Trigonometry Ratios

Trigonometry is one of the important branches of Mathematics. It is the study of triangles and its measurement. In this article, we come across the trigonometric ratios, graphs of trigonometric functions, identities, maximum and minimum values, main formulas and many more. 

Trigonometric Ratios

 

Trigonometric ratios

 

sinθ=ABAC=opphyp\sin \theta =\frac{AB}{AC}=\frac{opp}{hyp}

 

cosθ=BCAC=adjhyp\cos \theta =\frac{BC}{AC}=\frac{adj}{hyp}

 

tanθ=ABBC=oppadj\tan \theta =\frac{AB}{BC}=\frac{opp}{adj}

 

cotθ=BCAB\cot\theta =\frac{BC}{AB}

 

secθ=ACBC\sec\theta =\frac{AC}{BC}

 

cosecθ=ACAB\cos ec\theta =\frac{AC}{AB}

Trigonometric Circular Function

Trigonometric circular function

<AOP=arcAPr=lr<AOP=\frac{arcAP}{r}=\frac{l}{r}

Cos θ\theta= x

Sin θ\theta = y

tanθ=xy\tan\theta=\frac{x}{y}

Graphs of T Ratios

  • Sine

y = sin x

Graph of sine function

Domain R

Range (-1,1)

  • Cosine

y = cos x

Graph of cosine function

  • Tangent

y = tan x

Graph of tangent function

  • Co – tangent

y = cot x

Graph of co tangent function

 

  • Secant
y=secx,y=\sec x,

Graph of secant function

 

Domain:R{(2x+1)π2}Domain:\,\,\,\,\,R-\left\{ (2x+1)\,\frac{\pi }{2} \right\} and

Range:(1)(1,)Range:\,(-\infty -1)\, \cup \,(1,\infty )
  • Cosecant

y = cosec x

Graph of co secant function

 

Domain:R{nπ}Domain:\,R-\{n\pi \} and

Range:(1)(1,)Range:\,(-\infty -1)\, \cup \,(1,\infty )

Also Read:

Trigonometric Equations and its Solutions

Trigonometry JEE Main Previous Year Questions

Inverse Trig JEE Main Previous Year Questions

Trignometric Identities

sin2θ+cos2θ=1{{\sin }^{2 }\theta}+{{\cos }^{2 }\theta}=1.

1+tan2θ=sec2θ1+{{\tan }^{2 }\theta}=\,{{\sec }^{2 }\theta}.

1+cot2θ=cosec2θ1+{{\cot }^{2 }\theta}=\cos e{{c}^{2 }\theta}.

sin4θ+cos4=12sin2θcos2θ\sin ^{4}\theta +\cos^{4}=1-2\sin ^{2 }\theta\cos ^{2 }\theta.

sin6θ+cos6=13sin2θcos2θ{{\sin }^{6}}\theta +{{\cos }^{6}}=1-3{{\sin }^{2 }\theta}{{\cos }^{2 }\theta}

Remember:

secθtanθ=1secθ+tanθ\sec \theta -\tan \theta =\frac{1}{\sec \theta +\tan \theta } and

cosecθcotθ=1cosecθ+cotθ\cosec\theta -\cot \theta =\frac{1}{\cosec\theta +\cot \theta }

T-Ratio At Some Standard Angles:

T ratio of standard angles

 

T-Ratio At Some Specific Angles:

T-ratio

7127\frac{1{}^\circ }{2} 15o 221222\frac{1{}^\circ }{2} 18o
sinθ\sin\theta 42622\frac{\sqrt{4-\sqrt{2}-\sqrt{6}}}{2\sqrt{2}} 3122\frac{\sqrt{3}-1}{2\sqrt{2}} 1222\frac{1}{2}\sqrt{2-\sqrt{2}} 514\frac{\sqrt{5-1}}{4}
cosθ\cos\theta 4+2+622\frac{\sqrt{4+\sqrt{2}+\sqrt{6}}}{2\sqrt{2}} 3+122\frac{\sqrt{3}+1}{2\sqrt{2}} 122+2\frac{1}{2}\sqrt{2+\sqrt{2}} 1410+25\frac{1}{4}\sqrt{10+2\sqrt{5}}
tanθ\tan\theta (32)(21)\left( \sqrt{3}-\sqrt{2} \right)\left( \sqrt{2}-1 \right) 232-\sqrt{3} 21\sqrt{2-1} 25+10155\frac{\sqrt{25+10\sqrt{15}}}{5}

