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

Trigonometrical ratios

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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 600American racinggamefort. American Racing is another fun-addicting and challenging racing game developed by TurboNuke and you can play it online and for free on Silvergames.com. Live the american dream and go up against much opposing racing cars. Try to complete various tracks in best time and damage others to knock them out of the competition. American Racing 2. Super Drift 3D 2. Super Car Road Trip: Sport Car Simulator Racing Game. American Racing: 3D Car Game. Game description Compete for stock-car dominance on 12 different tracks in this power-packed racing game. American Racing. Powered by Create your own unique website with customizable templates. American Racing - maker of some of the toughest off-road wheels available – introduces its heavy truck collection of big rig wheel fitments. Built for the long haul, these wheels are designed with high performance strength to weight ratios and cutting edge aesthetics not found in your usual “over the road” wheel brands. \$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 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. 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

### 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

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