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Unit 10: Trigonometric Identities — Exercise 10 3

11th Class Mathematics · Unit 10: Trigonometric Identities

1.(i) Find the values of sin 2α, cos 2α and tan 2α, when sin α = 3/5, where 0 < α < π/2 (ii) Find the values of sin 2α, cos 2α and tan 2α, when cos α = 4/5, where 0 < α < π/2

Given

(i) Given sin α = 35, quad 0 < α < pi2

Therefore α lies in quadrant I

Formula

Pythagorean identity cos α = sqrt{1 - sin² α}

Working

Substituting = sqrt{1 - left(35right)²} = sqrt{25-925 = sqrt{1625 = 45

Formula

Double angle formula sin 2α = 2sin α cos α

Working

Substituting = 2 · 35 · 45 = 2425

Formula

Double angle formula cos 2α = 1 - 2sin² α

Working

Substituting = 1 - 2left(35right)² = 1 - 1825 = 725

Formula

Tangent ratio tan 2α = sin 2αcos 2α = 24/257/25 = 247

Result

sin 2α = 2425, quad cos 2α = 725, quad tan 2α = 247

Given

(ii) Given cos α = 45, quad 0 < α < pi2

Therefore α lies in quadrant I

Formula

Pythagorean identity sin α = sqrt{1 - cos² α}

Working

Substituting = sqrt{1 - left(45right)²} = sqrt{25-1625 = sqrt{925 = 35

Formula

Double angle formula sin 2α = 2sin α cos α

Working

Substituting = 2 · 35 · 45 = 2425

Formula

Double angle formula cos 2α = 1 - 2sin² α

Working

Substituting = 1 - 2left(35right)² = 1 - 1825 = 725

Formula

Tangent ratio tan 2α = sin 2αcos 2α = 24/257/25 = 247

Result

sin 2α = 2425, quad cos 2α = 725, quad tan 2α = 247

2.Prove the following identity: (i) $\cot\alpha - \tan\alpha = 2\cot 2\alpha$ (ii) $\dfrac{\sin 2\alpha}{1+\cos 2\alpha} = \tan\alpha$ (iii) $\dfrac{1-\cos\alpha}{\sin\alpha} = \tan\dfrac{\alpha}{2}$ (iv) $\dfrac{\cos\alpha-\sin\alpha}{\cos\alpha+\sin\alpha} = \sec 2\alpha - \tan 2\alpha$ (v) $\sqrt{\dfrac{1+\sin\alpha}{1-\sin\alpha}} = \dfrac{\sin\frac{\alpha}{2}+\cos\frac{\alpha}{2}}{\sin\frac{\alpha}{2}-\cos\frac{\alpha}{2}}$ (vi) $\dfrac{\csc\theta+2\csc 2\theta}{\sec\theta} = \cot\dfrac{\theta}{2}$ (vii) $1+\tan\alpha\tan 2\alpha = \sec 2\alpha$ (viii) $\dfrac{2\sin\theta\sin 2\theta}{\cos\theta+\cos 3\theta} = \tan 2\theta\tan\theta$ (ix) $\dfrac{\sin 3\theta}{\sin\theta} - \dfrac{\cos 3\theta}{\cos\theta} = 2$ (x) $\dfrac{\cos 3\theta}{\cos\theta} + \dfrac{\sin 3\theta}{\sin\theta} = 4\cos 2\theta$ (xi) $\dfrac{\tan\frac{\theta}{2}+\cot\frac{\theta}{2}}{\cot\frac{\theta}{2}-\tan\frac{\theta}{2}} = \sec\theta$ (xii) $\dfrac{\sin 3\theta}{\cos\theta} + \dfrac{\cos 3\theta}{\sin\theta} = 2\cot 2\theta$ (xiii) $\dfrac{3+\cos 4\theta}{1-\cos 4\theta} = \dfrac{1}{2}\left(\tan^2\theta+\cot^2\theta\right)$ (xiv) $\dfrac{1+\sin 2\theta}{1-\sin 2\theta} = \tan^2\left(\dfrac{\pi}{4}+\theta\right)$ (xv) $\cos^2\dfrac{\pi}{8}+\cos^2\dfrac{3\pi}{8}+\cos^2\dfrac{5\pi}{8}+\cos^2\dfrac{7\pi}{8} = 2$