Maximum and Minimum Value Of Standard Trigonometric Expressions

  • Maximum value of acosθ±bsinθ=a2+b2a\cos \theta \pm b\sin \theta =\sqrt{{{a}^{2}}+{{b}^{2}}}
  • Minimum value of acosθ±bsinθ=a2+b2a\cos \theta \pm b\sin \theta =-\sqrt{{{a}^{2}}+{{b}^{2}}}
  • Maximum value of acosθ±bsinθ+c=c+a2+b2a\cos \theta \pm b\sin \theta +c=c+\sqrt{{{a}^{2}}+{{b}^{2}}}
  • Minimum value of acosθ±bsinθ+c=ca2+b2a\cos \theta \pm b\sin \theta +c=c-\sqrt{{{a}^{2}}+{{b}^{2}}}

Problems on Trigonometry

Example 1: If sec4xcosec4x2sec2x+2cosec2x=154{{\sec }^{4}}x-{{{cosec}}^{4}}x-2{{\sec }^{2}}x+2\cos e{{c}^{2}}x=\frac{15}{4}. Find tan x.

Solution:

(sec4x2sec2x)(cosec4x2cosec2x)=154\left( {{\sec }^{4}}x-2{{\sec }^{2}}x \right)-\left( \cos e{{c}^{4}}x-2\cos e{{c}^{2}}x \right)=\frac{15}{4}

or

(sec4x2sec2x+1)(cosec4x2cosec2x+1)=154\left( {{\sec }^{4}}x-2{{\sec }^{2}}x+1 \right)-\left( \cos e{{c}^{4}}x-2\cos e{{c}^{2}}x+1 \right)=\frac{15}{4},

(sec2x1)2(cosec2x1)2=154{{\left( {{\sec }^{2}}x-1 \right)}^{2}}-{{\left( \cos e{{c}^{2}}x-1 \right)}^{2}}=\frac{15}{4},

(tan2x)2(cot2x)2=154{{\left( {{\tan }^{2}}x \right)}^{2}}-{{\left( {{\cot }^{2}}x \right)}^{2}}=\frac{15}{4},

tan4x1tan4x=154{{\tan }^{4}}x-\frac{1}{{{\tan }^{4}}x}=\frac{15}{4},

tan4x1tan4x=414{{\tan }^{4}}x-\frac{1}{{{tan }^{4}}x}=4-\frac{1}{4}

On comparing,

tan4x=4{{\tan }^{4}}x=4,

tan2x=2{{\tan }^{2}}x=2,

tanx=±2tan x=\pm \sqrt{2}

Example 2:

If sinθ,cosθ,tanθ\sin \theta ,\cos \theta ,\tan \theta are in G.P.

Then cos9θ+cos6θ+3cos3θcos2θ1=?{{\cos }^{9}}\theta +{{\cos }^{6}}\theta +3{{\cos }^{3}}\theta {{\cos }^{2}}\theta -1=?

Solution:

sinθ.tanθ=cos2θ\sin \theta .tan\theta =co{{s}^{2}}\theta,

sin2θ=cos3θ{{\sin }^{2}}\theta ={{\cos }^{3}}\theta

Now,

(cos3θ)3+(cos3θ)2+3cos3θcos2θ1{{\left( {{\cos }^{3}}\theta \right)}^{3}}+{{\left( {{\cos }^{3}}\theta \right)}^{2}}+3{{\cos }^{3}}\theta {{\cos }^{2}}\theta -1

=sin6θ+sin4θ+3sin2θcos2θ1{{\sin }^{6}}\theta +{{\sin }^{4}}\theta +3{{\sin }^{2}}\theta {{\cos }^{2}}\theta -1

=sin6θ+sin4θ(13sin2θcos2θ){{\sin }^{6}}\theta +{{\sin }^{4}}\theta -\left( 1-3{{\sin }^{2}}\theta {{\cos }^{2}}\theta \right)

