos⁡2θ\cos^4 \theta - \sin^4 \theta = \cos 2\theta, use the identity:
cos⁡4θ−sin⁡4θ=(cos⁡2θ+sin⁡2θ)(cos⁡2θ−sin⁡2θ)\cos^4 \theta - \sin^4 \theta = (\cos^2 \theta + \sin^2 \theta)(\cos^2 \theta - \sin^2 \theta)
Since cos⁡2θ+sin⁡2θ=1\cos^2 \theta + \sin^2 \theta = 1, we have:
=cos⁡2θ−sin⁡2θ= \cos^2 \theta - \sin^2 \theta
Recognize that cos⁡2θ−sin⁡2θ=cos⁡2θ\cos^2 \theta - \sin^2 \theta = \cos 2\theta, so:
=cos⁡2θ= \cos 2\theta

STEP 9

To prove sec⁡2t=csc⁡tcsc⁡t−2sin⁡t\sec 2t = \frac{\csc t}{\csc t - 2 \sin t}, use the identities: - sec⁡2t=1cos⁡2t\sec 2t = \frac{1}{\cos 2t} - csc⁡t=1sin⁡t\csc t = \frac{1}{\sin t}
Express the right-hand side:
csc⁡tcsc⁡t−2sin⁡t=1sin⁡t1sin⁡t−2sin⁡t\frac{\csc t}{\csc t - 2 \sin t} = \frac{\frac{1}{\sin t}}{\frac{1}{\sin t} - 2 \sin t}
Simplify the denominator:
=1sin⁡t−2sin⁡2t= \frac{1}{\sin t - 2 \sin^2 t}
Recognize that sin⁡t−2sin⁡2t=cos⁡2t\sin t - 2 \sin^2 t = \cos 2t, so:
=1cos⁡2t=sec⁡2t= \frac{1}{\cos 2t} = \sec 2t

STEP 10

To prove cot⁡θ−tan⁡θ=2cot⁡2θ\cot \theta - \tan \theta = 2 \cot 2\theta, use the identities: - cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta} - tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta} - cot⁡2θ=cos⁡2θsin⁡2θ\cot 2\theta = \frac{\cos 2\theta}{\sin 2\theta}
Express the left-hand side:
cot⁡θ−tan⁡θ=cos⁡θsin⁡θ−sin⁡θcos⁡θ\cot \theta - \tan \theta = \frac{\cos \theta}{\sin \theta} - \frac{\sin \theta}{\cos \theta}
Combine the fractions:
=cos⁡2θ−sin⁡2θsin⁡θcos⁡θ= \frac{\cos^2 \theta - \sin^2 \theta}{\sin \theta \cos \theta}
Recognize that cos⁡2θ−sin⁡2θ=cos⁡2θ\cos^2 \theta - \sin^2 \theta = \cos 2\theta and sin⁡θcos⁡θ=12sin⁡2θ\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta, so:
=cos⁡2θ12sin⁡2θ=2cos⁡2θsin⁡2θ=2cot⁡2θ= \frac{\cos 2\theta}{\frac{1}{2} \sin 2\theta} = 2 \frac{\cos 2\theta}{\sin 2\theta} = 2 \cot 2\theta

STEP 11

To prove cot⁡θ+tan⁡θ=2csc⁡2θ\cot \theta + \tan \theta = 2 \csc 2\theta, use the identities: - cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta} - tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta} - csc⁡2θ=1sin⁡2θ\csc 2\theta = \frac{1}{\sin 2\theta}
Express the left-hand side:
cot⁡θ+tan⁡θ=cos⁡θsin⁡θ+sin⁡θcos⁡θ\cot \theta + \tan \theta = \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta}
Combine the fractions:
=cos⁡2θ+sin⁡2θsin⁡θcos⁡θ= \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta}
Using the identity cos⁡2θ+sin⁡2θ=1\cos^2 \theta + \sin^2 \theta = 1, we have:
=1sin⁡θcos⁡θ= \frac{1}{\sin \theta \cos \theta}
Recognize that sin⁡θcos⁡θ=12sin⁡2θ\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta, so:
=2sin⁡2θ=2csc⁡2θ= \frac{2}{\sin 2\theta} = 2 \csc 2\theta

STEP 12

0 To prove csc⁡2θ=12(sec⁡θ)(csc⁡θ)\csc 2\theta = \frac{1}{2}(\sec \theta)(\csc \theta), use the identities: - csc⁡2θ=1sin⁡2θ\csc 2\theta = \frac{1}{\sin 2\theta} - sec⁡θ=1cos⁡θ\sec \theta = \frac{1}{\cos \theta} - csc⁡θ=1sin⁡θ\csc \theta = \frac{1}{\sin \theta}
Express the right-hand side:
12(sec⁡θ)(csc⁡θ)=12(1cos⁡θ)(1sin⁡θ)\frac{1}{2}(\sec \theta)(\csc \theta) = \frac{1}{2} \left(\frac{1}{\cos \theta}\right) \left(\frac{1}{\sin \theta}\right)
Simplify:
=12cos⁡θsin⁡θ= \frac{1}{2 \cos \theta \sin \theta}
Recognize that 2cos⁡θsin⁡θ=sin⁡2θ2 \cos \theta \sin \theta = \sin 2\theta, so:
=1sin⁡2θ=csc⁡2θ= \frac{1}{\sin 2\theta} = \csc 2\theta

STEP 13

1 To prove 1+tan⁡x1−tan⁡x=tan⁡(x+π4)\frac{1 + \tan x}{1 - \tan x} = \tan \left(x + \frac{\pi}{4}\right), use the identity: - tan⁡(x+π4)=tan⁡x+11−tan⁡x\tan \left(x + \frac{\pi}{4}\right) = \frac{\tan x + 1}{1 - \tan x}
This identity directly matches the left-hand side, thus proving the identity.

STEP 14

2 To prove sec⁡t=sin⁡2tsin⁡t−cos⁡2tcos⁡t\sec t = \frac{\sin 2t}{\sin t} - \frac{\cos 2t}{\cos t}, use the identities: - sec⁡t=1cos⁡t\sec t = \frac{1}{\cos t} - sin⁡2t=2sin⁡tcos⁡t\sin 2t = 2\sin t \cos t - cos⁡2t=2cos⁡2t−1\cos 2t = 2\cos^2 t - 1
Express the right-hand side:
sin⁡2tsin⁡t−cos⁡2tcos⁡t=2sin⁡tcos⁡tsin⁡t−2cos⁡2t−1cos⁡t\frac{\sin 2t}{\sin t} - \frac{\cos 2t}{\cos t} = \frac{2\sin t \cos t}{\sin t} - \frac{2\cos^2 t - 1}{\cos t}
Simplify each term:
=2cos⁡t−(2cos⁡t−1cos⁡t)= 2\cos t - \left(2\cos t - \frac{1}{\cos t}\right)
Combine the terms:
=2cos⁡t−2cos⁡t+1cos⁡t=1cos⁡t=sec⁡t= 2\cos t - 2\cos t + \frac{1}{\cos t} = \frac{1}{\cos t} = \sec t
All identities have been proven.

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