Basics Of Trigonometry
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Questions Based on Trigonometry is asked in almost all the competition exams like CAT,SSC,Banking, and many more. In this Post we will discuss this Topic from basics.
Questions Based on Trigonometry is asked in almost all the competition exams like CAT,SSC,Banking, and many more. In this Post we will discuss this Topic from basics.
TRIGNOMETRY
This is a branch of maths that
deals with the ratios between the sides of a right-angled triangle with
reference to acute angle or any angle in triangle. Trigonometry literally means
three angle measure.
Some Important Points To
remember:
Tan =
|
Perpendicular
|
P
|
|
Base.
|
B
|
||
SinΘ =
|
Perpendicular
|
P
|
|
Hypotenuse.
|
H
|
||
CosΘ =
|
Base
|
B
|
|
Hypotenuse.
|
H
|
Easy Way to remember the
above is
sinΘ
|
cosΘ
|
tanΘ
|
S
|
C
|
T
|
Sona
|
Chandi
|
Tol
|
P
|
B
|
P
|
H
|
H
|
B
|
Pandit
|
Badri
|
Prasad
|
Har
|
Har
|
Bhola
|
S C T
P B P
H H B
Theorms:
Sin2Θ + Cos2Θ = 1
Sec2Θ – Tan2Θ = 1
Cosec2Θ – Cot2Θ = 1
where,
Sec2Θ = 1/Cos2Θ
Cot2Θ = 1/Tan2Θ
Cosec2Θ = 1/Sin2Θ
Cot2Θ = 1/Tan2Θ
Cosec2Θ = 1/Sin2Θ
While solving
important points to keep in mind,
Sin(90°-Θ) = CosΘ,
Tan(90°-Θ) = CotΘ,
SIGN CONVENTIONS:
1 Sign
in First Quadrant i.e 0<Θ=<90, All will be Positive.
Sign
in Second Quadrant i.e 90<Θ=<180, only SinΘ and CosecΘ will be Positive,
3 Sign
in Third Quadrant i.e 180<Θ=<270, only TanΘ and CotΘ will be Positive,
4 Sign
in fourth Quadrant i.e 270<Θ=<360, only CosΘ and SecΘ will be positive,
As shown in
figure
|
S
|
A
|
T
|
C
|
In this Picture To
remember the above sign convention a easy and lucid way is give:
As shown in
figure
|
SCHOOL
|
ALL
|
TO
|
COLLEGE
|
As shown in figure
|
(only SinΘ and CosΘ +)
|
(all trigo function positive)
|
(only tanΘ and CotΘ +)
|
(only CosQ and SecQ +)
|
Below is a Table for
Value of Trigonometry function:
Angle ð
|
0°
|
30°
|
45°
|
60°
|
90°
|
||||||||||||||||
Trignometric Ratio
↓
|
(0°)
|
|
|
|
|
||||||||||||||||
Sin θ
|
0
|
|
|
|
1
|
||||||||||||||||
Cos θ
|
1
|
|
|
|
0
|
Complementary
Angles Trigonometry Concept
·
If you add two angles (A+B) and if their sum is 90, then they’re called
complimentary angles.
·
For example, 60 and 30 are complimentary because 60+30=90.
·
Same way (0,90)
·
Same way (45,45)
·
Same way (85,5), (12,78), (15,75) and so on…
·
If A and B are complimentary then A+B=90. Based on this we can also say
that,
·
B=90-A and
·
A=90-B.
In the table if you compare the rows of sin and cos. You’ll notice that
1. Sin0=cos90=0
2. Sin30=cos60=1/2
3. Sin45=cos45=1/root2
4. Sin60=cos30=root3/2
We can generalize this as: sinA=cos(90-A) and CosA=sin(90-A).
This generalization is true for all angles between 0 to 90 including 0
and 90.
Le’s try a question
Q. Find the value of cos18/sin72 NEXTGEN
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Approach
·
We know that cos and sin are complimentary.
·
CosA=sin(90-A)
·
So for Cos18, I can write cos18=sin(90-18)=sin72. Let’s use it
Cos18/sin72
=sin72/sin72 (because cos18=sin72).
=1 (because numerator and denominator are same so they’ll cancel each other.)
Important
1. You’ve to convert
only one of the two values. For example, in above sum we converted Cos into its
complimentary Sin. You could also solve it by converting the sin into its
complimentary cos. That is sin72=cos(90-72)=cos18. And then
cos18/sin72=cos18/cos18=1.
2. If you convert both
values then you’ll never get answer. ( that is cos18/sin72=> converted to
sin72/cos18). In that case you’ll run into an infinite loop!
3. In the exam, sin
and cos’s complimentary concept will help mostly for divisions and subtraction
cases.