=sin6θ+sin4θ(sin6θ+cos6θ){{\sin }^{6}}\theta +{{\sin }^{4}}\theta -\left( {{\sin }^{6}}\theta +{{\cos }^{6}}\theta \right)

=sin4θcos6θ{{\sin }^{4}}\theta -{{\cos }^{6}}\theta

=sin4θ(cos3θ)2{{\sin }^{4}}\theta -{{\left( {{\cos }^{3}}\theta \right)}^{2}}

=sin4θsin4θ=0{{\sin }^{4}}\theta -{{\sin }^{4}}\theta =0

Example 3:

If secxtanx=P\sec x-\tan x=P then sec x = ?

Solution:

Given: secx    tanx=P\sec x\;-\;tan x=P ….(1)

We know, secxtanx=1secx+tanx\sec x-\tan x=\frac{1}{\sec x+\tan x}

or secx  +  tanx=1P\sec x\;+\;tan x=\frac{1}{P} ….(2)

Adding (1) and (2)

2secx=P+1P{2\sec x=P+\frac{1}{P}}

or secx=P2+12P\sec x=\frac{{{P}^{2}}+1}{2P}

Example 4:

Prove that sin(420).cos(390)+cos(660)sin(330)=1\sin \left( -420 \right).\cos \left( 390 \right)+\cos \left( -660 \right)\sin \left( 330 \right)=-1

Solution:

sin(420)cos(390)+cos(660)sin330-\sin \left( 420 \right)\cos \left( 390 \right)+\cos \left( 660 \right)\sin 330

=sin(2π+60)cos(2π+30)+cos(4π60)sin(2π30)-\sin \left( 2\pi +60 \right)\cos \left( 2\pi +30 \right)+\cos \left( 4\pi -60 \right)\sin \left( 2\pi -30 \right)

=sin60cos30+cos60(sin30)-\sin 60\cos 30+\cos 60\left( -\sin 30 \right)

=32.3212.12-\frac{\sqrt{3}}{2}.\frac{\sqrt{3}}{2}-\frac{1}{2}.\frac{1}{2}

=3414=44=1.-\frac{3}{4}-\frac{1}{4}=-\frac{4}{4}=-1.

Trigonometric Ratios of Compound Angles

The algebraic sum of two or more angles are generally called compound angles and angles are known as constituent angles.

Some Important Results:

  1. sin(A±B)=sinAcosB±cosAsinB\sin \left( A\pm B \right)=\sin A\cos B\pm \cos \,A\,sin\,B
  2. cos(A±B)=cosAcosBsinAsinB\cos \left( A\pm B \right)=\cos A\cos B\mp \sin \,A\,sin\,B
  3. tan(A±B)=tanA±tanB1tanAtanB\tan \left( A\pm B \right)=\frac{\tan A\pm \tan B}{1\tan A\tan \,B}
  4. cot(A±B)=cotAcotB1cotB±cotA\cot \left( A\pm B \right)=\frac{\cot A\cot B\mp 1}{\cot B\pm \cot A}
  5. sin(A+B)sin(AB)=sin2Asin2B=cos2Bcos2A\sin \left( A+B \right)\sin \left( A-B \right)={{\sin }^{2}}A-{{\sin }^{2}}B={{\cos }^{2}}B-{{\cos }^{2}}A
  6. cos(A+B)cos(AB)=cos2Asin2B=cos2Bsin2A\cos \left( A+B \right)\cos \left( A-B \right)={{\cos }^{2}}A-{{\sin }^{2}}B={{\cos }^{2}}B-{{\sin }^{2}}A
  7. sin(A+B+C)=sinAcosBcosC+sinBcosAcosC+sinCcosAcosBsinAsinBsinC\sin \left( A+B+C \right)=\sin A\cos B\cos C+\sin B\cos A\cos C+\sin C\cos A\cos B-\sin A\sin B\sin C
  8. cos(A+B+C)=cosAcosBcosCcosAsinBsinCcosBsinAsinCcosCsinAsinB\cos \left( A+B+C \right)=\cos A\cos B\cos C-\cos A\sin B\sin C-\cos B\sin A\sin C-\cos C\sin A\sin B
  9. tan(A+B+C)=tanA+tanB+tanCtanAtanBtanC1tanAtanBtanBtanCtanCtanA\tan \left( A+B+C \right)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}

Remember:

If A+B+C = 0 then

tanA+tanB+tanC=tanAtanBtanC\tan A+\tan B+\tan C=\tan A\tan B\tan C

Let us solve some of the problems to understand the concept in a better way.

