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Trigonometry

Trigonometry concept

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Trigonometric Function

Trigonometric Functions (Right Triangle)

Special Angles

Trigonometric Function Values in Quadrants II, III, and IV

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)]

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

  1. sin2 X + cos2 X = 1
  2. 1 + tan2 X = cec2 X
  3. 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

Important Notes & Short Tricks on Trigonometric Identities

Today we will be covering a very important topic from the Advance Maths part of the Quantitative Aptitude section that is – Important Notes & Short Tricks on Trigonometric Identities.

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

Here are some important questions of Trigonometric identities.

(1)Value of image005 is

(a)image002

(b)image004

(c)image003

(d)None of these

Ans.(a)

image005 is equal to

image006

 

(2)If image007is acute and image008 then image009 is equal to

(a)

(b)3

(c)  2

(d)  4

Ans.  (c)

If sum of the inversely proportional value is 2

i.e if . image013 thenimage014

image015

 

so image009=2

or we can put image007= 45°

(3)The simplified value ofimage019 is

(a)-1

(b)0

(c)sec2x

(d)1

Ans. (d)

The simplified value of

image019is obtained by  putting x=y=45°

image022

(4) Find the value of image024

(a)  1

(b)  -1

(c)  2

(d)  -2

Ans. (c)

put image017

image026

(5) If image033 then image028 is equal

(a)7/4

(b) 7/2

(c)5/2

(d)5/4

Ans. (d)

image012 as we know thatimage034

j

on solving we get secimage007= 5/4

Note:if x+y=a

and x-y=b

then x=(a+b)/2 and y=(x-y)/2

 

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