11th Class Mathematics Exercise 10.2 Solutions

1.(i) Without using table find the value of sin 15° (ii) Without using table find the value of cos 15° (iii) Without using table find the value of tan 15° (iv) Without using table find the value of sin 105° (v) Without using table find the value of cos 105° (vi) Without using table find the value of tan 105°

Formula

(i) Difference formula sin 15° = sin(45° - 30°)

Working

Expanding = sin 45° cos 30° - cos 45° sin 30°

= left(frac{1}{sqrt{2}right)left(frac{sqrt{3}{2}right) - left(frac{1}{sqrt{2}right)left(12right)

= frac{sqrt{3}{2sqrt{2} - frac{1}{2sqrt{2}

Result

sin 15° = frac{sqrt{3} - 1}{2sqrt{2}

Formula

(ii) Difference formula cos 15° = cos(45° - 30°)

Working

Expanding = cos 45° cos 30° + sin 45° sin 30°

= left(frac{1}{sqrt{2}right)left(frac{sqrt{3}{2}right) + left(frac{1}{sqrt{2}right)left(12right)

= frac{sqrt{3}{2sqrt{2} + frac{1}{2sqrt{2}

Result

cos 15° = frac{sqrt{3} + 1}{2sqrt{2}

Formula

(iii) Difference formula tan 15° = tan(45° - 30°) = tan 45° - tan 30°1 + tan 45° tan 30°

Working

Substituting = frac{1 - frac{1}{sqrt{3}{1 + 1 · frac{1}{sqrt{3}

= frac{1 - frac{1}{sqrt{3}{1 + frac{1}{sqrt{3} = frac{frac{sqrt{3} - 1}{sqrt{3}{frac{sqrt{3} + 1}{sqrt{3}

= frac{sqrt{3} - 1}{sqrt{3} + 1}

Result

tan 15° = frac{sqrt{3} - 1}{sqrt{3} + 1}

Formula

(iv) Sum formula sin 105° = sin(60° + 45°)

Working

Expanding = sin 60° cos 45° + cos 60° sin 45°

= left(frac{sqrt{3}{2}right)left(frac{1}{sqrt{2}right) + left(12right)left(frac{1}{sqrt{2}right)

= frac{sqrt{3}{2sqrt{2} + frac{1}{2sqrt{2}

Result

sin 105° = frac{sqrt{3} + 1}{2sqrt{2}

Formula

(v) Sum formula cos 105° = cos(60° + 45°)

Working

Expanding = cos 60° cos 45° - sin 60° sin 45°

= left(12right)left(frac{1}{sqrt{2}right) - left(frac{sqrt{3}{2}right)left(frac{1}{sqrt{2}right)

= frac{1}{2sqrt{2} - frac{sqrt{3}{2sqrt{2}

Result

cos 105° = frac{1 - sqrt{3}{2sqrt{2}

Formula

(vi) Sum formula tan 105° = tan(60° + 45°) = tan 60° + tan 45°1 - tan 60° tan 45°

Working

Substituting = frac{sqrt{3} + 1}{1 - sqrt{3} · 1}

= frac{sqrt{3} + 1}{1 - sqrt{3}

Result

tan 105° = frac{sqrt{3} + 1}{1 - sqrt{3}

2.(i) Prove that sin(45° + α) = (1/√2)(sin α + cos α) (ii) Prove that cos(α + 45°) = (1/√2)(cos α − sin α)

Given

(i) To prove sin(45°+α) = frac{1}{sqrt{2}(sinα+cosα)

L.H.S = sin(45°+α)

Formula

= sin 45°cosα + cos 45°sinα

Working

= frac{1}{sqrt{2}cosα + frac{1}{sqrt{2}sinα

Result

= frac{1}{sqrt{2}(sinα+cosα) = R.H.S

Given

(ii) To prove cos(α+45°) = frac{1}{sqrt{2}(cosα-sinα)