Example 1:

If sinα=35&cosβ=941α,β\sin \alpha =\frac{3}{5}\And \cos \beta =\frac{9}{41}\alpha ,\beta \in I quadrant.

Find sin(αβ)\sin \left( \alpha -\beta \right)

Solution:

sinα=35\sin \alpha =\frac{3}{5},

sin2α+cos2α=1{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1,

cos2α=1925=1625{{\cos }^{2}}\alpha =1-\frac{9}{25}=\frac{16}{25},

cosα=45,45\cos \alpha =\frac{4}{5},\frac{-4}{5}as in first quadrant.

cosβ=941\cos \beta =\frac{9}{41},

cos2β+sin2β=1{{\cos }^{2}}\beta +{{\sin }^{2}}\beta =1,

sin2β=1811681=16001681{{\sin }^{2}}\beta =1-\frac{81}{1681}=\frac{1600}{1681},

sinβ=4041\sin \beta =\frac{40}{41},

sin(αβ)=sinαcosβcosαsinβ\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta,

=35×94145×4041=27160205=\frac{3}{5}\times \frac{9}{41}-\frac{4}{5}\times \frac{40}{41}=\frac{27-160}{205},

=133205.=\frac{-133}{205}.

Example 2:

Find value of sin(BC)cosBcosC+sin(CA)cosCcosA+sin(AB)cosAcosB=?\frac{\sin \left( B-C \right)}{\cos B\cos C}+\frac{\sin \left( C-A \right)}{\cos C\cos A}+\frac{\sin \left( A-B \right)}{\cos A\cos B}=?

Solution:

sinBcosCcosBsinCcosBcosC+sinCcosAcosCsinAcosCcosA+sinAcosBcosAsinBcosAcosB\frac{\sin B\cos C-\cos B\sin C}{\cos B\cos C}+\frac{\sin C\cos A-\cos C\sin A}{\cos C\cos A}+\frac{\sin A\cos B-\cos A\sin B}{\cos A\cos B},

tan  B    tan  C  +  tanC    tanA  +  tanA    tan  B=0tan \;B\;-\;tan \;C\;+\;tan C\;-\;tan A\;+\;tan A\;-\;tan\; B = 0

Transformation Formulae:

  1. 2sinAcosB=sin(A+B)+sin(AB)2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)
  2. 2sinBcosA=sin(A+B)sin(AB)2\sin B\cos A=\sin \left( A+B \right)-\sin \left( A-B \right)
  3. 2cosAcosB=cos(A+B)+cos(AB)2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)
  4. 2sinAsinB=cos(AB)cos(A+B)2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)
  5. sinC+sinD=2sin(C+D2)cos(CD2)\sin C+\sin D=2\sin \left( \frac{C+D}{2} \right)\cos \left( \frac{C-D}{2} \right)
  6. sinCsinD=2cos(C+D2)sin(CD2)\sin C-\sin D=2\cos \left( \frac{C+D}{2} \right)\sin \left( \frac{C-D}{2} \right)
  7. cosCcosD=2cos(C+D2)cos(CD2)\cos C-\cos D=2\cos \left( \frac{C+D}{2} \right)\cos \left( \frac{C-D}{2} \right)
  8. cosCcosD=2sin(C+D2)sin(DC2)\cos C-\cos D=2\sin \left( \frac{C+D}{2} \right)\sin \left( \frac{D-C}{2} \right)

