Trigonometry 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

<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 Ratio

  • Sine

Y= Sinx

Domain R

Range (-1,1)

  • Cosine

y = cos x

  • Tangent

y = tan x

 

 

 

  • Co – tangent

y = cot x

Domain

xR(nπ)x\sum{R-(n\pi )}

Range (,-\infty ,\infty)

  • Secant
y=secx,y=\sec x,

DomainR{(2x+1)π2}Domain\,\,\,\,\,R-\left\{ (2x+1)\,\frac{\pi }{2} \right\} Range(1)u(1,)Range\,(-\infty -1)\,u\,(1,\infty )
  • Cosecant

y = cosecx

.DomainR{nπ}Domain\,R-\{n\pi \} Range:(1)u(1,)Range:\,(-\infty -1)\,u\,(1,\infty )

Watch this Video for More Reference


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

Sign Of T-Ratios:

T-Ratio At Some 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 Trigonometric Expressions

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

Problems on Trignometry

Example:

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}

Adding and subtracting

(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} tan4x1cot4x=154{{\tan }^{4}}x-\frac{1}{{{\cot }^{4}}x}=\frac{15}{4} tan4x1cot4x=414{{\tan }^{4}}x-\frac{1}{{{\cot }^{4}}x}=4-\frac{1}{4}

On comparing,

tan4x=4{{\tan }^{4}}x=4 tan4x=2{{\tan }^{4}}x=2 tan4x=±2{{\tan }^{4}}x=\pm \sqrt{2}

Example:

If sinx+sin2x=1\sin x+{{\sin }^{2}}x=1 then

cos8x+2cos6x+cos4x=?{{\cos }^{8}}x+2{{\cos }^{6}}x+{{\cos }^{4}}x=?

Solution:

cos2x=sinx{{\cos }^{2}}x=\sin x sin4x+2sin3x+sin2x{{\sin }^{4}}x+2{{\sin }^{3}}x+{{\sin }^{2}}x (sin2x+sinx)2{{\left( {{\sin }^{2}}x+\sin x \right)}^{2}} (1)2=1{{\left( 1 \right)}^{2}}=1

Example:

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

Then cos9θ+cos6θ+3cos5θ1=?{{\cos }^{9}}\theta +{{\cos }^{6}}\theta +3{{\cos }^{5}}\theta -1=?

Solution:

sinθ.tanθ=cos2θ\sin \theta .tan\theta =co{{s}^{2}}\theta sinθcosθ=cos2θ\frac{\sin \theta }{\cos \theta }=co{{s}^{2}}\theta sin2θ=cos3θ{{\sin }^{2}}\theta ={{\cos }^{3}}\theta (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:

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

Solution:

secxtanx=1secx+tanx\sec x-\tan x=\frac{1}{\sec x+\tan x} secx  +  tanx=1P\sec x\;+\;tan x=\frac{1}{P} secx    tanx=P\sec x\;-\;tan x=P 2sec=P+1P\overline{2\sec =P+\frac{1}{P}} secx=P2+12P\sec x=\frac{{{P}^{2}}+1}{2P}

Example:

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.

Example:

  1. sin1>0T/F?\sin 1{}^\circ >0T/F?

sin 57o70

Hence True.

  1. cos1>0T/F?\cos 1{}^\circ >0T/F?

cos 57o70

Hence True.

Trignometric 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

Example:

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

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

Example:

Find value of tan2(2θ)tan2θ1tan2(2θ)tan2θ=?\frac{{{\tan }^{2}}\left( 2\theta \right)-{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\left( 2\theta \right){{\tan }^{2}}\theta }=?

Solution:

(tan2(2θ)+tan2θ)(tan2θtanθ)(1+tan2θtanθ)(1tan2θtanθ)\frac{\left( {{\tan }^{2}}\left( 2\theta \right)+{{\tan }^{2}}\theta \right)\left( \tan 2\theta -\tan \theta \right)}{\left( 1+\tan 2\theta \tan \theta \right)\left( 1-\tan 2\theta \tan \theta \right)} tan3θ.tanθ\tan 3\theta .\tan \theta

Example:

Prove that tan8θtan6θtan2θ=tan8θ.tan6θ.tan2θ.\tan 8\theta -\tan 6\theta -\tan 2\theta =\tan 8\theta .\tan 6\theta .\tan 2\theta . tan8θtan2θtan6θtan8θ=tan2θ+tan6θ\tan 8\theta -\tan 2\theta \tan 6\theta \tan 8\theta =\tan 2\theta +\tan 6\theta tan8θtan6θtanθ=tan2θtan6θtan8θ.\tan 8\theta -\tan 6\theta -\tan \theta =\tan 2\theta \tan 6\theta \tan 8\theta .