L.H.S = cos(α+45°)

Formula

= cosαcos 45° - sinαsin 45°

Working

= (cosα)left(frac{1}{sqrt{2}right) - (sinα)left(frac{1}{sqrt{2}right)

Result

= frac{1}{sqrt{2}(cosα-sinα) = R.H.S

3.(i) Prove that tan(45° + A) tan(45° − A) = 1 (ii) Prove that tan(π/4 − α) + tan(3π/4 + α) = 0 (iii) Prove that sin(θ + π/6) + cos(θ + π/3) = cos θ (iv) Prove that (sin θ − cos θ tan(θ/2)) / (cos θ + sin θ tan(θ/2)) = tan(θ/2) (v) Prove that (1 − tan θ tan φ) / (1 + tan θ tan φ) = cos(θ+φ) / cos(θ−φ)

Given

(i) To prove tan(45°+A)tan(45°-A) = 1

L.H.S = tan(45°+A)tan(45°-A)

Formula

= left(tan 45°+tan A1-tan 45°tan Aright)left(tan 45°-tan A1+tan 45°tan Aright)

Working

= left(1+tan A1-tan Aright)left(1-tan A1+tan Aright)

Result

= 1 = R.H.S

Given

(ii) To prove tanleft(pi4-αright) + tanleft(3pi4+αright) = 0

L.H.S = tanleft(pi4-θright) + tanleft(3pi4+θright)

Formula

= left(frac{tanpi4-tanθ}{1+tanpi4tanθ}right) + left(frac{tan3pi4+tanθ}{1-tan3pi4tanθ}right)

Working

= left(1-tanθ1+tanθright) + left(-1+tanθ1-(-1)tanθright)

= left(1-tanθ1+tanθright) + left(-1+tanθ1+tanθright)

= 1-tanθ-1+tanθ1+tanθ = 01+tanθ

Result

= 0 = R.H.S

Given

(iii) To prove sinleft(θ+pi6right) + cosleft(θ+pi3right) = cosθ

L.H.S = sinleft(θ+pi6right) + cosleft(θ+pi3right)

Formula

= left(sinθcospi6+cosθsinpi6right) + left(cosθcospi3-sinθsinpi3right)

Working

= sinθ · frac{sqrt{3}{2} + cosθ · 12 + cosθ · 12 - sinθ · frac{sqrt{3}{2}

= 12cosθ + 12cosθ = 2 × 12cosθ

Result

= cosθ = R.H.S

Given

(iv) To prove frac{sinθ - cosθtanθ2{cosθ + sinθtanθ2 = tanθ2

L.H.S = dfrac{sinθ - cosθtanθ2{cosθ + sinθtanθ2

Working

= dfrac{sinθ - cosθ × dfrac{sinθ2{cosθ2{cosθ + sinθ × dfrac{sinθ2{cosθ2

= dfrac{dfrac{sinθcosθ2 - cosθsinθ2{cosθ2{dfrac{cosθcosθ2 + sinθsinθ2{cosθ2

= dfrac{sinθcosθ2 - cosθsinθ2{cosθcosθ2 + sinθsinθ2

Formula

= dfrac{sinleft(θ-θ2right)}{cosleft(θ-θ2right)}

Result

= dfrac{sinθ2{cosθ2 = tanθ2 = R.H.S

Given

(v) To prove 1-tanθtanphi1+tanθtanphi = cos(θ+phi)cos(θ-phi)

L.H.S = 1-tanθtanphi1+tanθtanphi

Working

= dfrac{1-sinθcosθ × sinphicosphi{1+sinθcosθ × sinphicosphi

= dfrac{cosθcosphi - sinθsinphicosθcosphi{cosθcosphi + sinθsinphicosθcosphi

= cosθcosphi - sinθsinphicosθcosphi + sinθsinphi

Result

= cos(θ+phi)cos(θ-phi) = R.H.S

4.Show that cos(α+β) cos(α−β) = cos²α − sin²β = cos²β − sin²α

Given

To prove cos(α+β)cos(α-β) = cos²α - sin²β = cos²β - sin²α

cos(α+β)cos(α-β)