T-Ratio Of Multiple Angles

  1. sin2A=2sinAcosA=2tanA1+tan2A\sin 2A=2\sin A\cos A=\frac{2\tan A}{1+{{\tan }^{2}}A}
  2. cos2A=2cos2A1=12sin2A=1tan2A1+tan2A=cos2Asin2A\cos 2A=2{{\cos }^{2}}A-1=1-2{{\sin }^{2}}A=\frac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A
  3. tan2A=2tanA1tan2A\tan 2A=\frac{2\tan A}{1-{{\tan }^{2}}A}
  4. sin3A=3sinA4sin3A\sin 3A=3\sin A-4{{\sin }^{3}}A
  5. cos3A=4cos3A3cosA\cos 3A=4{{\cos }^{3}}A-3\cos A
  6. tan3A=3tanAtan3A13tan2A\tan 3A=\frac{3\tan A-{{\tan }^{3}}A}{1-3{{\tan }^{2}}A}

Remember:

1+cos2A=2cos2A1+\cos 2A=2{{\cos }^{2}}A

 

1cos2A=2sin2A1-\cos 2A=2{{\sin }^{2}}A

 

1cos2A1+cos2A=tan2A\frac{1-\cos 2A}{1+\cos 2A}={{\tan }^{2}}A

Values Of T-Ratios At Some Useful Angles

  1. sin75=3+122=cos15\sin 75{}^\circ =\frac{\sqrt{3}+1}{2\sqrt{2}}=\cos 15{}^\circ
  2. cos75=3122=sin15\cos 75{}^\circ =\frac{\sqrt{3}-1}{2\sqrt{2}}=\sin 15{}^\circ
  3. tan75=2+3=cot15\tan 75{}^\circ =2+\sqrt{3}=\cot 15{}^\circ
  4. cot75=23=tan15\cot 75{}^\circ =2-\sqrt{3}=\tan 15{}^\circ
  5. sin9=3+5554=cos81\sin 9{}^\circ =\frac{\sqrt{3+\sqrt{5}}-\sqrt{5-\sqrt{5}}}{4}=\cos 81{}^\circ
  6. cos9=3+5554=sin81\cos 9{}^\circ =\frac{\sqrt{3+\sqrt{5}}-\sqrt{5-\sqrt{5}}}{4}=\sin 81{}^\circ

Important Results:

sinθsin(60θ)sin(60+θ)=14sin3θ\sin \theta \sin \left( 60-\theta \right)\sin \left( 60+\theta \right)=\frac{1}{4}\sin 3\theta,

cosθcos(60θ)cos(60+θ)=14cos3θ\cos \theta \cos \left( 60-\theta \right)\cos \left( 60+\theta \right)=\frac{1}{4}\cos3 \theta,

tanθtan(60θ)tan(60+θ)=tan3θ\tan \theta \tan \left( 60-\theta \right)\tan \left( 60+\theta \right)=\tan 3\theta

Lets solve few more examples:

Example 1: Find value of sinAsin2A+sin3Asin6A+sin4Asin13A2sinAcos2A+sin3Acos6A+sin4Acos13A=?\frac{\sin A\sin 2A+\sin 3A\sin 6A+\sin 4A\sin 13A}{2\sin A\cos 2A+sin3Acos6A+sin4Acos13A}=?

Solution:

cos(A)cos(3A)+cos3Acos9A+cos9Acos17Asin3A+sin(A)+sin9A+sin(3A)+sin(17A)+sin(9A)\frac{\cos \left( -A \right)-{\cos \left( 3A \right)}+{\cos 3A}-{\cos 9A}+{\cos 9A}-\cos 17A}{{\sin 3A}+\sin \left( -A \right)+{\sin 9A}+{sin\left( -3A \right)}+\sin \left( 17A \right)+{\sin \left( -9A \right)}}

=cosAcos17Asin17AsinA\frac{\cos A-\cos 17A}{\sin 17A-\sin A}

=2sin9Asin8A2cos9Asin8A\frac{{2}\sin 9A{\sin 8A}}{{2}\cos 9A{\sin 8A}}

=tan 9A

Example 2:

sin25+sin210+sin215+.+sin290=?{{\sin }^{2}}5{}^\circ +{{\sin }^{2}}10{}^\circ +{{\sin }^{2}}15{}^\circ +……….+{{\sin }^{2}}90{}^\circ =?