Division
cases
|
Substation
cases
|
Cos18/sin72=sin72/sin72=1.
|
Cos18-sin72=sin72-sin72=0
|
Here
they numerator and denominator cancels each other.
|
Here
too they’ll cancel each other.
|
But suppose if there is cos18+sin72, this doesn’t lead to any value.
(because In the exam, sin and cos’s complimentary concept usually does not
help in multiplication or addition cases.
Multiplication cases
|
Addition cases
|
Cos18 x sin72 = sin72 x sin72 = sin272. But this is dead end because we don’t know the
value of sin72
|
Cos18 + sin72=sin72 + sin72= 2sin72. Again dead
end because we don’t know the value of sin72.
|
Point being: in the exam, after doing some steps, if you run into above
situations, there is a chance that you’ve made some mistake in calculation or
you’ve moved into the wrong direction. Although there can be exceptional
situation where it could help, when they’ve framed equation using both
trigonometry + algebraic equations (That is Type#4 questions, we’ll see about
it in next article).
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Just like sin and
cos, we can see that sec and cosec are also complimentary. ( in the table you
can see that for (30,60) and (45,45) the values of sec and cosec are same.
Therefore,
1. secA=cosec(90-A).
but this is not valid for 90 degrees because sec90 is not defined.
2. cosecA=sec(90-A).
but this is not valid for 0 degree because cosec0 is not defined.
Q. find value of sec70 x sin20 – cos20 x cosec70
If you convert all four (sin, cos, cosec, and sec) into their
complimentaries (cos, sin, sec and cosec) then you’ll run into infinite loop.
Part A
minus
|
Part b
|
Sec70 x
sin20-
|
Cos20 x
cosec70
|
In the part A, if I convert sin into its complimentary cos, then sec x
cos =1 because sec and cos are inverse of each other. Same way in part B if I
convert cosec into its complimentary sec, that too will lead to 1. Let’s see
Part A
–
|
Part b
|
Sec70 x
cos(90-20)-
|
Cos20 x
cosec(90-70)
|
Sec70 x
cos70-
|
Cos20 x
sec20
|
1-
|
1
|
=1-1
=0 is the final answer.
In the table, you can see that
1. Tan30=cot60=1/root3
2. Tan45=cot45=1
3. Tan60=cot30=root.
So, tan and cot are complimentary
1. tanA=cot(90-A) but
not valid for 90 degrees because tan90 is not defined.
2. cotA=tan(90-A) but
this is not valid for 0 degree because cot0 is not defined.
The tan and cot’s complimentary relationship is also import for
multiplication chain questions because tan and cot are also inverse of each
other. That is tan A x cot A= tan A x 1/tan A=1.
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Q. Find value of tan48 x tan23 x tan42 x tan67
When you get this type multiplication chain question, you’ve to find out
the pair of complimentary angles. Here 48 + 42=90 and 23 +67=90 so I’ll club
them together
Part A
x
|
Part B
|
Tan48 x
tan 42x
|
Tan23 x
tan67
|
In both parts, we’ll convert any one tan into cot, then tan x cot = 1
(because they’re inverse of each other).
Part A
x
|
Part B
|
Tan48 x
cot (90-42)x
|
Tan23 x
cot(90-67)
|
=tan48
x cot48
|
Tan23 x
cot23
|
=1x
|
1
|
Final answer =1.
Let’s try another one
Q. Find value of tan1 x tan 2 x tan3 x ….x…tan88 x tan89
Approach: you make pairs of complimentary numbers: 1+89=90, 2+88=90….
There is only one angle left who doesn’t get a pair (45)
So it’ll look like this
= (tan1 x ta89) (tan2 x tan88)x…xtan45
In each of those pairs, you convert one tan into its complimentary cot.
=(tan1 x tan(90-89)) x (tan2 x cot(90-88))…..x1 ; because tan45=1
=(tan1xcot1) x (tan2xcot2)x…..x1
=1 because tan and cot are inverse of each other.
Q. given that 4A is an acute angle and sec4A=cosec(A-20), what is the
value of A?
Ans. NEXTGEN CAREER COACHING –
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LHS
|
RHS
|
Sec4A
|
Cosec(A-20)=sec(90-(A-20))=sec(90-A+20)
|
Sec4A
|
=sec(110-A)
|
Since LHS=RHS so we can say that on both sides the angles are equal
4A=110-A
5A=110
A=110/5=22 degrees.
Q. in triangle ABC, if A, B and C are
acute angles, then Cos(B+C)/2=sin(A/2).
