TRIGONOMETRY
Compound Angle
• An angle made up of the algebraic sum of two or more angles is called compound angle
i) sin(A+ B)= sinA cosB + cosA sinB
ii) sin(A- B)= sinA cosB - cosA sinB
iii) cos(A+ B)= cosA cosB - sinA sinB
iv) cos(A- B)= cosA cosB + sinA sinB
v) tan(A+ B)= (tanA + tanB)/(1 - tanA tanB)
vi) tan(45+ A)= (1 + tanA)/(1 - tanA)
vii) tan(A - B)= (tanA - tanB)/(1 + tanA tanB)
viii) tan(45- A)= (1 - tanA)/(1 + tanA)
ix) cot(A+ B)= (cotA cotB -1)/(cotA+ cotB)
x) cot(A- B)= (cotA cotB +1)/(cotA- cotB)
xi) sin(A+ B) sin(A - B)= sin²A - sin²B
= Cos²B - cos²A
xii) cos(A+ B) cos(A - B)= cos²A - sin²B
= Cos²B - sin²A
xiii) tan(A+ B + C)= (tanA + tanB+ tanC - tanA tanB tanC)/(1 - tanA tanB - tanB tanC - tanC tanA).
MULTIPLE AND SU MULTIPLE
1) Sin2A= 2 sinA cosA = 2 tanA/(1+ tan²A)
2) cos2A= cos²A- sin²A = 1- 2 sin²A= 2 cos²A -1= (1- tan²A/(1+ tan²A).
3) 1+ cos2A= 2 cos²A , 1- cos2A= 2 sin²A
4) tan2A= 2tanA/(1- tan²A), where A≠ (2n +1)π/4
5) sin3A = 3SinA - 4sin²A
= 4 sin(60°-A)sin(60° +A)
6) cos3A= 4cos³A - 3 cosA
= 4 cos(60°-A) cosA cos(60°+A)
7) tan3A= (3tanA - tan³A)/((1- 3tan²A)
= tan(60° -A) tanA tan(60° +A)
PRODUCT
1) 2 sinA cosB = sin(A + B) + sin(A - B)
2) 2 coA sinB = sin(A + B) - sin(A - B)
3) 2 cosA cosB = cos(A + B)+ cos(A - B)
4) 2 sinA sinB = cos(A - B) - cos(A + B)
SUM
1) SinC + sinD= 2 sin{(C+ D)/2} cos{(C - D)/2}
2) SinC - sinD= 2 cos{(C+ D)/2} sin{(C - D)/2}
3) cosC + cosD= 2 cos{(C+ D)/2} cos{(C - D)/2}
4) cosC - cosD= - 2 sin{(C+ D)/2} sin{(C - D)/2}
5) tanA+ tanB= sin(A+ B)/(cosA cosB)
6) tanA- tanB= sin(A- B)/(cosA cosB) where A, B ≠ np + p/2
MAXIMUM AND MINIMUM VALUE OF
f(0)= a cosθ + b sinθ, θ ∈ R
Let a= r sinα, b = r cosα so that r= √(a²+ b²), r ∈R
Also, a cosθ + b sinθ = r(cosθ sinα+ sinθ cosα)= r sin(θ +α)
Now the maximum and minimum values of sin(θ +α) are 1 and -1 respectively
Hence - r≤ r sin(θ +α)≤ r
=> - √(a²+ b²) ≤ a cosθ + b sinθ ≤ √(a²+ b²)
Hence the maximum value is √(a²+ b²) and minimum value is - √(a²+ b²)
Conditional Identities
When the angels A, B, C satisfy a given relation, many interesting identities can be established connecting the trigonometric functions of these angles, In providing these identities, we require the properties of complementary and supplementary angles. For example, if A+ B+ C=π, then
• sin(B+ C)= sinA, cosB = - cos(C+ A).
• Cos(A+ B)= - cosC, sinC= sin(A+ B)
• tan(C+ A)= - tanB, cotA = - cot(B+ C)
• cos{(A+B)/2= sin(C/2), cos(C/2)= = sin{A+B)/2}
• Sin{(C+ A)/2}= cos(B/2), sun(A/2) = cos{(B+C)/2}
• tan{B+ C)/2}= cot A, tan(B/2)= cot{(C+A)/2}
Some Important Identities
If A, B, C are angles of a triangle (or A+ B+ C)=π):
• tanA + tanB + tanC= tanA tanB tanC
• cotA + cotB + cotC= tanA tanB tanC
• tan(A/2)tan(B/2) + tan(B/2)tan(C/2) + tan(C/2) tan(A/2) = 1
• cot(A/2) + cot(B/2) + cot(C/2) = cot(C/2) cor(B/2) cot(C/2).
• sin2A + sin2B + sin2C= 4 sinA sinB sinC.
• cos2A cos2B + cos2C= -1 - 4 cosA cosB cosC.
• sinA + sinB + sinC= 4 cos(A/2) cos(B/2) cos(C/2).
• cosA + cosB + cosC= 1+ 4 sin(A/2) sin(B/2) sin(C/2).