Solution:

sin25+sin210+.+sin285+sin290{{\sin }^{2}}5{}^\circ +{{\sin }^{2}}10{}^\circ +……….+{{\sin }^{2}}85{}^\circ +{{\sin }^{2}}90{}^\circ

=sin25+sin210+.+cos25+sin290{{\sin }^{2}}5{}^\circ +{{\sin }^{2}}10{}^\circ +……….+{{\cos }^{2}}5{}^\circ +{{\sin }^{2}}90{}^\circ

=(sin25+cos25)+.+sin245+sin290\left( {{\sin }^{2}}5{}^\circ +{{\cos }^{2}}5{}^\circ \right)+………. +{{\sin }^{2}}45{}^\circ+{{\sin }^{2}}90{}^\circ

=[1+1+…+1] + (1/√2)2 + 1

=8×1+1+128\times 1+1+\frac{1}{2}

=8+1+12=9128+1+\frac{1}{2}=9\frac{1}{2}

Example 3:

Prove that sinα+sin(α+2π3)+sin(α+4π3)=0\sin \alpha +sin\left( \alpha +\frac{2\pi }{3} \right)+\sin \left( \alpha +\frac{4\pi }{3} \right)=0

Solution:

2sin(2α2+4π6)cos(4π6)+sin(α+2π3)2\sin \left( \frac{2\alpha }{2}+\frac{4\pi }{6} \right)\cos \left( -\frac{4\pi }{6} \right)+\sin \left( \alpha +\frac{2\pi }{3} \right) 2sin(α+2π3)cos(2π3)+sin(α+2π3)2\sin \left( \alpha +\frac{2\pi }{3} \right)\cos \left( \frac{2\pi }{3} \right)+\sin \left( \alpha +\frac{2\pi }{3} \right) 2×(12)sin(α+2π3)+sin(α+2π3)2\times \left( -\frac{1}{2} \right)\sin \left( \alpha +\frac{2\pi }{3} \right)+\sin \left( \alpha +\frac{2\pi }{3} \right) sin(α+2π3)+sin(α+2π3)-\sin \left( \alpha +\frac{2\pi }{3} \right)+\sin \left( \alpha +\frac{2\pi }{3} \right)

= 0

Example 4:

If asinθ=bsin(θ+2π3)=csin(θ+4π3)a\sin \theta =b\sin \left( \theta +\frac{2\pi }{3} \right)=c\sin \left( \theta +\frac{4\pi }{3} \right).

Prove that ab + bc + ca = 0.

Solution:

asinθ=bsin(θ+2π3)=csin(θ+4π3)=ka\sin \theta =b\sin \left( \theta +\frac{2\pi }{3} \right)=c\sin \left( \theta +\frac{4\pi }{3} \right)=k,

sinθ=ka\sin \theta =\frac{k}{a},

sin(θ+2π3)=kb\sin \left( \theta +\frac{2\pi }{3} \right)=\frac{k}{b},

sin(θ+4π3)=kc\sin \left( \theta +\frac{4\pi }{3} \right)=\frac{k}{c}

We know

sinθ+sin(θ+2π3)+sin(θ+4π3)=0\sin \theta +\sin \left( \theta +\frac{2\pi }{3} \right)+\sin \left( \theta +\frac{4\pi }{3} \right)=0,

ka+kb+kc=0\frac{k}{a}+\frac{k}{b}+\frac{k}{c}=0,

k(bc+ac+ab)abc=0\frac{k\left( bc+ac+ab \right)}{abc}=0,

ab+bc+ca=0ab+bc+ca=0

Summation Of Series In Trigonometry

1. sinα+sin(α+β)+sin(α+2β)+.+sin(α+(n1)β)=sin[α+(n1)2β]sinnβ2sinβ2\sin \alpha +\sin \left( \alpha +\beta \right)+\sin \left( \alpha +2\beta \right)+……….+\sin \left( \alpha +\left( n-1 \right)\beta \right)=\frac{\sin \left[ \alpha +\frac{\left( n-1 \right)}{2}\beta \right]\sin \frac{n\beta }{2}}{\sin \frac{\beta }{2}}

When the angles of sine are in A.P. the sum of sine series is given by above formula.