1. This statement is
always correct
2. This statement is
always Incorrect
3. Depends on values
of A,B and C
4. None of above. NEXTGEN CAREER COACHING – 9462900411 https://www.facebook.com/groups/NextGenCareers
Approach
We know that in any triangle ABC, the sum of three angle A+B+C=always
180
·
B+C=180-A ; (I’m taking c on right side)
·
(B+C)/2=(180-A)/2 ;(I’m dividing both sides by 2)
·
(B+C)/2=(180/2 MINUS A/2)
·
(B+C)/2=[90-(A/2)]……let’s call this equation 1.
Now back to our complimentary formulas. we already know that for angles
between 0 to 90, sinX=cos(90-x) this is definitely correct. So instead of “x”,
I’ll now write A/2
Sin(A/2)=cos[90-(A/2)]…..eq2
But according to equation 1, [90-(A/2)]= (B+C)/2
Putting that back in eq2
Sin(A/2)=cos(B+C)/2. Therefore answer is (A) always correct.
COMPLIMENTARY(90-A)
|
INVERSE
(1/A)
|
|
SIN
|
COS
|
COSEC
|
COS
|
SIN
|
SEC
|
TAN
|
COT
|
COT
|
COSEC
|
SEC
|
SIN
|
SEC
|
COSEC
|
COS
|
COT
|
TAN
|
TAN
|
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These complimentary are valid for acute angles (between 0 to 90) and
some of them are not valid for 0 or 90 if they’re “not defined”.
·
But what about angles greater than 90 degree. For example, sin150? For
that we’ve to look at the quadrants and change the signs accordingly. While
that concept is important for CAT, but as far as SSC is concerned, there is not
much point into going that deep.
·
In the SSC exam, if you’re given a lengthy trigonometry equation and
asked to find its value or angle then what to do? Well three situations can
happen:
1. If you’re given an
equation that contains 0, 30, 45, 60 and or 90 angles of sin/cos/tan/cosec/sec
or cot then you just have to plug in the values from trig.table and simplify
the equation. You’ll get the answer. (the complimentary case usually doesn’t
apply in such situations).
2. If you’re given an
equation that contains odd looking angles such as 12,15, 43, 85 or anything
that is not from the (0/30/45/60/90) group, then you’ve to think about the
application of complimentary angles.
1. You’ve to make
“pairs” of complimentary angles (e.g. sin10/cos80) and then convert ANY ONE in
the given pair. If you convert both into their complimentary, then you’ll never
get the answer.
2. Complimentary pair
(sin|cos), (cosec|sec), usually leads to division or subtraction then most of
the stuff in the equation gets eliminated and you get “good looking number” as
answer for example 1,2,3,4,1/2, 1/root3 etc. (if such pair leads to
multiplication case then perhaps their inverse is present (sin x cosec) or (cos
x sec).
3. Complimentary pair
(tan|cot), usually leads to multiplication. Because tan and cot are also
inverse of each other (tan x cot = 1), and then most of the things in the
equations get eliminated. But keep in mind tan and cot are inverse only for
same angle. E.g. tan33 x cot33=1. But tan33xcot34=not equal to 1.
If above situation 1 and 2 doesn’t help, then it is most likely a
question involving trigonometry formulas like
0. sin2A+cos2A=1, or
1. sec2A-tan2A=1 or
2. cosec2A-cot2A=1 (or all three
of them).
3. and or the
Application of algebraic formulas such as x2-y2=(x-y)(x+y).
4. we’ll see about
such scenarios in the next article.
SOLVED
PROBLEMS ON TRIGONOMETRY:
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Q.
Find the value of sin2a+[1/(1+tan2a)] ?
A.
sec2a – tan2a=1.
=> sec2a=1+tan2a.
=> 1/sec2a=cos2a.
therefore, sin2a+cos2a = 1.
Hence, our Answer is 1.
=> sec2a=1+tan2a.
=> 1/sec2a=cos2a.
therefore, sin2a+cos2a = 1.
Hence, our Answer is 1.
Q.
if 3cos2A+7sin2A=4 then find value of cotA, given that A
is an acute angle?
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A.
3cos2A+7sin2A=4, _______(1)
=> cos2A+sin2A=1,
Therefore, 3cos2A+3sin2A=3, ________(2)
Putting eq(2) in eq(1),
We get 4sin2A=1,
Therefore sinA=1/2.=> A=30 degree,
Therefore cot30 = 1/root 3.
=> cos2A+sin2A=1,
Therefore, 3cos2A+3sin2A=3, ________(2)
Putting eq(2) in eq(1),
We get 4sin2A=1,
Therefore sinA=1/2.=> A=30 degree,
Therefore cot30 = 1/root 3.
Q.
if cos2a+cos4a=1, what is the value of tan2a+tan4a?
