Grade 11 Mathematics Note
Trigonometric Equations and General Values.
Sinθ = $\frac{{\rm{p}}}{{\rm{h}}}$
cosθ = $\frac{{\rm{b}}}{{\rm{h}}}$
tanθ = $\frac{{\rm{P}}}{{\rm{b}}}$
cotθ = $\frac{{\rm{b}}}{{\rm{P}}}$
secθ = $\frac{{\rm{h}}}{{\rm{b}}}$
cosecθ = $\frac{{\rm{h}}}{{\rm{p}}}$
Reciprocal Properties:
Cotθ = $\frac{1}{{{\rm{tanx}}}}$
Cosθ = $\frac{1}{{{\rm{sinx}}}}$
Secθ = $\frac{1}{{{\rm{cosx}}}}$
tanθ cotθ = 1
sinθ cosθ = 1
cosθ secθ = 1
Quotient Properties:
Tanθ = $\frac{{{\rm{sin}}\theta }}{{{\rm{cos}}\theta }}$
Cotθ =$\frac{{{\rm{cos}}\theta }}{{{\rm{sin}}\theta }}$
Odd/Even Identities
1. sin (-θ ) = -sin θ
2. cos (-θ ) = cos θ
3. tan (-θ ) = -tan θ
4. cos (-θ ) = -cos θ
5. sec (-θ ) = sec θ
6. cot (-θ ) = -cot θ
Co- function identity
Sin (${\rm{\s: }}\frac{{\rm{\pi }}}{2}$ - θ ) = cosθ ,
sin (90° - θ ) = cosθ ,
Cos(${\rm{\: }}\frac{{\rm{\pi }}}{2}$ - θ ) =sinθ
cos( 90° –θ ) = sinθ
tan (${\rm{\: }}\frac{{\rm{\pi }}}{2}$ - θ ) = cotθ ,
tan ( 90°${\rm{\: }}$- θ ) = cotθ ,
Cot(${\rm{\: }}\frac{{\rm{\pi }}}{2}$ - θ ) =tanθ
Cot (90° - θ ) = tanθ
Addition and subtraction formula
Sin (A+ B) = sinACosB + cosACosB
Cos (A +B) = cosA cosB - SinASinB
Sin (A –B) = sinACosB - cosASinB
Cos (A - B) = cosA cosB + SinASinB
Tan ( A +B) = $\frac{{{\rm{tanA}} + {\rm{tanB}}}}{{1 + {\rm{\: tanA\: tanB}}}}{\rm{\: }}$
Tan (A - B) = $\frac{{{\rm{tanA}} - {\rm{tanB}}}}{{1 + {\rm{\: tanAtanB\: }}}}$
Cot (A – B) =$\frac{{{\rm{cotAcotB}} + 1{\rm{\: }}}}{{{\rm{cotB}} - {\rm{cotA\: }}}}$
Cot (A + B) =$\frac{{{\rm{cotAcotB}} - {\rm{\: }}1{\rm{\: }}}}{{{\rm{cotB}} + {\rm{\: cotA\: }}}}$
Multiple and Sub –multiple angle formulae:
Double angle formula
Sin2A = 2sinAcosB
Cos2A = cos2 A – sin2 A = 2 cos2 A -1 = 1 – 2sin2 A
Tan2A = $\frac{{2{\rm{tanA}}}}{{1 - {{\tan }^2}{\rm{A}}}}$
Sin2A = $\frac{{2{\rm{tanA}}}}{{1 + {\rm{\: }}{{\tan }^2}{\rm{A}}}}$
And cos2A = $\frac{{1 - {\rm{ta}}{{\rm{n}}^2}{\rm{A}}}}{{1 + {{\tan }^2}{\rm{A}}}}$
Half angle formula
SinA = 2sin $\frac{{\rm{A}}}{2}$ cos${\rm{\: }}\frac{{\rm{B}}}{2}$
Cos A = cos2 $\frac{{\rm{A}}}{2}$ – sin2 $\frac{{\rm{A}}}{2}$ = 2 cos2 $\frac{{\rm{A}}}{2}$ - 1 = 1 – 2sin2 $\frac{{\rm{A}}}{2}$
Tan A = $\frac{{2\tan \frac{{\rm{A}}}{2}{\rm{\: }}}}{{1 - {{\tan }^2}\frac{{\rm{A}}}{2}}}$
Sin A = $\frac{{2\tan \frac{{\rm{A}}}{2}}}{{1 + {\rm{\: }}{{\tan }^2}\frac{{\rm{A}}}{2}}}$
