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Unit 10: Trigonometric Identities — Exercise 10 1

11th Class Mathematics · Unit 10: Trigonometric Identities

1.(i) Without using the tables, find the value of cos(-1230°) (ii) Without using the tables, find the value of tan(-1035°) (iii) Without using the tables, find the value of sec(1140°) (iv) Without using the tables, find the value of cosec(-690°) (v) Without using the tables, find the value of cot(1320°) (vi) Without using the tables, find the value of cos(-240°)

Given

(i) Angle -1230°

Reduce angle -1230° + 4(360°) = -1230° + 1440° = 210°

Express in standard form 210° = 180° + 30°

Formula

Apply identity cos(180° + θ) = -cos(θ)

Working

Substitute cos(210°) = -cos(30°) = -frac{sqrt{3}{2}

Result

Result cos(-1230°) = -frac{sqrt{3}{2}

Given

(ii) Angle -1035°

Reduce angle -1035° + 3(360°) = -1035° + 1080° = 45°

Formula

Evaluate tan(45°) = 1

Result

Result tan(-1035°) = 1

Given

(iii) Angle 1140°

Reduce angle 1140° - 3(360°) = 1140° - 1080° = 60°

Formula

Evaluate sec(60°) = 1cos(60°) = 11/2 = 2

Result

Result sec(1140°) = 2

Given

(iv) Angle -690°

Reduce angle -690° + 2(360°) = -690° + 720° = 30°

Formula

Evaluate cosec(30°) = 1sin(30°) = 11/2 = 2

Result

Result cosec(-690°) = 2

Given

(v) Angle 1320°

Reduce angle 1320° - 3(360°) = 1320° - 1080° = 240°

Express in standard form 240° = 180° + 60°

Formula

Apply identity cot(180° + θ) = cot(θ)

Working

Substitute cot(240°) = cot(60°) = frac{1}{sqrt{3}

Result

Result cot(1320°) = frac{1}{sqrt{3}

Given

(vi) Angle -240°

Use even property cos(-θ) = cos(θ), so cos(-240°) = cos(240°)

Reduce angle 240° = 180° + 60°

Formula

Apply identity cos(180° + θ) = -cos(θ)

Working

Substitute cos(240°) = -cos(60°) = -12

Result

Result cos(-240°) = -12

2.(i) Express cos168° as a trigonometric function of an angle of positive degree measure of less than 45° (ii) Express sin192° as a trigonometric function of an angle of positive degree measure of less than 45° (iii) Express cos333° as a trigonometric function of an angle of positive degree measure of less than 45° (iv) Express tan213° as a trigonometric function of an angle of positive degree measure of less than 45° (v) Express cos(-435°) as a trigonometric function of an angle of positive degree measure of less than 45° (vi) Express sin219° as a trigonometric function of an angle of positive degree measure of less than 45° (vii) Express tan(-597°) as a trigonometric function of an angle of positive degree measure of less than 45° (viii) Express cos(-111°) as a trigonometric function of an angle of positive degree measure of less than 45° (ix) Express sin(-390°) as a trigonometric function of an angle of positive degree measure of less than 45°

Given

(i) Angle 168°

Express relative to 180° 168° = 180° - 12°

Formula

Apply identity cos(180° - θ) = -cos(θ)

Working

Substitute cos(168°) = -cos(12°)

Result

Result -cos(12°)

Given

(ii) Angle 192°

Express relative to 180° 192° = 180° + 12°

Formula

Apply identity sin(180° + θ) = -sin(θ)

Working

Substitute sin(192°) = -sin(12°)

Result

Result -sin(12°)

Given

(iii) Angle 333°

Express relative to 360° 333° = 360° - 27°

Formula

Apply identity cos(360° - θ) = cos(θ)

Working

Substitute cos(333°) = cos(27°)

Result

Result cos(27°)

Given

(iv) Angle 213°

Express relative to 180° 213° = 180° + 33°

Formula

Apply identity tan(180° + θ) = tan(θ)

Working

Substitute tan(213°) = tan(33°)

Result

Result tan(33°)

Given

(v) Angle -435°

Use even property and add 360° cos(-435°) = cos(435° - 360°) = cos(75°)

Express as complementary angle 75° = 90° - 15°

Formula

Apply cofunction identity cos(90° - θ) = sin(θ)

Working

Substitute cos(75°) = sin(15°)

Result

Result sin(15°)

Given

(vi) Angle 219°

Express relative to 180° 219° = 180° + 39°

Formula

Apply identity sin(180° + θ) = -sin(θ)