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A.
sin2a+cos2a=1 …………..eq(1)
In the question, we are given
that
cos2a+cos4a=1
taking
cos2a on the right hand side
cos4a=1-cos2a
taking value from eq(1)
cos4a=sin2a
cos2a x cos2a = sin2a ; laws of surds and indices
cos2a=(sin2a/cos2a)
cos2a=tan2a……………….eq(2)
now lets move to the next part.
in the question, we have to find the value of
tan2a+tan4a
=tan2a+(tan2a)2 ;
because (a2)2=a2×2=4
= cos2a+cos4a ;applying value
from eq2
=1 ;this was already given in
the first part of the question itself.
Final answer=1.
Q.find
value of sin4a-cos4a
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We know that
(a2-b2)=(a+b)(a-b)
So instead
of sin4a-cos4a, I can write
(sin2a)2-(cos2a)2 ; because a4=2×2 =(a2)2
=(Sin2a+cos2a)(Sin2a-cos2a)
=1 x
(Sin2a-cos2a) ; because (a2-b2)=(a+b)(a-b)
=(Sin2a-cos2a)
Q.
if sin4a-cos4a=-2/3 then what is the value of 2cos2a-1?
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A. sin4a-cos4a=-2/3,
a2-b2=(a+b)(a-b)………..eq(1)
From eq(1),
sin4a-cos4a=(sin2-cos2)(sin2+cos2)
and from identities, sin2+cos2=1,
sin2-cos2=-2/3,
2cos2a-1=2/3. NEXTGEN CAREER COACHING – 9462900411 https://www.facebook.com/groups/NextGenCareers
a2-b2=(a+b)(a-b)………..eq(1)
From eq(1),
sin4a-cos4a=(sin2-cos2)(sin2+cos2)
and from identities, sin2+cos2=1,
sin2-cos2=-2/3,
2cos2a-1=2/3. NEXTGEN CAREER COACHING – 9462900411 https://www.facebook.com/groups/NextGenCareers
Now, you can solve the following Questions out of Test Paper No161A-E20 and submit answers in the form of 1a,2b,3c,4d,5e,6b in the Comments Section. You will get your scorecard, answer key, and detailed explanation of each question. Please submit your email address, WhatsApp# to get full Practice Test Paper No161A-E20 and similar other practice test papers. You can also download Lessons and Test Papers from https://www.facebook.com/groups/NextGenCareers/
1. Find value of Cos2a+(1/1+cot2a)
1. 0
2. 1
3. 2
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2. Find value of 2sin2a+4sec2a+5cot2a+2cos2a-4tan2a-5cosec2a
1. 0
2. 1
3. 2
4. 3
3. Find the value of
(cosA-sinA)2+(cosA+sinA)2
1. 0
2. 1
3. 2
4. 3 NEXTGEN CAREER COACHING – 9462900411
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4. Find the value of
Cot2A x (sec2A-1)
1. 0
2. 1
3. 2
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5. Find the value of
(secA x cotA)2 – (cosec A x cosA)2
1. 0
2. 1
3. 2
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6. secA/(cotA+tanA)
will be equal to
1. cosA
2. cosecA
3. sinA
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7. (1+tanA+secA)(1+cotA-cosecA)
will be equal to
1. 0
2. 1
3. 2
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8. Cos6A+sin6A=
1. 1-3(cosA x sinA)2
2. 1+3(cosA x sinA)2
3. 1-3(cosA x sinA)3
4. 1-3(cosA x sinA)6 NEXTGEN
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9. (sin2A x cos2B) – (cos2A x sin2B) will be equal
to
1. Sin2A-cos2A
2. Sin2A+cos2A
3. Cos2A – cos2A
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10. (tanA+cotA)(secA-cosA)(cosecA-sinA)=
1. 0
2. 1
3. 2
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11. If Tan2A+Tan4A=1 then what is
the value of cos2A+cos4A, given that A is an acute angle?
1. 0
2. 1
3. 2
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12. (cosecA-cotA)2=
1. (1-cosA)/(2+cosA)
2. (1+cosA)/(1-cosA)
3. (1-cosA)/(1+cosA)
4. (1-cosA)x(1+cosA) NEXTGEN CAREER COACHING – 9462900411
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13. [(tanA+secA)2-1]/[(tanA+secA)2+1], will be
equal to
1. tanA
2. secA
3. sinA
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14. [(sin220+sin270)/(sec250-cos240)]+2cosec258-2(cot58xtan32)-(4tan13 x tan37 x tan45 x tan53 x
tan77)
1. 0
2. -1
3. 1
4. None of above NEXTGEN CAREER COACHING – 9462900411
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15. If cotA=root 7,
what is the value of (cosec2A-sec2A)/( cosec2A+sec2A)
1. 3/2
2. 3/4
3. 4/3
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16. (sinA+cosecA)2+(cosA+secA)2 will be
equal to
1. Tan2A+cot2A+4
2. Tan2A-cot2A+5
3. Tan2A-cot2A
4. Tan2A+cot2A+7
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