Math 10 Honoursmr. B. Quast's Websitemount Douglas Secondary

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Trigonometrical ratios

Douglas Math 1130 Assignment 7. Douglas Math 1130 Assignment 8 Douglas College math MATH 1130 - Fall 2010 Register Now Douglas Math 1130 Assignment 8. Math Tests Questions & Answers. Showing 1 to 3 of 3 View all. Consider the function f: AB, where A, B R or Z, and f(x)=x^2. Siyavula's open Mathematics Grade 10 textbook. We use this information to present the correct curriculum and to personalise content to better meet the needs of our users.


SinA = $frac{{rm{p}}}{{rm{h}}}$, cosA = $frac{{rm{b}}}{{rm{h}}}$, tanA = $frac{{rm{p}}}{{rm{b}}}$, cotA = $frac{{rm{b}}}{{rm{p}}}$, SecA = $frac{{rm{h}}}{{rm{b}}}$ and cosecA = $frac{{rm{h}}}{{rm{p}}}.$

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Where, p = perpendicular, b = base and h = hypotenuse and h2 = p2 + b2.

2. SinA * cosecA = 1, cosA * secA = 1, tanA * cotA = 1.

tanA = $frac{{{rm{sinA}}}}{{{rm{cosA}}}}$, cotA = $frac{{{rm{cosA}}}}{{{rm{sinA}}}}$

3. sin2A + cos2A = 1, sec2A – tan2A = 1, cosec2A – cot2A = 1

4. The trigonometric values of different degree(0° - 180°) are listed below:

Angles Ratios

Sin

Cos

Tan

Cosec

Sec

Cot

00

0

1

0

1

300

$frac{1}{2}$

$frac{{sqrt 3 }}{2}$

$frac{1}{{sqrt 3 }}$

2

$frac{2}{{sqrt 3 }}$

$sqrt 3 $

450

$frac{1}{{sqrt 2 }}$

$frac{1}{{sqrt 2 }}$

1

$sqrt 2 $

$sqrt 2 $

1

600

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$frac{{sqrt 3 }}{2}$

$frac{1}{2}$

$sqrt 3 $

$frac{2}{{sqrt 3 }}$

2

$frac{1}{{sqrt 3 }}$

900

1

0

1

0

1200

$frac{{sqrt 3 }}{2}$

$ - frac{1}{2}$

$ - sqrt 3 $

$frac{2}{{sqrt 3 }}$

$ - 2{rm{ }}$

$ - frac{1}{{sqrt 3 }}{rm{ }}$

1350

$frac{1}{{sqrt 2 }}$

$ - frac{1}{{sqrt 2 }}$

-1

$sqrt 2 $

$ - sqrt 2 $

-1

1500

$frac{1}{2}$

$ - frac{{sqrt 3 }}{2}$

$ - frac{1}{{sqrt 3 }}$

2

$ - frac{2}{{sqrt 3 }}$

$ - sqrt 3 $

1800

0

-1

0

-1

5.With angle is (-A),

Sin(-A) = -sinA, cos(-A) = cosA. Tan(-A) = - tanA, etc.

6. When angle is (90° - A) and (90° + A)

Sin(90° - A) = cosA sin(90°+A) = cosA

Cos(90° - A) = sinA cos(90° + A) = -sinA

Tan(90° - A) = cotA tan(90°+A) = -cotA etc.

7. With angle (180° - A) and (180° + A)

Sin(180° - A) = sinA sin(180°+A) = -sinA

Cos(180° - A) = -cosA cos(180° + A) = -cosA

Tan(180° - A) = -tanA tan(180°+A) = tanA etc.

8. When angle is (270° - A) and (270° + A)

Sin(270° - A) = -cosA sin(270°+A) = -cosA

Cos(270° - A) = -sinA cos(270° + A) = sinA

Math 10 honoursmr. b. quast

Tan(270° - A) = cotA tan(270°+A) = -cotA etc.

9. With angle (360° - A) and (360° + A)

Sin(360° - A) = -sinA sin(360°+A) = sinA

Cos(360° - A) = cosA cos(360° + A) = cosA

Tan(360° - A) = -tanA tan(360°+A) = tanA etc.

Trigonometric Ratios of compound angles:

1. sin(A + B) = sinA.cosB + cosA.sinB

2. cos(A + B) = cosA.cosB – sinA.sinB

3. Tan (A + B) = $frac{{{rm{tanA}} + {rm{tanB}}}}{{1 - {rm{tanA}}.{rm{tanB}}}}$

4. sin (A – B) = sinA.cosB - cosA.sinB

5. cos(A – B) = cosA.cosB + sinA.sinB

6. tan(A – B) = $frac{{{rm{tanA}} - {rm{tanB}}}}{{1 + {rm{tanA}}.{rm{tanB}}}}$

Trigonometric ratios of multiple angles:

1.