Formula for General Solutions
• tan²θ = tan²α = θ = nπ ± α
Where α ∈ [i.π/2), n ∈ I
• sinθ = 0<=> θ= nπ, n ∈ I
• cosθ = 0 <=> θ= (2n +1)ⁿπ/2, n ∈ I
• tanθ = 0 <=> θ= nπ, n ∈ I
• sinθ = 1 <=> θ= (4n +1)π/2, n ∈ I
• sinθ = -1 <=> θ= (4n -1)π/2, n ∈ I
• cosθ = 1 <=> θ= 2nπ/2, n ∈ I
• cosθ = -1 <=> θ= (2n +1)π, n ∈ I
• sinθ = sin α => θ= nπ +(-1)ⁿα, α ∈ [-π/2, π/2]
• cosθ = cos α => θ= 2nπ ± α, α ∈ [0, π] n ∈ I
SALIENT POINTS
* The general solution should be given unless the solution is required in a specified interval or range.
* While solving a trigonometric equation the equation at any steps should be avoided as far as possible. If squaring is necessary, check the solution for extraneous values.
* Never cancel terms containing unknown quantities on the two sides, which are in product. It may cause loss of genuine solution.
* The answer should not contain such values of angles, which make any of the terms undefined or infinite.
* Domain should not be changed. if it changed, necessary corrections must be incorporated.
* Check, that the denominator is not zero at any stage while solving equations.
* While solving trigonometric equations you may get same set of solution repeated in your answer. It is necessary for you to exclude these repetitions. eg., nπ+ π/2, forms a part of kπ/5 + π/10, k ∈ I the second part of second set of solution (you can check by putting k= 5m +2 (m∈ I). Hence the final answer is kπ/5 + π/10 , k ∈ I.
* Sometimes the two solutions set consist partly of common values. In all such cases the common part must be represented only once.
Definition
If f: A--> B is one to one and onto function and g is a rule under which for every element y ∈ B there exists and unique element x ∈ A then g : B --> A is called inverse function of f : A--> A, i.e., g= f⁻¹
So x= g(y) => x = f⁻¹(y)
So y= f(x) and x = g(y) such that then f(g(y))= y and x= g(f(x)) then f and g are said to be inverse function of each other.
* f⁻¹(x) ≠ [f(x)]⁻¹
* are sinx = sin⁻¹x
* arc sinx = 2nπ + arc sinx, n ∈ I
* Only one one onto function has an inverse function.
S.n Function Domain Range
1. sinx x ∈ R y ∈ [-1,1]
2. cosx x ∈ R y ∈ [-1,1]
3. tanx x ∈ R-(2n+1)π/2; n∈I y ∈ R
4. Cotx x ∈ R- nπ; n∈I y ∈ R
5. cosecx x ∈ R- nπ; n∈I y ∈ (-∞,-1]U[1,∞)
6. secx x ∈ R- (2n+1)π/2; n∈I y ∈ (-∞,-1]U[1,∞)
Domain, Range and Graph of Inverse Trigonometric functions
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[i.π/2), n ∈ I
= θ = nπ ± α
∈
tanA tanB tanC
+α)α)
θ
3- D vectors, plane etc
• Vector have magnitude and direction denoted by bar AB whereas scalars have only magnitude. The magnitude of vector is the length of the line segment AB denoted by |AB|
Types of Vectors
• Zero vector: Initial point and terminal point coincide as bar AA
• Unit vector: magnitude is unity as cap a
• Coinitial vectors: Vectors having same initial point as bar OA, OC, OD
• Collinear vectors: Vectors which are parallel to the same vector. .--<-.->--> .
• Equal vectors: vectors having some magnitude and same direction bar a= bar b.
• Nagative of a vector: vector having some magnitude but opposite direction. bar BA= - bar AB
• Free vectors: vectors whose initial point is not specified.
Laws of vectors
a) Addition of vectors
i) Triangle law of Addition: AC= AB+ BC
ii) Parallelogram law of Addition: OP+ OQ= OR
Properties of Addition of vectors
i) a+ b = b+ a (cumulative)
ii) a+ (b + c)= (a+ b)+ c (Associative)
iii) a+ 0= a. (Additive identity)
iv) a+ (-a)= 0 (additive inverse)
Multiplication of a vector by a scalar
• Let a be a vector and m a scalar, then ma is multiplication of vector a by scalar m.
|ma|= |m| a
Properties Multiplication of a vector by a scalar
i) m(-a)= -(ma)
ii) -m)(-a)= ma
iii) m(na)= (nm)a
iv) (m+ n)a= ma+ na
v) m(a+ b)= ma+ nb
• Two vectors a and b are colllinear or parallel iff a= mb for some nonzero scalar m.
• position vector: Position vector of a point P(x,y,z) is given as bar OP= xi + yj + zk aits magnitude as |OP|= √(x²+ y²+ z²), where O is the origin.