Formula

= (cosαcosβ-sinαsinβ)(cosαcosβ+sinαsinβ)

= (cosαcosβ)² - (sinαsinβ)²

= cos²αcos²β - sin²αsin²β

Working

= cos²α(1-sin²β) - (1-cos²α)sin²β

= cos²α - cos²αsin²β - sin²β + cos²αsin²β

Result

= cos²α - sin²β

Further it can be written as = (1-sin²α) - (1-cos²β)

= 1 - sin²α - 1 + cos²β

= cos²β - sin²α

5.Show that [sin(α+β) + sin(α−β)] / [cos(α+β) + cos(α−β)] = tan α

Given

To prove sin(α+β)+sin(α-β)cos(α+β)+cos(α-β) = tanα

L.H.S = sin(α+β)+sin(α-β)cos(α+β)+cos(α-β)

Working

= sinαcosβ+cosαsinβ+sinαcosβ-cosαsinβcosαcosβ-sinαsinβ+cosαcosβ+sinαsinβ

= 2sinαcosβ2cosαcosβ = sinαcosα

Result

= tanα = R.H.S

6.(i) Show that sin²(α + β/2) − sin²(α − β/2) = sin2α sinβ (ii) Show that sin²α + sin²β + cos²(α+β) + 2 sinα sinβ cos(α+β) = 1

Given

(i) To prove sin²left(α+β2right) - sin²left(α-β2right) = sin 2αsinβ

L.H.S = sin²left(α+β2right) - sin²left(α-β2right) = left[sinleft(α+β2right)+sinleft(α-β2right)right]left[sinleft(α+β2right)-sinleft(α-β2right)right]

Working

= left[sinαcosβ2+cosαsinβ2+sinαcosβ2-cosαsinβ2right]left[sinαcosβ2+cosαsinβ2-sinαcosβ2+cosαsinβ2right]

= left(2sinαcosβ2right)left(2cosαsinβ2right) = (2sinαcosα)left(2sinβ2cosβ2right)

Result

= sin 2αsinβ = R.H.S

Given

(ii) To prove sin²α + sin²β + cos²(α+β) + 2sinαsinβcos(α+β) = 1

L.H.S = sin²α + sin²β + cos²(α+β) + 2sinαsinβcos(α+β) quad ...(i)

Adding and subtracting sin²α sin²β = sin²α+sin²β-sin²αsin²β+sin²αsin²β+cos²(α+β)+2sinαsinβcos(α+β)

= sin²α+sin²β-sin²αsin²β+left[(sinαsinβ)²+cos²(α+β)+2sinαsinβcos(α+β)right]

= sin²α+sin²β-sin²αsin²β+left[sinαsinβ+cos(α+β)right]²

Working

= sin²α+sin²β-sin²αsin²β+left[sinαsinβ+cosαcosβ-sinαsinβright]²

= sin²α+sin²β-sin²αsin²β+(cosαcosβ)² = sin²α+sin²β-sin²αsin²β+cos²αcos²β

= sin²α+sin²β-sin²αsin²β+(1-sin²α)(1-sin²β)

= sin²α+sin²β-sin²αsin²β+1-sin²β-sin²α+sin²αsin²β

Result

= 1 = R.H.S

7.(i) Show that cos(α−β) = (1 + tanα tanβ) / (secα secβ) (ii) Show that sin(α+β) = (1 + cotα tanβ) / (cosecα secβ) (iii) Show that cot(α−β) = (cotα cotβ + 1) / (cotβ − cotα) (iv) Show that (tanα + tanβ) / (tanα − tanβ) = sin(α+β) / sin(α−β) (v) Show that cot(α+β) = (cotα cotβ − 1) / (cotα + cotβ)

