Trigonometry Tricks

1. किसी भी समिोण (Right angle) िे लऱये सूत्र (formula) – िणण2 = ऱम्ब2 + आधार2

2. अब याद रखिये LAL/KKA, (ऱाऱ/ िक्िा) L- ऱम्ब, A- आधार, K- िणण 3. अब इनिा क्रम sin θ , cos θ, tan θ, तथा cot θ, sec θ, cosec θ इनिे ठीि उल्टे होते हैं a. Sin θ= ऱम्ब / िणण, cosec θ = िणण / ऱम्ब

b. cos θ= आधार / िणण, sec θ= िणण / आधार

c. tan θ = ऱम्ब / आधार , cot θ = आधार/ ऱम्ब

Pythagorean Identities

sin2 θ + cos2 θ = 1 tan2 θ + 1 = sec2 θ cot2 θ + 1 = csc2 θ

Negative of a Function

sin (–x) = –sin x cos (–x) = cos x tan (–x) = –tan x csc (–x) = –csc x sec (–x) = sec x cot (–x) = –cot x

If A + B = 90o, Then

Sin A = Cos B Sin2A + Sin2B = Cos2A + Cos2B = 1 Tan A = Cot B Sec A = Csc B

For example:

If tan (x+y) tan (x-y) = 1, then find tan (2x/3)?

Solution:

Tan A = Cot B, Tan A*Tan B = 1

So, A +B = 90o

(x+y)+(x-y) = 90o, 2x = 90o , x = 45o

Tan (2x/3) = tan 30o = 1/√3

If A - B = 90o, (A › B) Then

Sin A = Cos B Cos A = - Sin B Tan A = - Cot B

If A ± B = 180o, then

Sin A = Sin B Cos A = - Cos B

If A + B = 180o

Then, tan A = - tan B

If A - B = 180o

Then, tan A = tan B

For example:

Find the Value of tan 80o + tan 100o ?

Solution:Since 80 + 100 = 180

Therefore, tan 80o + tan 100o = 1

If A + B + C = 180o, then

Tan A + Tan B +Tan C = Tan A * Tan B *Tan C

sin θ * sin 2θ * sin 4θ = ¼ sin 3θ

cos θ * cos 2θ * cos 4θ = ¼ cos 3θ

For Example:What is the value of cos 20o cos 40o cos 60o cos 80o?

Solution: We know cos θ * cos 2θ * cos 4θ = ¼ cos 3θ

Now, (cos 20o cos 40o cos 80o ) cos 60o

¼ (Cos 3*20) * cos 60o

¼ Cos2 60o = ¼ * (½)2 = 1/16

If a sin θ + b cos θ = m & a cos θ - b sin θ = n

then a2 + b2 = m2 + n2

For Example:

If 4 sin θ + 3 cos θ = 2 , then find the value of 4 cos θ - 3 sin θ:

Solution:

Let 2 cos θ - 3 sin θ = x

By using formulae a2 + b2 = m2 + n2

42 + 32 = 22 + x2

16 + 9 = 4 + x2

X = √21

If

sin θ + cos θ = p & csc θ - sec θ = q

then P – (1/p) = 2/q

For Example:

If sin θ + cos θ = 2 , then find the value of csc θ - sec θ:

Solution:

By using formulae:

P – (1/p) = 2/q

2-(1/2) = 3/2 = 2/q

Q = 4/3 or csc θ - sec θ = 4/3

If

a cot θ + b csc θ = m & a csc θ + b cot θ = n

then b2 - a2 = m2 - n2

If

cot θ + cos θ = x & cot θ - cos θ = y

then x2 - y2 = 4 √xy

If

tan θ + sin θ = x & tan θ - sin θ = y

then x2 - y2 = 4 √xy

If

y = a2 sin2x + b2 csc2x + c

y = a2 cos2x + b2 sec2x + c

y = a2 tan2x + b2 cot2x + c

then,

ymin = 2ab + c

ymax = not defined

For Example:

If y = 9 sin2 x + 16 csc2 x +4 then ymin is:

Solution:

