# 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

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.

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