However if α=β\alpha =\beta in above case.

Then sinα+sin2α+sin3α+.+sinα=sin(n+12)αsinnα2sinα2\sin \alpha +\sin 2\alpha +\sin 3\alpha +……….+\sin \alpha =\frac{\sin \left( \frac{n+1}{2} \right)\alpha \sin \frac{n\alpha }{2}}{\sin \frac{\alpha }{2}}

2. When cosine angles are in AP. cosα+cos2α+cosα+.+cosnα=cos(n+12)αsinnα2sinα2cos\alpha +cos2\alpha +\cos \alpha +……….+\cos n\alpha =\frac{\cos \left( \frac{n+1}{2} \right)\alpha \sin \frac{n\alpha }{2}}{\sin \frac{\alpha }{2}}

Remember:

If A,B,C are angles of Δ\Delta

  1. sin(B+C)=sinA\sin \left( B+C \right)=\sin A
  2. cos(B+C)=cosA\cos \left( B+C \right)=-\cos A
  3. sin(B+C2)=cosA2\sin \left( \frac{B+C}{2} \right)=\cos \frac{A}{2}
  4. cos(B+C2)=sinA2\cos \left( \frac{B+C}{2} \right)=\sin \frac{A}{2}
  5. sin2A+sin2B+sin2C=4sinAsinBsinC\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C
  6. cos2A+cos2B+cos2C=14cosAcosBcosC\cos 2A+\cos 2B+\cos 2C=1-4\cos A\cos B\cos C
  7. tanA+tanB+tanC=tanAtanBtanC\tan A+\tan B+\tan C=\tan A\tan B\tan C

Trigonometric Equations

A solution of trigonometric equation is the value of the unknown angle that satisfy the equation.

Following are general solutions of trigonometric equation in the standard form:

Equation General Solution
sinθ=0\sin \theta =0 θ=nπnz\theta =n\pi n\in z
cosθ=0\cos \theta =0 θ=(2n+1)π2nz\theta =\left( 2n+1 \right)\frac{\pi }{2}n\in z
tanθ=0\tan \theta =0 θ=nπ,nz\theta =n\pi ,n\in z
sinθ=sinα\sin \theta =\sin \alpha θ=nπ+(1)nα,nz\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha ,n\in z
cosθ=cosα \cos \theta =\cos \alpha  θ=2nπ±α,nz\theta =2n\pi \pm \alpha ,n\in z
tanθ=tanα\tan \theta =\tan \alpha θ=nπ+α,nz\theta =n\pi +\alpha ,n\in z
sin2θ=sin2αcos2θ=cos2αtan2θ=tan2α}                                                          θ=nπ±α,nz\left. \begin{matrix} {{\sin }^{2}}\theta ={{\sin }^{2}}\alpha \\ {{\cos }^{2}}\theta ={{\cos }^{2}}\alpha \\ {{\tan }^{2}}\theta ={{\tan }^{2}}\alpha \\ \end{matrix} \right\} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \theta =n\pi \pm \alpha ,n\in z

Example 1:

Solve 4sinxsin2xsin4x=sin3x4\sin x\sin 2x\sin 4x=\sin 3x

Solution:

4sinxsin(3xx)sin(3x+x)=sin3x4\sin x\sin \left( 3x-x \right)\sin \left( 3x+x \right)=\sin 3x,

4sinx(sin23xsin2x)=sin3x4\sin x\left( {{\sin }^{2}}3x-{{\sin }^{2}}x \right)=\sin 3x,

4sinx.sin23x4sin3x=3sinx4sin3x4\sin x.{{\sin }^{2}}3x-4{{\sin }^{3}}x=3\sin x-4{{\sin }^{3}}x,

4sinxsin23x3sinx=04\sin x{{\sin }^{2}}3x-3\sin x=0,

sinx(4sin23x3)=0\sin x\left( 4{{\sin }^{2}}3x-3 \right)=0,

sinx=0    and    sin23x=34\sin x=0\;\; and \;\;{{\sin }^{2}}3x=\frac{3}{4},

nzn\in z,

sin23x=34{{\sin }^{2}}3x=\frac{3}{4},

sin23x=(34)2=sin2π