Working

Substitute sin(219°) = -sin(39°)

Result

Result -sin(39°)

Given

(vii) Angle -597°

Add 2 × 360° to reduce -597° + 2(360°) = -597° + 720° = 123°

Express relative to 90° 123° = 90° + 33°

Formula

Apply identity tan(90° + θ) = -cot(θ)

Working

Substitute tan(123°) = -cot(33°)

Result

Result -cot(33°)

Given

(viii) Angle -111°

Use even property cos(-111°) = cos(111°)

Express relative to 90° 111° = 90° + 21°

Formula

Apply identity cos(90° + θ) = -sin(θ)

Working

Substitute cos(111°) = -sin(21°)

Result

Result -sin(21°)

Given

(ix) Angle -390°

Use odd property and add 360° sin(-390°) = -sin(390°) = -sin(390° - 360°) = -sin(30°)

Result

Result -sin(30°)

3.(i) Prove that: sin(180° + α) sin(90° − α) = −sin α cos α (ii) Prove that: sin810° sin630° + cos135° sin225° = −1/2 (iii) Prove that: tan150° cot330° − 2 sec135° cosec225° = −3 (iv) Prove that: sin210° + cos240° + tan225° + cot225° = 1

Given

(i) To prove sin(180°+α)sin(90°-α) = -sinαcosα

L.H.S L.H.S = sin(180°+α)sin(90°-α)

Working

Apply identities = [-sinα][cosα]

Simplify = -sinαcosα

Result

Result = R.H.S

Given

(ii) To prove sin810°sin630° + cos135°sin225° = -12

L.H.S L.H.S = sin810°sin630° + cos135°sin225°

Reduce angles = sinbig(810°-2(360°)big)sinbig(630°-2(360°)big) + cos(180°-45°)sin(180°+45°)

Working

Apply identities = sin(810°-720°)sin(630°-720°) + [-cos45°][-sin45°]

Substitute values = sin(90°)sin(-90°) + left(1sqrt2right)left(1sqrt2right)

Simplify = (1)(-1) + 12 = -1+12 = -12

Result

Result = R.H.S

Given

(iii) To prove tan150°cot330° - 2sec135°csc225° = -3

L.H.S L.H.S = tan150°cot330° - 2sec135°csc225°

Express in standard form = tan(180°-30°)cot(360°-30°) - 2sec(180°-45°)csc(180°+45°)

Working

Apply identities = [-tan30°][-cot30°] - 2[-sec45°][-csc45°]

Substitute values = left(-1sqrt3right)left(-sqrt3right) - 2(-sqrt2)(-sqrt2)

Simplify = 1 - 4 = -3

Result

Result = R.H.S

Given

(iv) To prove sin210°+cos240°+tan225°+cot225° = 1

L.H.S L.H.S = sin210°+cos240°+tan225°+cot225°

Express in standard form = sin(180°+30°)+cos(270°-30°)+tan(180°+45°)+cot(180°+45°)

Working

Apply identities = -sin30°-sin30°+tan45°+cot45°

Simplify = -12-12+1+1 = 1

Result

Result = R.H.S

4.(i) Prove that: [tan(180°+α) cot(90°−α)] / [sin(360°−α) cos(270°+α)] = −sec²α (ii) Prove that: [sin²(π+θ) tan(3π/2+θ)] / [cot²(3π/2−θ) cos²(π−θ) cosec(2π−θ)] = cosθ (iii) Prove that: [cos(90°+θ) sec(−θ) tan(180°−θ)] / [sec(360°−θ) sin(180°+θ) cot(90°−θ)] = −1

Given

(i) To prove tan(180°+α)cot(90°-α)sin(360°-α)cos(270°+α) = -sec²α

L.H.S L.H.S = tan(180°+α)cot(90°-α)sin(360°-α)cos(270°+α)

Formula

Allied angles tan(180°+α)=tanα, cot(90°-α)=tanα, sin(360°-α)=-sinα, cos(270°+α)=sinα

Working

The book prints = [tanα][tanα][-sinα][-sinα] = tan²αsin²α

cos(270° + α) = +sin α, not −sin α, so the denominator is (−sin α)(sin α) = −sin²α. Followed through, the book's line ends at sec²α instead of the required −sec²α.