And cos A = $\frac{{1 - {\rm{ta}}{{\rm{n}}^2}\frac{{\rm{A}}}{2}{\rm{\: }}}}{{1 + {{\tan }^2}\frac{{\rm{A}}}{2}}}$
1- cosA = 2sin2 $\frac{{\rm{A}}}{2}$
1 + cosA = 2cos2 $\frac{{\rm{A}}}{2}$
Triple angle formula
Sin3A= 3sinA – 4sin3 A
Cos3A = 4cos3 A - 3cosA
Tan3A = 3cosA
Tan3A = $\frac{{{\rm{\: }}3{\rm{tanA}} - {{\tan }^3}{\rm{A\: \: }}}}{{1 - 3{{\tan }^2}{\rm{A}}}}$
Transformation formulae
SinC + SinD = 2sin${\rm{\: }}\frac{1}{2}$ (C+D ) cos $\frac{1}{2}$ (C - D)
SinC - sinD = 2cos${\rm{\: }}\frac{1}{2}$ (C+D ) sin $\frac{1}{2}$ (C - D)
cosC + cosD = 2cos${\rm{\: }}\frac{1}{2}$ (C+D ) sin $\frac{1}{2}$ (C - D)
cosC - cosD = 2cos${\rm{\: }}\frac{1}{2}$ (C+D ) sin $\frac{1}{2}$ (D - C)
Conditional identities
If A+B+C = π
A+B = π -C
Sin (A+B) = sin (π - c) = sinC
cos(A+B) = cos(π - c) = - cosC
Tan (A+B) = tan (π - c) = - tanC
If A+B+C =${\rm{\: }}\frac{{\rm{\pi }}}{2}$ then.$\frac{{{\rm{A}} + {\rm{B}}}}{2} = {\rm{\: }}\frac{{\rm{\pi }}}{2}$
Sin (${\rm{\: }}\frac{{\rm{A}}}{2}$ +$\frac{{\rm{B}}}{2}{\rm{\: }}$) = sin (${\rm{\: }}\frac{{{\rm{\: \pi }}}}{2}$ -$\frac{{\rm{c}}}{2}{\rm{\: }}$) = cos${\rm{\: }}\frac{{\rm{C}}}{2}$
cos($\frac{{{\rm{\: A}}}}{2}$+$\frac{{\rm{B}}}{2}{\rm{\: }}$) = cos(${\rm{\: }}\frac{{\rm{\pi }}}{2}$ - $\frac{{\rm{c}}}{2}{\rm{\: }}$) = sin $\frac{{\rm{C}}}{2}$
Tan (${\rm{\: }}\frac{{\rm{A}}}{2}$+$\frac{{\rm{B}}}{2}{\rm{\: }}$) = tan (${\rm{\: }}\frac{{\rm{\pi }}}{{{\rm{\: }}2}}{\rm{\: }}$- $\frac{{\rm{c}}}{2}{\rm{\: }}$) = cot${\rm{\: }}\frac{{\rm{C}}}{2}$
Examples
1.
$\frac{{{\rm{tanA}} - {\rm{secA}} + 1}}{{{\rm{tanA}} + {\rm{secA}} - 1}}$ = $\frac{{{\rm{tanA}} - {\rm{secA}} + 1}}{{{\rm{tanA}} + {\rm{secA}} - \left( {{{\sec }^2}{\rm{A}} - {{\tan }^2}{\rm{A}}} \right)}}$ = $\frac{{{\rm{cosA}}}}{{1 + {\rm{sinA}}}}$
soln
L.H.S. = $\frac{{{\rm{tanA}} - {\rm{secA}} + 1}}{{{\rm{tanA}} + {\rm{secA}} - 1}}$ = $\frac{{{\rm{tanA}} - {\rm{secA}} + 1}}{{{\rm{tanA}} + {\rm{secA}} - \left( {{{\sec }^2}{\rm{A}} - {{\tan }^2}{\rm{A}}} \right)}}$
= $\frac{{{\rm{tanA}} - {\rm{secA}} + 1}}{{\left( {{\rm{tanA}} + {\rm{secA}}} \right) - \left( {{\rm{secA}} - \tan {\rm{A}}} \right)\left( {{\rm{secA}} + {\rm{tanA}}} \right)}}$
= $\frac{1}{{{\rm{tanA}} + {\rm{secA}}}}$
= $\frac{1}{{\frac{{{\rm{sinA}}}}{{{\rm{cosA}}}} + \frac{1}{{{\rm{cosA}}}}}}$
= $\frac{{{\rm{cosA}}}}{{1 + {\rm{sinA}}}}$ = R.H.S.