(i)sin2A = 2sinA.cosA

(ii)cos2A = cosA2A – sin2A = 2cos2A – 1 = 1 – 2sin2A

(iii)tan2A = $frac{{2{rm{tanA}}}}{{1 - {{tan }^2}{rm{A}}}}$

(iv)sin2A = $frac{{2{rm{tanA}}}}{{1 + {{tan }^2}{rm{ A}}}}$

(v)cos2A = $frac{{1 - {{tan }^2}{rm{A}}}}{{1 + {{tan }^2}{rm{A}}}}$

2.

(i) sin3A = 3sinA – 4sin3A

(ii) cos3A = 4cos3A – 3cosA

(iii) tan3A = $frac{{3{rm{tanA}} - {{tan }^3}{rm{A}}}}{{1 - 3{{tan }^2}{rm{A}}}}{rm{ }}$

Trigonometric ratios of Sub-multiple angles:

1.

(i)sinA = 2sin$frac{{rm{A}}}{2}$.cos$frac{{rm{A}}}{2}$.

(ii)cosA = cosA2$frac{{rm{A}}}{2}$ – sin2$frac{{rm{A}}}{2}$. = 2cos2$frac{{rm{A}}}{2}$. – 1 = 1 – 2sin2$frac{{rm{A}}}{2}$.

(iii)tanA = $frac{{2{rm{tan}}frac{{rm{A}}}{2}.}}{{1 - {{tan }^2}frac{{rm{A}}}{2}.}}$

(iv) sinA = 3sin$frac{{rm{A}}}{3}$ – 4sin3$frac{{rm{A}}}{3}$

(v) cosA = 4cos3$frac{{rm{A}}}{3}$ – 3cos$frac{{rm{A}}}{3}$

(vi) tanA = $frac{{3{rm{tan}}frac{{rm{A}}}{3} - {{tan }^3}frac{{rm{A}}}{3}}}{{1 - 3{{tan }^2}frac{{rm{A}}}{3}}}$

Transformation of Trigonometric Formulae:

1.2sinA.cosB = sin(A + B) + sin(A – B)

2. 2cosA.sinB = sin(A+B) – sin(A – B)

3. 2cosA.cosB = cos(A + B) + cos(A – B)

4. 2sinA.sinB = cos(A + B) – cos(A + B)

5. sinC + sinD = 2sin $frac{{{rm{C}} + {rm{D}}}}{2}.$ cos $frac{{{rm{C}} - {rm{D}}}}{2}$

6. sinC – sinD = 2cos $frac{{{rm{C}} + {rm{D}}}}{2}.$ sin $frac{{{rm{C}} - {rm{D}}}}{2}$

7. cosC + cosD = 2cos $frac{{{rm{C}} + {rm{D}}}}{2}$. cos $frac{{{rm{C}} - {rm{D}}}}{2}$

8. cosC – cosD = -2sin${rm{ }}frac{{{rm{C}} + {rm{D}}}}{2}$.sin $frac{{{rm{C}} - {rm{D}}}}{2}$

Where, A = $frac{{{rm{C}} + {rm{D}}}}{2}$ and B = $frac{{{rm{C}} - {rm{D}}}}{2}$

Conditional Trigonometric Identities:

If A + B + C = π

1.sinA = sin[ π– (B + C)] = sin(B + C)

2. cosA = cos[π – (B + C)] = -cos(B + C)

3. tanA = tan[π – (B + C)] = -tan(B + C)etc.

If, $frac{{rm{A}}}{2}{rm{ }}$+ $frac{{rm{B}}}{2}$ + $frac{{rm{C}}}{2}$ = $frac{{rm{pi }}}{2}$

1.sin$frac{{rm{A}}}{2}{rm{ }}$= sin[$frac{{rm{pi }}}{2}$ – ($frac{{rm{B}}}{2}$ + $frac{{rm{C}}}{2}$)] = cos($frac{{rm{B}}}{2}$ + $frac{{rm{C}}}{2}$)

2. cos$frac{{rm{A}}}{2}{rm{ }}$ = cos[$frac{{rm{pi }}}{2}$ – ($frac{{rm{B}}}{2}$ + $frac{{rm{C}}}{2}$))] = sin($frac{{rm{B}}}{2}$ + $frac{{rm{C}}}{2}$))

3. tan$frac{{rm{A}}}{2}{rm{ }}$ = tan[$frac{{rm{pi }}}{2}$ – ($frac{{rm{B}}}{2}$ + $frac{{rm{C}}}{2}$))] = cot($frac{{rm{B}}}{2}$ + $frac{{rm{C}}}{2}$))etc.