Given

(i) To prove cos(α-β) = 1+tanαtanβsecαsecβ

L.H.S = cos(α-β)

Formula

= cosαcosβ + sinαsinβ = cosαcosβ+sinαsinβ1

Divide numerator and denominator by cosα cosβ = dfrac{cosαcosβcosαcosβ+sinαsinβcosαcosβ{1cosαcosβ

Result

= 1+tanαtanβsecα · secβ = R.H.S

Given

(ii) To prove sin(α+β) = frac{1+cotαtanβ}{cosecαsecβ}

L.H.S = sin(α+β)

Formula

= sinαcosβ + cosαsinβ = sinαcosβ+cosαsinβ1

Divide numerator and denominator by sinα cosβ = dfrac{sinαcosβsinαcosβ+cosαsinβsinαcosβ{1sinαcosβ

Result

= dfrac{1+cotαtanβ}{cosecαsecβ} = R.H.S

Given

(iii) To prove cot(α-β) = cotαcotβ+1cotβ-cotα

L.H.S = cot(α-β) = 1tan(α-β)

Formula

= dfrac{1}{tanα-tanβ1+tanαtanβ = 1+tanαtanβtanα-tanβ

Working

= dfrac{1+1cotαcotβ{1cotα-1cotβ

= dfrac{cotαcotβ+1cotαcotβ{cotβ-cotαcotαcotβ

Result

= cotαcotβ+1cotβ-cotα = R.H.S

Given

(iv) To prove tanα+tanβtanα-tanβ = sin(α+β)sin(α-β)

L.H.S = tanα+tanβtanα-tanβ

Working

= dfrac{sinαcosα+sinβcosβ{sinαcosα-sinβcosβ

= dfrac{sinαcosβ+cosαsinβcosαcosβ{sinαcosβ-cosαsinβcosαcosβ

= sinαcosβ+cosαsinβsinαcosβ-cosαsinβ

Result

= sin(α+β)sin(α-β) = R.H.S

Given

(v) To prove cot(α+β) = cotαcotβ-1cotα+cotβ

L.H.S = cot(α+β) = 1tan(α+β) = dfrac{1}{tanα+tanβ1-tanαtanβ

= 1-tanαtanβtanα+tanβ

Working

= dfrac{1-1cotαcotβ{1cotα+1cotβ

= dfrac{cotαcotβ-1cotαcotβ{cotβ+cotαcotαcotβ

Result

= cotαcotβ-1cotα+cotβ = R.H.S

8.If sin α = 24/25 and cos β = 20/29 where 0 < α < π/2 and 0 < β < π/2 show that sin(α - β) = 333/725

Given

sin α = 2425, quad 0 < α < pi2

cos β = 2029, quad 0 < β < pi2

cos α = sqrt{1 - sin² α} = sqrt{1 - left(2425right)²} = sqrt{625 - 576625 = sqrt{49625 = 725

sin β = sqrt{1 - cos² β} = sqrt{1 - left(2029right)²} = sqrt{841 - 400841 = sqrt{441841 = 2129

Formula

Difference formula sin(α - β) = sin α cos β - cos α sin β

Working

Substituting = 2425 · 2029 - 725 · 2129

= 480725 - 147725

Result

sin(α - β) = 333725

9.(i) If sin α = -8/17 and cos β = -4/5 where 3π/2 < α < 2π and π < β < 3π/2 find sin(α + β) (ii) If sin α = -8/17 and cos β = -4/5 where 3π/2 < α < 2π and π < β < 3π/2 find cos(α + β) (iii) If sin α = -8/17 and cos β = -4/5 where 3π/2 < α < 2π and π < β < 3π/2 find tan(α + β) (iv) If sin α = -8/17 and cos β = -4/5 where 3π/2 < α < 2π and π < β < 3π/2 find sin(α - β) (v) If sin α = -8/17 and cos β = -4/5 where 3π/2 < α < 2π and π < β < 3π/2 find cos(α - β) (vi) If sin α = -8/17 and cos β = -4/5 where 3π/2 < α < 2π and π < β < 3π/2 find tan(α - β)