2.
tan6°.tan42°.tan66°.tan78° = 1
Soln:
L.H.S. = tan6°.tan42°.tan66°.tan78°
= $\frac{{{\rm{sin}}6^{\circ} }}{{{\rm{cos}}6^{\circ} }}$ .${\rm{\: }}\frac{{{\rm{sin}}42^{\circ} }}{{{\rm{cos}}42^{\circ} }}$ .${\rm{\: }}\frac{{{\rm{sin}}66^{\circ} }}{{{\rm{cos}}66^{\circ} }}$.${\rm{\: }}\frac{{{\rm{sin}}78^{\circ} }}{{{\rm{cos}}78^{\circ} }}$
= $\frac{{\left( {2{\rm{sin}}66^{\circ} .{\rm{sin}}6^{\circ} } \right)\left( {2{\rm{sin}}78^{\circ} {\rm{sin}}42^{\circ} } \right)}}{{\left( {2{\rm{cos}}66^{\circ} .{\rm{cos}}6^{\circ} } \right)\left( {2{\rm{cos}}78^{\circ} .{\rm{cos}}42^{\circ} } \right)}}$
= $\frac{{\left\{ {\cos \left( {66^{\circ} - 6^{\circ} } \right) - \cos \left( {66^{\circ} + 6^{\circ} } \right)} \right\}\left\{ {\cos \left( {78^{\circ} - 42^{\circ} } \right) - \cos \left( {78^{\circ} + 42^{\circ} } \right)} \right\}}}{{\left\{ {\cos \left( {66^{\circ} + 6^{\circ} } \right) + \cos \left( {66^{\circ} - 6^{\circ} } \right)} \right\}\left\{ {\cos \left( {78^{\circ} + 42^{\circ} } \right) + \cos \left( {78^{\circ} - 42^{\circ} } \right)} \right\}}}{\rm{\: \: }}$
= $\frac{{\left( {\frac{1}{2} - {\rm{cos}}72^{\circ} } \right)\left( {{\rm{cos}}36^{\circ} + \frac{1}{2}} \right)}}{{\left( {{\rm{cos}}72^{\circ} + \frac{1}{2}} \right)\left( { - \frac{1}{2} + {\rm{cos}}36^{\circ} } \right)}}$
= $\frac{{\left( {\frac{1}{2} - \frac{{\sqrt 5 - 1}}{4}} \right)\left( {\frac{{\sqrt 5 + 1}}{4} + \frac{1}{2}} \right)}}{{\left( {\frac{{\sqrt 5 - 1}}{4} + \frac{1}{2}} \right)\left( { - \frac{1}{2} + \frac{{\sqrt 5 + 1}}{4}} \right)}}$
= $\frac{{2 - \sqrt 5 + 1}}{{\sqrt 5 - 1 + 2}},\frac{{\sqrt 5 + 1 + 2}}{{\sqrt 5 {\rm{\: }} + 1 - 2}}$
= $\frac{{\left( {3 - \sqrt 5 } \right)\left( {3 + \sqrt 5 } \right)}}{{\left( {\sqrt 5 + 1} \right)\left( {\sqrt 5 - 1} \right)}}$
= $\frac{{9 - 5}}{{5 - 1}}$
= $\frac{4}{4}$ = 1 = R.H.S.
3.
If tanx =ktany show that
(k- 1) sin(x+y) = (k+1)sin(x- y)
Soln:
tanx = k.tany.