For, y min = 2* √9 * √16 + 4

= 2*3*4 + 20 = 24 + 4 = 28

If

y = a sin x + b cos x + c

y = a tan x + b cot x + c

y = a sec x + b csc x + c

then, ymin = + [√(a2+b2)] + c

ymax = - [√(a2+b2)] + c

For Example:

If y = 1/(12sin x + 5 cos x +20) then ymax is:

Solution:

For, y max = 1/x min

= 1/- (√122 +52) +20 = 1/(-13+20) = 1/7

Sin2 θ, maxima value = 1, minima value = 0

Cos2 θ, maxima value = 1, minima value = 0

Trigonometric Function

Trigonometric Functions (Right Triangle)

Special Angles

Trigonometric Function Values in Quadrants II, III, and IV

Examples:

Example2:

Example: 3:

Unit Circle

Addition Formulas:

cos(X+Y) = cosXcoxY – sinXsinY

cos(X-Y) = cosXcoxY + sinXsinY

sin(X+Y) = sinXcoxY + cosXsin

sin(X-Y) = sinXcoxY – cosXsinY

tan(X+Y) = [tanX+tanY]/ [1– tanXtanY]

tan(X-Y) = [tanX-tanY]/ [1+ tanXtanY]

cot(X+Y) = [cotX+cotY-1]/ [cotX+cotY]

cot(X-Y) = [cotX+cotY+1]/ [cotX-cotY]

Sum to Product Formulas:

cosX + cosY = 2cos [(X+Y) / 2] cos[(X-Y)/2]

sinX + sinY = 2sin [(X+Y) / 2] cos[(X-Y)/2]

Difference to Product Formulas

cosX - cosY = - 2sin [(X+Y) / 2] sin[(X-Y)/2]

sinX + sinY = 2cos [(X+Y) / 2] sin[(X-Y)/2]

Product to Sum/Difference Formulas

cosXcosY = (1/2) [cos (x-Y) + cos (X+Y)]

sinXcoxY = (1/2) [sin (x+Y) + sin (X-Y)]

cosXsinY = (1/2) [sin (x+Y) + sin (X-Y)]

sinXsinY = (1/2) [cos (x-Y) + cos (X+Y)]

Double Angle Formulas

sin (2X) = 2 sin X cos X

cos (2X) = 1 – 2sin2X= 2cos2X – 1

tan(2X) = 2tanX/[1-tan2X]

Multiple Angle Formulas

More half-angle formulas

Law of Sines

a/sinA = b/sinB= c/sinC

Law of Cosines

a2 = b2 +c2 – 2bcCosA

b2 = a2 + c2 – 2ac CosB

c2 = a2 + b2 – 2abCosC

Pythagorean Identities

a. sin2 X + cos2 X = 1

b. 1 + tan2 X = cec2 X

a. 1 + cot2 X = csc2 X

Given Three Sides and no Angles (SSS) Given three segment lengths and no angle measures, do the following:

Use the Law of Cosines to determine the measure of one angle.

Use the Law of Sines to determine the measure of one of the two remaining angles.

Subtract the sum of the measures of the two known angles from 180˚ to obtain the

measure of the remaining angle.

Given Two Sides and the Angle between Them (SAS)

Given two segment lengths and the measure of the angle that is between them, do the

following: Use the Law of Cosines to determine the length of the remaining leg.

Use the Law of Sines to determine the measure of one of the two remaining angles.

Subtract the sum of the measures of the two known angles from 180˚ to obtain the measure

of the remaining angle.

Given One Side and Two Angles (ASA or AAS)

Given one segment length and the measures of two angles, do the following:

Subtract the sum of the measures of the two known angles from 180˚ to obtain the measure

of the remaining angle.

Use the Law of Sines to determine the lengths of the two remaining legs.