Math 10 Honoursmr. B. Quast

Example 1

Prove:

cos2(A – 120°) + cos2A + cos2(A – 120°) = $frac{3}{2}$

L.H.S. = cos2(A – 120°) + cos2A + cos2(A – 120°)

= {cos2(A – 120°)}2 + cos2 A + {cos2(A + 120°)}2

= (cosA.cos120°+sinA.sin120°) + cos2A + (cosA.cos120°– sinA.sin120°)

= ${left( { - frac{1}{2}{rm{: cosA}} + frac{{sqrt 3 }}{2}{rm{sinA}}} right)^2}$+ cos2A + ${left( { - frac{1}{2}{rm{: cosA}} + frac{{sqrt 3 }}{2}{rm{sinA}}} right)^2}$

= $frac{1}{4}$cos2A – 2.$frac{1}{2}$.$frac{{sqrt 3 }}{2}$ cosA.sinA + $frac{3}{4}{rm{: }}$sin2A + cos2A + $frac{1}{4}{rm{: }}$cos2A + 2.$frac{1}{2}$.$frac{{sqrt 3 }}{2}$ cosA.sinA +$frac{3}{4}{rm{: }}$sin2A

= $frac{1}{4}$cos2 A + $frac{3}{4}$ sin2A + cos2A + $frac{1}{4}$ cos2A + $frac{3}{4}$ sin2A

= $frac{2}{4}$ cos2A + $frac{6}{4}$sin2A + cos2A.

= $frac{1}{4}$ (2cos2A + 6sin2A + 4cos2A) = $frac{1}{4}$(6cos2A + 6sin2A)

Math 10 honoursmr. b. quast

= $frac{6}{4}$(cos2A + sin2A) = $frac{3}{2}$ = R.H.S.

Example 2

If A + B+ C = π prove that cos2A + cos2B + cos2C = 1 – 2cosA.cosB.cosC

Soln:

L.H.S. =cos2A + cos2B + cos2C

= $frac{{1 + {rm{cos}}2{rm{A}}}}{2}$ + $frac{{1 + {rm{cos}}2{rm{B}}}}{2}$ + cos2C.

= $frac{1}{2} + frac{1}{2} + frac{1}{2}$ (cos2A + cos2B) + cos2C.

= 1 + $frac{1}{2}$. 2 cos $frac{{2{rm{A}} + 2{rm{B}}}}{2}$. Cos $frac{{2{rm{A}} - 2{rm{B}}}}{2}$ + cos2C.

= 1 + cos(A + B).cos(A – B) + cos2C.

= 1 – cosC.cos(A – B) + cos2C[A + B = π – C, So, cos(A+B) = cos(π – C) = – cosC]

= 1 – cosC {cos(A – B) – cosC}

= 1 – cosC {cos(A – B) + cos(A + B)}

= 1 – cosC.(cosA.cosB + sinA.sinB + cosA.cosB – sinA.sinB)

= 1 – cosC.(2cosA.cosB) = 1 – 2cosA.cosB.cosC = R.H.S.

Example 3

Solve:

2 sin2x – sinx = 0

Soln:

Here, 2 sin2x = sinx

Or, 4sinx cosx – sinx = 0

Or, sinx (4cosx – 1) = 0

Either, sinx = 0

So, x = nπ

Or, 4cosx – 1 = 0

Math 10 Honoursmr. B. Quast's Websitemount Douglas Secondary Intention

Or, cosx = 1414

Or, cosx = cosα [cosx = 1414]

So, x = 2nπ ±± α.

So, x = 2nπ ±± cos–11414 n ԑ Z.

Height and Distance

Angle of elevation: The angle of elevation is defined as angle between the line of sight and horizontal line made by the observer when the observer observes the object above the horizontal line.

Example 4

Math 10 Honoursmr. B. Quast's Websitemount Douglas Secondary Hyperparathyroidism

A man observes the top of a pole 52cm height situated in front of him and finds the angle of elevation to be 30° .If the distance between man and pole is 86m .Find the height of that man.

Soln

Math 10 Honoursmr. B. Quast's Websitemount Douglas Secondary Student

Tan30° =$frac{{{rm{ED}}}}{{{rm{AD}}}}$=$frac{{{rm{ED}}}}{{86}}$

$frac{1}{{sqrt 3 }}$=$frac{{{rm{ED}}}}{{86}}$

Honoursmr.

Or, ED = $frac{{86}}{{sqrt 3 }}$

Or, ED = 49.65cm

Math 10 Honoursmr. B. Quast's Websitemount Douglas Secondary Education

Height of man (DC) = EC – ED = 50- 47.65 =2.35 m

Example 5

The angle of elevation from the roof of a house to the top of atop of tree is found to be 30°. If the height of the house and tree are 8 m and 20m respectively, find the distance between the house and the tree.

Soln

ED = EC - DC = 20- 8 = 12 cm

Tan 30°=$frac{{{rm{ED}}}}{{{rm{AD}}}}$

$frac{1}{{sqrt 3 }}$=$frac{{12}}{{{rm{AD}}}}$

Math 10 Honoursmr. B. Quast's Websitemount Douglas Secondary Tertiary

AD =20.78 cm

Math 10 Honoursmr. B. Quast's Websitemount Douglas Secondary Sources

∴ Distance between tree and house is20.78 cm