Given

(i) To prove cotα - tanα = 2cot 2α

L.H.S. L.H.S. = cotα - tanα

Working

= cosαsinα - sinαcosα = cos²α - sin²αsinαcosα

= cos 2αsinαcosα = 2cos 2α2sinα · cosα

Result

= 2cos 2αsin 2α = 2cot 2α = R.H.S.

Given

(ii) To prove sin 2α1+cos 2α = tanα

L.H.S. L.H.S. = sin 2α1+cos 2α

Working

= 2sinαcosα2cos²α = sinαcosα

Result

= tanα = R.H.S.

Given

(iii) To prove 1-cosαsinα = tanα2

L.H.S. L.H.S. = 1-cosαsinα

Working

= frac{2sin²α2{2sinα2cosα2 = frac{sinα2{cosα2

Result

= tanα2 = R.H.S.

Given

(iv) To prove cosα-sinαcosα+sinα = sec 2α - tan 2α

R.H.S. R.H.S. = sec 2α - tan 2α

Working

= 1cos 2α - sin 2αcos 2α

= 1-sin 2αcos 2α

= sin²α+cos²α-2sinαcosαcos²α-sin²α

= (cosα-sinα)²(cosα-sinα)(cosα+sinα)

Result

= cosα-sinαcosα+sinα = L.H.S.

Alternate Solution, L.H.S. L.H.S. = cosα-sinαcosα+sinα

Working

= cosα-sinαcosα+sinα × cosα-sinαcosα-sinα

= (cosα-sinα)²cos²α-sin²α

= cos²α+sin²α-2sinαcosαcos 2α

= 1-sin 2αcos 2α

Result

= 1cos 2α - sin 2αcos 2α = sec 2α - tan 2α = R.H.S.

Given

(v) To prove sqrt{1+sinα1-sinα = frac{sinα2+cosα2{sinα2-cosα2

L.H.S. L.H.S. = sqrt{1+sinα1-sinα

Working

= sqrt{frac{sin²α2+cos²α2+2sinα2cosα2{sin²α2+cos²α2-2sinα2cosα2

= sqrt{frac{left(sinα2+cosα2right)²}{left(sinα2-cosα2right)²} = frac{sinα2+cosα2{sinα2-cosα2

Result

= R.H.S.

Given

(vi) To prove cscθ+2csc 2θsecθ = cotθ2

L.H.S. L.H.S. = cscθ+2csc 2θsecθ

= cscθsecθ + 2csc 2θsecθ = cosθsinθ + 2cosθsin 2θ

Working

= cosθsinθ + 2cosθ2sinθcosθ = cosθsinθ + 1sinθ

= 1+cosθsinθ

= frac{2cos²θ2{2sinθ2cosθ2

Result

= frac{cosθ2{sinθ2 = cotθ2 = R.H.S.

Given

(vii) To prove 1+tanαtan 2α = sec 2α

L.H.S. L.H.S. = 1+tanαtan 2α

Working

= 1+sinαcosα · sin 2αcos 2α

= cos 2αcosα+sin 2α · sinαcosαcos 2α

= cos(2α-α)cosαcos 2α = cosαcosαcos 2α

Result

= 1cos 2α = sec 2α = R.H.S.

Given

(viii) To prove 2sinθsin 2θcosθ+cos 3θ = tan 2θtanθ

L.H.S. L.H.S. = 2sinθsin 2θcosθ+cos 3θ

Working

= 2sinθsin 2θcosθ+left(4cos³θ-3cosθright)

= 2sinθsin 2θcosθleft[1+4cos²θ-3right]

= 2sinθsin 2θcosθleft(4cos²θ-2right)

= 2sinθsin 2θ2cosθleft(2cos²θ-1right)

= 2sinθsin 2θ2cosθ · cos 2θ = sinθcosθ · sin 2θcos 2θ

Result

= tanθ · tan 2θ = R.H.S.