Some Important Tricks

Remember Useful Point :

tan1. tan2. ……… tan89 = 1

cot1. cot2 ……. Cot890 = 1

cos10.cos20…… cos900 = 0

cos10.cos20…… to (greater than cos900) = 0

sin10.sin20.sin30 ……… sin1800 = 0

sin10. sin20 sin30 ….. to (greater than sin1800) = 0

प्रश्न संख्या -1

sin 43 + cos 19 – 8cos260

cos 47 sin 71

हऱ :- sin 43 + cos 19 – 8cos260

cos (90-43) sin (90-19)

= sin 43 + cos 19 – 8(1/2)2

sin 43 cos 19

= 1+1 – 2

= 0 उत्तर

प्रश्न संख्या -2

1+ 1 – sec227 + 1 – cosec227

cot263 sin263

= 1 + tan263 – sec2(90- 63) + cosec263 – cosec2(90-63)

= sec263 – cosec263 + cosec263 – sec263

= 0 उत्तर

प्रश्न संख्या- 3

यदद x = cos@ तो 1- sin@

cos@ िा मान क्या होगा ??

1+sin@

हऱ: x = cos@

1- sin@

1 = 1- sin@

x cos @

= (1- sin@) (1+ sin@)

cos @ (1+ sin@)

= 1 – sin2@

cos @ (1+ sin@)

= cos2@

cos @ (1+ sin@)

= cos@ उत्तर

(1+ sin@)

प्रश्न संख्या 4-

यदद tan @ + cot @ = 2 तो @ िा मान क्या होगा?

इसे हऱ िरने िे लऱये हम @ िा एि ऐसा मान सोचेंगे जिसिा मान tan तथा cot िे लऱये 1 हो और ऐसा 45 ऩर सम्भव है

tan 45 + cot 45 = 1

अत: @ = 1

प्रश्न संख्या -5

tan2@+3 = 3 sec @ तो @ िा मान क्या होगा ?

हऱ:

tan2@+3= 3sec@

sec2@ – 1 +3 = 3 sec @

sec2@ – 3sec@ +2 = 0

गुणनिण्ड िरने ऩर

sec2@ – 2sec@ – sec @ +2 = 0

sec@(sec@-2) -1(sec@-2)=0

(sec@-1)(sec@-2) = 0 अत: sec@ =1 तथा @= 0

या sec@=2 तथा @ = 60

प्रश्न संख्या-6

sin265+sin225+cos235+ cos255 िा मान क्या होगा?

हऱ: sin265+sin2(90-65)+cos2(90-55)+ cos255)

= sin265+cos265+sin255+ cos255

= 1+1

= 2

प्रश्न संख्या-7

1-2sin2Q+sin4Q िा मान ज्ञात िीजिये ?

हऱ:

1-2sin2Q+sin4Q

= 1 – 2sin2Q+sin2Q.sin2Q

= 1- sin2Q – sin2Q+sin2Q.sin2Q

= cos2Q –sin2Q + (1- cos2Q.1- cos2Q)

= cos2Q – (1-cos2Q) + (1- cos2Q.1- cos2Q)

= cos2Q – 1 + cos2Q + (1- 2cos2Q + cos4Q)

= cos2Q – 1 + cos2Q + 1- 2cos2Q + cos4Q

= 2cos2Q – 1 + 1- 2cos2Q + cos4Q

= cos4Q उत्तर

प्रश्न संख्या-8

त्रत्रभुि ABC में Sin (A+B)/2 िा मान किसिे बराबर है?

हऱ:

किसी त्रत्रभुि ABC में –

A+B+C = 180

A+B+C = 90

2

A/2 + B/2 + C/2 = 90

A/2 + B /2 = 90 – C/2

A+B = 90- C/2

2

Sin (A+B) = Sin 90- C/2

2

= Cos C/2 उत्तर

प्रश्न संख्या-9

tan Q+Cot Q = /3

तो tan3 Q + cot3 Q िा मान क्या होगा ?

हऱ:

दोनों ओर घन िरने ऩर

(tan Q + cot Q)3 = (/3)3

tan3 Q + cot3 Q+ 3 tanQ.cotQ(tan Q+Cot Q) = 3 ./3

tan3 Q + cot3 Q+ 3.(/3) = 3 ./3

tan3 Q + cot3 Q = 3 ./3 – 3 ./3

tan3 Q + cot3 Q= 0 उत्तर

प्रश्न संख्या-10

Cot 40 – 1 { cos 35}

tan 50 2 {sec 55}

हऱ:

Cot 40 – 1 { cos 35}

tan (90-40) 2 {sec (90-35)}

Cot 40 – 1 { cos 35}

Cot 40 2 {cos 35}

1 – 1

2

= 1/2