Grade 3 » Introduction Print this page. In Grade 3, instructional time should focus on four critical areas: (1) developing understanding of multiplication and division and strategies for multiplication and division within 100; (2) developing understanding of fractions, especially unit fractions (fractions with numerator 1); (3) developing understanding of the structure of rectangular arrays. QUAST'S WEBSITE - MOUNT DOUGLAS SECONDARY. Home Chemistry 11. Math 10 Honours Period 1 Room 106. Click Here for Course Calendar Click Here for Waterloo Math. Here is a list of all of the maths skills students learn in secondary 3! These skills are organised into categories, and you can move your mouse over any skill name to preview the skill. To start practising, just click on any link. IXL will track your score, and the questions will automatically increase in difficulty as you improve! Chapter 8: Trigonometry - Non-Right Triangles. 8.1 The Sine Law 8.2 The Cosine Law 8.3 Finding Angles Using the Cosine Law 8.4 Solve Problems Using Trigonometry. Chapter Review Practice Test.

Trigonometrical ratios

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 = $frac{{rm{h}}}{{rm{p}}}.$

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:

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

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 = $frac{{1 - {{tan }^2}{rm{A}}}}{{1 + {{tan }^2}{rm{A}}}}$ software, free download sites.

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

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

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.

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)

= $frac{6}{4}$(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

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

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

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

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

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

Or, ED = 49.65cm

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°=$frac{{{rm{ED}}}}{{{rm{AD}}}}$

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

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

AD =20.78 cm

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

∴ Distance between tree and house is20.78 cm