Given

(ix) To prove sin 3θsinθ - cos 3θcosθ = 2

L.H.S. L.H.S. = sin 3θsinθ - cos 3θcosθ

= sin 3θcosθ - cos 3θsinθsinθcosθ

= sin(3θ-θ)sinθcosθ = sin 2θsinθcosθ

Working

= 2sinθcosθsinθcosθ

Result

= 2 = R.H.S.

Given

(x) To prove cos 3θcosθ + sin 3θsinθ = 4cos 2θ

L.H.S. L.H.S. = cos 3θcosθ + sin 3θsinθ

= cos 3θsinθ + sin 3θcosθcosθsinθ

= sin 3θcosθ + cos 3θsinθsinθcosθ

= sin(3θ+θ)sinθcosθ = sin 4θsinθcosθ

Working

= 2sin 2θcos 2θsinθcosθ

Multiply and divide by 2 = 2 × 2sin 2θcos 2θ2sinθcosθ = 4sin 2θcos 2θsin 2θ

Result

= 4cos 2θ = R.H.S.

Given

(xi) To prove frac{tanθ2+cotθ2{cotθ2-tanθ2 = secθ

L.H.S. L.H.S. = frac{tanθ2+cotθ2{cotθ2-tanθ2

Working

= dfrac{dfrac{sinθ2{cosθ2+dfrac{cosθ2{sinθ2{dfrac{cosθ2{sinθ2-dfrac{sinθ2{cosθ2

= frac{sin²θ2+cos²θ2{cos²θ2-sin²θ2

= 1cosθ

Result

= secθ = R.H.S.

Given

(xii) To prove sin 3θcosθ + cos 3θsinθ = 2cot 2θ

L.H.S. L.H.S. = sin 3θcosθ + cos 3θsinθ

= sin 3θsinθ + cos 3θ · cosθcosθ · sinθ

= cos(3θ-θ)cosθsinθ = cos 2θcosθsinθ

Working

= 2cos 2θ2sinθcosθ = 2cos 2θsin 2θ

Result

= 2cot 2θ = R.H.S.

Given

(xiii) To prove 3+cos 4θ1-cos 4θ = 12left(tan²θ+cot²θright)

R.H.S. R.H.S. = 12left(tan²θ+cot²θright)

Working

= 12left(sin²θcos²θ+cos²θsin²θright)

= 12left(sin⁴θ+cos⁴θsin²θcos²θright)

= left(sin²θright)²+left(cos²θright)²+2left(sin²θright)left(cos²θright)-2left(sin²θright)left(cos²θright)2sin²θcos²θ

= left(sin²θ+cos²θright)²-2sin²θcos²θ2sin²θcos²θ

= (1)²-2sin²θcos²θ2sin²θcos²θ

= 12sin²θcos²θ - 2sin²θcos²θ2sin²θcos²θ

= 24sin²θcos²θ - 1 = 2left(2sinθcosθright)² - 1

= 2sin² 2θ - 1 = frac{2}{1-cos 4θ2 - 1 = 41-cos 4θ - 1

Result

= 4-(1-cos 4θ)1-cos 4θ = 3+cos 4θ1-cos 4θ = R.H.S.

Given

(xiv) To prove 1+sin 2θ1-sin 2θ = tan²left(pi4+θright)

L.H.S. L.H.S. = 1+sin 2θ1-sin 2θ

Working

= frac{1+cosleft(pi2-2θright)}{1-cosleft(pi2-2θright)} qquad because cosleft(pi2-θright)=sinθ

= frac{2cos²left[dfrac{left(pi2-2θright)}{2}right]}{2sin²left[dfrac{left(pi2-2θright)}{2}right]}

= cot²left(pi4-θright) qquad because tanleft(pi2-θright)=cotθ

= tan²left(pi2-left(pi4-θright)right)

Result

= tan²left(pi4+θright) = R.H.S.