Given

(i) sin α = -817, quad 3pi2 < α < 2pi

cos β = -45, quad pi < β < 3pi2

cos α (in quad IV, +ve) = sqrt{1 - left(-817right)²} = sqrt{289 - 64289 = sqrt{225289 = 1517

sin β (in quad III, -ve) = -sqrt{1 - left(-45right)²} = -sqrt{25 - 1625 = -sqrt{925 = -35

Formula

Sum formula sin(α + β) = sin α cos β + cos α sin β

Working

Substituting = left(-817right)left(-45right) + left(1517right)left(-35right)

= 3285 - 4585

Result

sin(α + β) = -1385

Formula

(ii) Sum formula cos(α + β) = cos α cos β - sin α sin β

Working

Substituting = left(1517right)left(-45right) - left(-817right)left(-35right)

= -6085 - 2485

Result

cos(α + β) = -8485

Formula

(iii) tan(α + β) = sin(α + β)cos(α + β)

Working

Substituting = frac{-1385{-8485

Result

tan(α + β) = 1384

Formula

(iv) Difference formula sin(α - β) = sin α cos β - cos α sin β

Working

Substituting = left(-817right)left(-45right) - left(1517right)left(-35right)

= 3285 + 4585

Result

sin(α - β) = 7785

Formula

(v) Difference formula cos(α - β) = cos α cos β + sin α sin β

Working

Substituting = left(1517right)left(-45right) + left(-817right)left(-35right)

= -6085 + 2485

Result

cos(α - β) = -3685

Formula

(vi) tan(α - β) = sin(α - β)cos(α - β)

Working

Substituting = frac{7785{-3685

Result

tan(α - β) = -7736

10.(i) Find sin(α + β) and cos(α + β) given that tan α = 3/4, cos β = 5/13 and neither the terminal sides of the angle of measure α nor that of β is in the quadrant I. (ii) Find sin(α + β) and cos(α + β) given that tan α = -15/8 and sin β = -7/25 and neither the terminal side of the angle of measure α nor that of β is in the quadrant II.

Given

(i) tan α = 34

α in quad III (not quad I) sec α = sqrt{1 + tan² α} = sqrt{1 + 916 = sqrt{2516 = 54

cos α = -45, sin α = tan α · cos α = 34 · left(-45right) = -35

cos β = 513

β in quad IV (not quad I) sin β = -sqrt{1 - cos² β} = -sqrt{1 - 25169 = -sqrt{144169 = -1213

Formula

sin(α + β) = sin α cos β + cos α sin β

Working

Substituting = left(-35right)left(513right) + left(-45right)left(-1213right)

= -1565 + 4865

Result

sin(α + β) = 3365

Formula

cos(α + β) = cos α cos β - sin α sin β

Working

Substituting = left(-45right)left(513right) - left(-35right)left(-1213right)

= -2065 - 3665

Result

cos(α + β) = -5665

Given

(ii) tan α = -158

α in quad II (not quad II) - so quad II sec α = sqrt{1 + tan² α} = sqrt{1 + 22564 = sqrt{28964 = 178

sec α = -178, cos α = -817, sin α = tan α · cos α = left(-158right) · left(-817right) = 1517

sin β = -725

β in quad III (not quad II) cos β = -sqrt{1 - sin² β} = -sqrt{1 - 49625 = -sqrt{576625 = -2425

Formula

sin(α + β) = sin α cos β + cos α sin β

Working

Substituting = left(1517right)left(-2425right) + left(-817right)left(-725right)

= -360425 + 56425

Result

sin(α + β) = -304425

Formula

cos(α + β) = cos α cos β - sin α sin β

Working

Substituting = left(-817right)left(-2425right) - left(1517right)left(-725right)