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+12)cos(CD2)\sin C+\sin D=2\sin \left( \frac{C+1}{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)=14sinθ\sin \theta \sin \left( 60-\theta \right)\sin \left( 60+\theta \right)=\frac{1}{4}\sin \theta cosθcos(60θ)cos(60+θ)=14cosθ\cos \theta \cos \left( 60-\theta \right)\cos \left( 60+\theta \right)=\frac{1}{4}\cos \theta tanθtan(60θ)tan(60+θ)=tan3θ\tan \theta \tan \left( 60-\theta \right)\tan \left( 60+\theta \right)=\tan 3\theta

Example:

If A+B=225A+B=225{}^\circ (cotA1+cotA)(cotB1+cotB)=?\left( \frac{\cot A}{1+\cot A} \right)\left( \frac{\cot B}{1+\cot B} \right)=?.

Solution:

A+B=225=180+45A+B=225=180+45{}^\circ tan(A+B)=tan(ˉ+45)\tan \left( A+B \right)=\tan \left( \bar{\wedge }+45{}^\circ \right) tan(A+B)=tan45\tan \left( A+B \right)=\tan 45{}^\circ A+B=tan45A+B=\tan 45{}^\circ (1tanA1+1tanA)(1tanB1+1tanB)\left( \frac{\frac{1}{\tan A}}{1+\frac{1}{\tan A}} \right)\left( \frac{\frac{1}{\tan B}}{1+\frac{1}{\tan B}} \right) 1(1+tanA)(1+tanB)=11+tanAtanB+tanB+tanA\frac{1}{\left( 1+\tan A \right)\left( 1+{tanB} \right)}=\frac{1}{1+\tan A\tan B+\tan B+\tan A}

1(1+tanA)(1+tanB)=11+tanAtanB+tanB+tanA\frac{1}{\left( 1+\tan A \right)\left( 1+{tanB} \right)}=\frac{1}{1+\tan A\tan B+\tan B+\tan A} =12=\frac{1}{2}

Example:

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:

Prove that cosAsin(BC)+cosBsin(CA)+cosCsin(AB)=0\cos A\sin \left( B-C \right)+\cos B\sin \left( C-A \right)+\cos C\sin \left( A-B \right)=0

Solution:

sin(A+BC)sin(AB+C)+sin(B+CA)sin(BC+A)+sin(C+AB)sin(CA+B)\sin {\left( A+B-C \right)}-\sin {\left( A-B+C \right)}+\sin {\left( B+C-A \right)}-\sin {\left( B-C+A \right)}+\sin {\left( C+A-B \right)}-\sin {\left( C-A+B \right)}

Example:

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)+.+sin290+sin245\left( {{\sin }^{2}}5{}^\circ +{{\cos }^{2}}5{}^\circ \right)+……….+{{\sin }^{2}}90{}^\circ +{{\sin }^{2}}45{}^\circ 8×1+1+128\times 1+1+\frac{1}{2} 8+1+12=9128+1+\frac{1}{2}=9\frac{1}{2}

Example:

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:

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) P.Tab+bc+ca=0P.Tab+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. AP.cosα+cos2α+cosα+.+cosnα=cos(n+12)αsinnα2sinα2AP.\,\,cos\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

Trignometric Equations

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

Consider sinθ=12\sin \theta =\frac{1}{2} θ=π6,5π6\theta =\frac{\pi }{6},\frac{5\pi }{6} etc.

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

The equation acosθ+bsinθ=ca\cos \theta +b\sin \theta =c is solvable for ca2+b2\left| c \right|\sqrt{{{a}^{2}}+{{b}^{2}}}

Example:

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=sin2π3{{\sin }^{2}}3x={{\left( \frac{\sqrt{3}}{4} \right)}^{2}}={{\sin }^{2}}\frac{\pi }{3} 3x=mπ±π