Apply identities = [tanα][tanα][-sinα][sinα] = tan²α-sin²α

Simplify = -sin²αcos²α · sin²α = -1cos²α

Result

Result = -sec²α = R.H.S

Given

(ii) To prove frac{sin²(pi+θ)tanleft(3pi2+θright)}{cot²left(3pi2-θright)cos²(pi-θ)csc(2pi-θ)} = cosθ

L.H.S L.H.S = frac{sin²(pi+θ)tanleft(3pi2+θright)}{cot²left(3pi2-θright)cos²(pi-θ)csc(2pi-θ)}

Working

Apply identities = frac{[sin(pi+θ)]²[-cotθ]}{left[cotleft(3pi2-θright)right]²[cos(pi-θ)]²(-cscθ)}

Apply identities = [sinθ]²[-cotθ][-tanθ]²[-cosθ]²(-cscθ)

Simplify = sin²θ · (-cotθ)tan²θ · cos²θ · (-cscθ)

Write in sin, cos = frac{sin²θ · cosθsinθ{sin²θcos²θ · cos²θ · 1sinθ = sinθcosθsinθ

Result

Result = cosθ = R.H.S

Given

(iii) To prove cos(90°+θ)sec(-θ)tan(180°-θ)sec(360°-θ)sin(180°+θ)cot(90°-θ) = -1

L.H.S L.H.S = cos(90°+θ)sec(-θ)tan(180°-θ)sec(360°-θ)sin(180°+θ)cot(90°-θ)

Working

Apply identities = [-sinθ][secθ][-tanθ][secθ][-sinθ][tanθ]

Simplify = (-1)(-1)(-1)

Result

Result = -1 = R.H.S

5.Show that: sec(3π/2 − θ) sec(5π/2 − θ) − tan(3π/2 − θ) tan(5π/2 + θ) = −1

Given

To prove secleft(3pi2-θright)secleft(5pi2-θright) - tanleft(3pi2-θright)tanleft(5pi2+θright) = -1

L.H.S L.H.S = secleft(3pi2-θright)secleft(5pi2-θright) - tanleft(3pi2-θright)tanleft(5pi2+θright)

Working

Apply identities = [-cscθ][cscθ] - [cotθ][-cotθ]

The book prints = -csc²θ - cot²θ

−[cot θ][−cot θ] = +cot²θ, so this line is −cosec²θ + cot²θ. The answer −1 is unchanged.

Simplify = -csc²θ + cot²θ

Formula

Pythagorean identity because 1+cot²θ=csc²θ ⇒ cot²θ-csc²θ=-1

Result

Result = -1 = R.H.S

6.If α, β, γ are the angles of a triangle ABC, then prove that: (i) sin(α+β) = sinγ (ii) sec((α+β)/2) = cosec(γ/2) (iii) cosecα = 1 / sin(β+γ) (iv) tan(α+β) + tanγ = 0

Given

(i) To prove sin(α+β) = singamma

Given because α,β,gamma are the angles of triangle ABC

Formula

Angle sum property therefore α+β+gamma = 180°

Rearrange α+β = 180°-gamma

Taking sine Taking sine

Working

Substitute sin(α+β) = sin(180°-gamma)

Result

Result sin(α+β) = sin(180°-gamma) = singamma

Given

(ii) To prove secleft(α+β2right) = cscgamma2

Given because α,β,gamma are the angles of triangle ABC

Formula

Angle sum property therefore α+β+gamma = 180°

Rearrange α+β = 180°-gamma

Dividing both sides by 2 α+β2 = 180°-gamma2

Taking secant Taking secant

Working

Substitute secleft(α+β2right) = secleft(180°-gamma2right)

Simplify secleft(α+β2right) = secleft(90°-gamma2right)

Result

Result secleft(α+β2right) = cscgamma2

Given

(iii) To prove cscα = 1sin(β+gamma)

Given because α,β,gamma are the angles of triangle ABC

Formula

Angle sum property therefore α+β+gamma = 180°

Rearrange β+gamma = 180°-α

Taking sine Taking sine

Working

Substitute sin(β+gamma) = sin(180°-α)

Simplify sin(β+gamma) = sinα

Taking reciprocal Taking reciprocal

Substitute 1sin(β+gamma) = 1sinα

Result

Result 1sin(β+gamma) = cscα

Given

(iv) To prove tan(α+β) + tangamma = 0

Given because α,β,gamma are the angles of triangle ABC

Formula

Angle sum property therefore α+β+gamma = 180°

Rearrange α+β = 180°-gamma

Taking tangent Taking tangent

Working

Substitute tan(α+β) = tan(180°-gamma)

Simplify tan(α+β) = -tangamma

Result

Result tan(α+β) + tangamma = 0

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Unit 10: Trigonometric Identities — Exercise 10 1