= 192425 + 105425

Result

cos(α + β) = 297425

11.Prove that (cos19° + sin19°) / (cos19° − sin19°) = tan64°

Given

To prove cos 19°+sin 19°cos 19°-sin 19° = tan 64°

R.H.S = tan 64° = tan(45°+19°)

Formula

= tan 45°+tan 19°1-tan 45°tan 19°

Working

= 1+tan 19°1-(1)tan 19° = dfrac{1+sin 19°cos 19°{1-sin 19°cos 19°

= dfrac{cos 19°+sin 19°cos 19°{cos 19°-sin 19°cos 19°

Result

= cos 19°+sin 19°cos 19°-sin 19° = L.H.S

12.Prove that cos(60°+θ) cos(60°−θ) + sin(60°+θ) sin(60°−θ) = cos2θ

Given

To prove cos(60°+θ)cos(60°-θ) + sin(60°+θ)sin(60°-θ) = cos 2θ

L.H.S = cos(60°+θ)cos(60°-θ) + sin(60°+θ)sin(60°-θ)

Formula

= cosleft[(60°+θ)-(60°-θ)right] = cos 2θ

Result

L.H.S = R.H.S

13.If α, β, γ are the angles of a triangle ABC, show that cot(α/2) + cot(β/2) + cot(γ/2) = cot(α/2) cot(β/2) cot(γ/2)

Given

To prove cotα2+cotβ2+cotgamma2 = cotα2cotβ2cotgamma2

It is given that α, β, γ are the angles of triangle ABC, so α+β+gamma = 180°

α+β = 180°-gamma

α2+β2 = 90°-gamma2

Formula

cotleft(α2+β2right) = cotleft(90°-gamma2right)

Working

dfrac{cotα2cotβ2-1}{cotα2+cotβ2 = tangamma2

dfrac{cotα2+cotβ2{cotα2cotβ2-1} = cotgamma2

cotα2+cotβ2 = cotgamma2left[cotα2cotβ2-1right]

cotα2+cotβ2 = cotα2cotβ2cotgamma2 - cotgamma2

Result

cotα2+cotβ2+cotgamma2 = cotα2cotβ2cotgamma2

14.If α + β + γ = 180° show that cotα cotβ + cotβ cotγ + cotγ cotα = 1

Given

To prove cotαcotβ+cotβcotgamma+cotgammacotα = 1

Here α+β+gamma = 180°

α+β = 180°-gamma

Formula

cot(α+β) = cot(180°-gamma)

Working

cotαcotβ-1cotα+cotβ = -cotgamma

cotαcotβ-1 = -cotgammaleft[cotα+cotβright]

cotαcotβ-1 = -cotgammacotα - cotgammacotβ

Result

cotαcotβ+cotβcotgamma+cotgammacotα = 1

15.(i) Express 24 sinθ + 7 cosθ in the form r sin(θ+φ) or r sin(θ−φ), where the terminal sides of the angles of measure θ and φ are in the first quadrant (ii) Express 12 sinθ − 5 cosθ in the form r sin(θ+φ) or r sin(θ−φ) (iii) Express sinθ − cosθ in the form r sin(θ+φ) or r sin(θ−φ) (iv) Express 8 sinθ − 6 cosθ in the form r sin(θ+φ) or r sin(θ−φ) (v) Express (1/2) sinθ + (√3/2) cosθ in the form r sin(θ+φ) or r sin(θ−φ) (vi) Express 13 sinθ − 84 cosθ in the form r sin(θ+φ) or r sin(θ−φ)

(i) Let 24 = rcosphi quad ...(i)

7 = rsinphi quad ...(ii)

Squaring and adding (i) and (ii) 24²+7² = r²cos²phi+r²sin²phi

Working

576+49 = r²(cos²phi+sin²phi)