Components of a vector in Two dimension
If a point P in a plane has coordinates (x,y) then
i) OP= xi + yj
ii) |OP|= √(x²+ y²)
iii) The component of OP along x-axis is a vector xi, whose magnitude is |x| and whose direction is along OX or OX' according as x is positive or negative.
• Intercept form of the equation of a plane
x/a + y/b + z/c= 1
• Equation of the plane passing through the intersection of two given planes
i) Vector form: r: (a₁ + λa₂) = d₁ + λd₂
ii) Cartesian form: (a₁x+ b₁y + c₁z - d₁) + λ(a₁x + b₁y + c₁z - d₂)= 0
• Coplanarity of two vectors
i) Vector form: (a₂ - a₁). (b₁ x b₂)= 0
ii) Cartesian form:
|x₂ - x₁ y₂ - y₁ z₂ - z₁
a₁ b₁ c₁ =0
a₂ b₂ c₂
• Angle between two planes
i) Vector form: cosθ= |(a₁. a₂)/+|n₁| |n₂|)|
ii) Cartesian form:
cosθ=|(a₁a₂ + b₁b₂ + c₁c₂)/√(a₁²+ b₁²+ c₁²)√(a₂² + b₂²+ c₂²)|
• Distance of a plane from a plane
i) Vector form: |(a.N - d)|/|N| , where N is normal to the plane.
ii) Cartesian form: |(ax₁+ by₁ + cz₁- d)/√(a²+ b² + c²)|
• Angle between a line and a plane
i) Vector form:
φ = sin⁻¹|(b.n)/(|b| |n||
SAP- 1
Very short Type (1)
1) Find the equation of lines parallel to y-axis and passing through origin. x/0= y/1= z/0
2) Show that the vector 3i + 5j + 2k, 2i - 3j - 5k and 5i+ 2j - 3k form the sides of an equilateral triangle.
3) Find the area of the parallelogram determined by the vector i + 2j + 3k and 3i - 2j + k. 8√3 square. units
4) Direction ratioa of a line are 1,-2,3, find the direction cosines. 1.√14,-2/√14,3/√14
5) Write the vector equation of (x+5)/3 = (y -4)/2= (6- z)/2. r= -5i+ 4j + 6k + λ(3i + 2j - 3k)
Short Answer type (4)
6) Find the equation of plane containing the lines of intersection of the planes x+ y + z -6=0 and 2x + 3y + 4z +5=0 and passing through (1,1,1). 20x + 23y + 26z - 69=0
7) In ∆ OAB, E is the midpoint of OB and D is a point on AB such that AD: DB= 2:1. If OD AE intersect at P, determine the ratio OP: PD using vector methods. 3:2
8) Given three points whose position vectors are xi + yj + zk, i+ zj and - i - j. Find the condition for the point to be Collinear. x - 2y = 1
9) If a,b,c are three non zero vectors such that a x b = c are b x c = a, show that a,b,c are mutually at right angles and |b|= 1 and |c|= |a|.
10) ABCD is a quadrilateral such that AB= b, AD= d, AC = mb + pd. Show that the area of the quadrilateral ABCD is (1/2) |m+ p| |b x d|.
11) Find the direction cosines of the two lines which are connected by the relations l - 5m + 3n=0 and 7l²+ 5m²- 3n²=0. ±1.√14,-±2/√14,±3/√14 and ±1/√6, ±1/√6, ±2/√6
12) Show that the lines (x -1)/3= (y +1)/2 = (z -1)/5 and (x +2)/4 = (y -1)/3 = (z +1)/-2 do not intersect.
13) Show that the line x= ay + b, z= cy+ d and x= a'y + B', z= c'y + d' are perpendicular if aa' + cc' + 1=0
14) Find the angle between the lines
x - 2y + z= 0= x + 2y - 2z and x + 2y + z= 0= 3x + 9y + 5z. cos⁻¹(8/√406)
15) Find the equation of the plane through the points (1,0,-1), (3,2,2) and parallel to the line (x -1)/1 = (y +1)/-2 = (z -2)/3. 4x - y - 2z -6=0
Long Answer type (6)
16) Find the equation of the plane passing through the point (-1,2,1) and perpendicular to the line joining the points (-3,1,2) and (2,3,4). Find also the perpendicular distance of the origin from this plane. r.(5i+ 2j + 2k)= 1, 1/√33
17) Find the shortest distance between the lines whose vector equation are
r= (i + 2j + 3k) + λ(2i + 3j + 4k) and r= (2i + 4j + 5k) + μ(4i + 6j + 8k).. √5/√29
18) Find the distance of the point P(i + j + k) from the plane through the point A(2i + j + k), B(i + 2j + k) and C(i + j + 2k). Also, find the position vector of the foot of perpendicular from P on this plane. 1/√3, (4/3) (i + j + k)
19) Show (x +1)/3= (y +3)/5 = (z +5)/7 and (x -2)/1 = (y -4)/4 = (z -6)/7 are coplanar. Also, find the plane containing these two lines.
20) Find the image of the point (3,1,2) in the plane 2x - y + z= 4. (1,2,1)
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