Or, $\frac{{{\rm{tanx}}}}{{{\rm{tany}}}}$ = $\frac{{\rm{k}}}{1}$
Or, $\frac{{{\rm{tanx}} + {\rm{tany}}}}{{{\rm{tanx}} - {\rm{tany}}}}$ = $\frac{{{\rm{k}} + 1}}{{{\rm{k}} - 1}}$ [By componendo and dividendo]
Or, $\frac{{\frac{{{\rm{sinx}}}}{{{\rm{cosx}}}} + \frac{{{\rm{siny}}}}{{{\rm{cosy}}}}}}{{\frac{{{\rm{sinx}}}}{{{\rm{cosx}}}} - \frac{{{\rm{siny}}}}{{{\rm{cosy}}}}}}$ = $\frac{{{\rm{k}} + 1}}{{{\rm{k}} - 1}}$
Or, $\frac{{{\rm{sinx}}.{\rm{cosy}} + {\rm{cosx}}.{\rm{siny}}}}{{{\rm{sinx}}.{\rm{cosy}} - {\rm{cosx}}.{\rm{siny}}}}$ = $\frac{{{\rm{k}} + 1}}{{{\rm{k}} - 1}}$
Or, $\frac{{{\rm{sin}}\left( {{\rm{x}} + {\rm{y}}} \right)}}{{\sin \left( {{\rm{x}} - {\rm{y}}} \right)}} = $$\frac{{{\rm{k}} + 1}}{{{\rm{k}} - 1}}$
So, (k – 1)sin(x + y) = (k + 1)sin(x – y).
Trigonometric equation
A Trigonometric equation is an equation that contains trigonometric function of unknown variable.
Examples
1. sin2x + sinx = 0
Soln:
Here, sin2x + sinx = 0
Or, 2 sinx.cosx + sinx = 0
Or, sinx(2cosx + 1) = 0
Either, sinx = 0
So, x = nπ.
Or, 2cosx + 1 = 0
Or, cosx = $ - \frac{1}{2}$ = cos$\frac{{2{\rm{\pi }}}}{3}$
So, x = 2nπ $ \pm $$\frac{{2{\rm{\pi }}}}{3}$ = (6n$ \pm $ 2) $\frac{{\rm{\pi }}}{3}$
Hence, x = nπ, (6n$ \pm $ 2) $\frac{{\rm{\pi }}}{3}$, n ԑ Z.
2.
tanθ + tan2θ = tan 3θ.
Soln:
Here, tan θ + tan 2θ = tan 3θ.
Or, tanθ + tan2θ – tan(θ + 2θ) = 0
Or, (tanθ + tan2θ) – $\frac{{{\rm{tan}}\theta + {\rm{tan}}2\theta }}{{1 - {\rm{tan}}\theta .{\rm{tan}}2\theta }}{\rm{\: \: }}$= 0
Or, (tanθ + tan2θ) (1 – tanθ.tan2θ – 1) = 0
So, tanθ.tan2θ(tanθ + tan2θ) = 0
Either, tan θ = 0 = tan 0.
So, θ = nπ.
Or, tan 2θ = 0 = tan 0.
So, 2θ = nπ.
So, θ = $\frac{{{\rm{n\pi }}}}{2}$.
Or, tan θ + tan2θ = 0.
Or, tan θ = –tanθ = tan(–θ)
So, 2θ = nπ + (–θ)
SO, 2θ + θ = nπ
SO, θ = $\frac{{{\rm{n\pi }}}}{3}$.
Hence, θ = nπ, $\frac{{{\rm{n\pi }}}}{2}$, $\frac{{{\rm{n\pi }}}}{3}$, n ԑ Z
3.
Sin2x tanx + 1 = sin2x + tanx
Soln:
Given equation is:
Sin2x tanx + 1 = sin2x + tanx
Or, sin2x tanx – sin2x – tanx + 1 = 0
Or, sin2x(tanx – 1) – 1(tanx – 1) = 0
Or, (tanx – 1)(sin2x – 1) = 0
Either, tanx – 1 =0
Or, tanx = 1 = tan $\frac{{\rm{\pi }}}{4}$.
So, x = nπ + $\frac{{\rm{\pi }}}{4}$ = (4n + 1)$\frac{{\rm{\pi }}}{4}$.
OR, sin2x – 1 = 0
Or, sin2x = 1
So, 2x = (4n + 1) $\frac{{\rm{\pi }}}{2}$
SO, x = (4n + 1) $\frac{{\rm{\pi }}}{4}$
Hence, x = (4n + 1) $\frac{{\rm{\pi }}}{4}$, n ԑ Z.