Given

(xv) To prove cos²pi8+cos²3pi8+cos²5pi8+cos²7pi8 = 2

L.H.S. L.H.S. = cos²pi8+cos²3pi8+cos²5pi8+cos²7pi8

Working

= cos²pi8 + left[cosleft(pi2-pi8right)right]² + left[cosleft(pi2+pi8right)right]² + left[cosleft(pi-pi8right)right]²

= cos²pi8 + sin²pi8 + sin²pi8 + cos²pi8

Result

= 1+1 = 2 = R.H.S.

3.Show that: $2\cos\theta = \sqrt{2+\sqrt{2+2\cos 4\theta}}$

Given

To prove 2cosθ = sqrt{2+sqrt{2+2cos 4θ}

R.H.S. R.H.S. = sqrt{2+sqrt{2+2cos 4θ}

= sqrt{2+sqrt{2left(1+cos 4θright)}

Working

= sqrt{2+sqrt{2left(2cos² 2θright)}

= sqrt{2+2cos 2θ}

= sqrt{2left(1+cos 2θright)} = sqrt{2left(2cos²θright)}

Result

= 2cosθ = L.H.S.

4.Reduce $\sin^4\theta$ to an expression involving only function of multiples of $\theta$ raised to the first power.

Given

Here sin⁴θ = left(sin²θright)²

Formula

Half-angle formula sin²θ = 1-cos 2θ2

Working

sin⁴θ = left(1-cos 2θ2right)² = 14left(1-cos 2θright)²

= 14left(1+cos² 2θ - 2cos 2θright)

Formula

Half-angle formula cos² 2θ = 1+cos 4θ2

Working

= 14left(1+1+cos 4θ2 - 2cos 2θright)

= 14left(1+12+cos 4θ2 - 2cos 2θright) = 14left(32+cos 4θ2 - 2cos 2θright)

= 38+cos 4θ8 - 2cos 2θ4

Result

= 3-4cos 2θ+cos 4θ8

5.Find the value of sin θ and cos θ without using table or calculator, when θ is (i) 18° (ii) 36° (iii) 54° (iv) 72°. Hence prove that: cos 36° cos 72° cos 108° cos 144° = 1/16

Result

For θ = 18° sin 18° = frac{sqrt{5}-1}{4}, quad cos 18° = frac{sqrt{10+2sqrt{5}{4}

For θ = 36° sin 36° = frac{sqrt{10-2sqrt{5}{4}, quad cos 36° = frac{sqrt{5}+1}{4}

For θ = 54° sin 54° = frac{sqrt{5}+1}{4}, quad cos 54° = frac{sqrt{10-2sqrt{5}{4}

For θ = 72° sin 72° = frac{sqrt{10+2sqrt{5}{4}, quad cos 72° = frac{sqrt{5}-1}{4}

Angle conversions cos 108° = cos(180° - 72°) = -cos 72° = -frac{sqrt{5}-1}{4}

cos 144° = cos(180° - 36°) = -cos 36° = -frac{sqrt{5}+1}{4}

Formula

The product cos 36° cos 72° cos 108° cos 144°

Working

Substituting values = frac{sqrt{5}+1}{4} · frac{sqrt{5}-1}{4} · left(-frac{sqrt{5}-1}{4}right) · left(-frac{sqrt{5}+1}{4}right)

= frac{(sqrt{5}+1)(sqrt{5}-1)}{16} · frac{(sqrt{5}-1)(sqrt{5}+1)}{16}

Difference of squares (sqrt{5}+1)(sqrt{5}-1) = 5 - 1 = 4

= 416 · 416 = 14 · 14

Result

Therefore cos 36° cos 72° cos 108° cos 144° = 116

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Unit 10: Trigonometric Identities — Exercise 10 3