625 = r² implies 25 = r

Divide (ii) by (i) 724 = rsinphircosphi

724 = tanphi implies tan^-1724 = phi

Now 24sinθ+7cosθ = (rcosphi)sinθ+(rsinphi)cosθ

Formula

= rleft[sinθcosphi+cosθsinphiright]

Result

= rsin(θ+phi), quad where r=25 and phi=tan^-1724

(ii) Let 12 = rcosphi quad ...(i)

And 5 = rsinphi quad ...(ii)

Squaring and adding (i) and (ii) 12²+5² = r²cos²phi+r²sin²phi

Working

144+25 = r²(cos²phi+sin²phi)

169 = r² implies 13 = r

Divide (ii) by (i) 512 = rsinphircosphi

512 = tanphi implies tan^-1512 = phi

Now 12sinθ-5cosθ = (rcosphi)sinθ-(rsinphi)cosθ

Formula

= rleft[sinθcosphi-cosθsinphiright]

Result

= rsin(θ-phi), quad where r=13 and phi=tan^-1512

(iii) Let 1 = rcosphi quad ...(i)

and 1 = rsinphi quad ...(ii)

squaring and adding (i) and (ii) 1²+1² = r²cos²phi+r²sin²phi

Working

2 = r²(cos²phi+sin²phi)

2 = r² implies r=sqrt{2}

Divide (ii) by (i) 11 = rsinphircosphi

1 = tanphi implies phi=tan^-1(1)

Now sinθ-cosθ = (1)sinθ-(1)cosθ

= rcosphisinθ - rsinphicosθ

Formula

= rleft[sinθcosphi-cosθsinphiright]

Result

= rsin(θ-phi), quad where r=sqrt{2} and phi=tan^-1(1) = pi4

(iv) Let 8 = rcosphi quad ...(i)

and 6 = rsinphi quad ...(ii)

squaring and adding (i) and (ii) 8²+6² = r²cos²phi+r²sin²phi

Working

64+36 = r²(cos²phi+sin²phi)

100 = r² implies 10 = r

Divide (ii) by (i) 68 = rsinphircosphi

34 = tanphi implies tan^-134 = phi

Now 8sinθ-6cosphi = (rcosphi)sinθ-(rsinphi)cosθ

Formula

= rleft[sinθcosphi-cosθsinphiright]

Result

= rsin(θ-phi), quad where r=10 and phi=tan^-134

(v) Let 12 = rcosphi quad ...(i)

and frac{sqrt{3}{2} = rsinphi quad ...(ii)

squaring and adding (i) and (ii) left(12right)²+left(frac{sqrt{3}{2}right)² = r²cos²phi+r²sin²phi

Working

14+34 = r²(cos²phi+sin²phi)

1 = r² implies 1 = r

Divide (ii) by (i) dfrac{frac{sqrt{3}{2}{12 = rsinphircosphi implies sqrt{3}=tanphi

tan^-1sqrt{3} = phi

Now 12sinθ+frac{sqrt{3}{2}cosθ = rcosphisinθ+rsinphicosθ

Formula

= rleft[sinθcosphi+cosθsinphiright] = rsin(θ+phi)

Result

Where r=1 and phi=tan^-1sqrt{3}, quad phi=pi3

(vi) Let 13 = rcosphi quad ...(i)

and 84 = rsinphi quad ...(ii)

Squaring and adding (i) and (ii) 13²+84² = r²cos²phi+r²sin²phi

Working

169+7056 = r²(cos²phi+sin²phi)

7225 = r² implies 85 = r

Divide (ii) by (i) 8413 = rsinphircosphi

8413 = tanphi implies tan^-18413 = phi

Now 13sinθ-84cosθ = (rcosphi)sinθ-(rsinphi)cosθ

Formula

= rleft[sinθcosphi-cosθsinphiright]

Result

= rsin(θ-phi), quad where r=85 and phi=tan^-18413

Getting there and back

In school

At the gate

Fees & pay

Unit 10: Trigonometric Identities — Exercise 10 2