Math Assignment Class VIII | Square & Square Root

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Extra questions of chapter 3 class 11 Trigonometric Functions with answer and hints to the difficult questions. Important and useful math. assignment for the students of class 11
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ASSIGNMENT FOR XI STANDARD TRIGONOMETRY
Find the degree measure for the following radian measure
Question 2
Find the radian measure for the following degree measure
Question 3
Find the magnitude, radian and degree, of the interior angles of a regular
Solution Hint:
Note: Each interior angle of a regular polygon is given by :
Where n is the number of sides.
Question 4
If sin θ =12/13 and θ lie in the second quadrant, then find the value of sec θ + tan θ. Ans. [- 5]
Question 5
Prove the followings
i) cos24o + cos 55o + cos 125o + cos 204o + cos 300o = 1/2
ii) sin 600o tan(-690o) + sec 840o cot(-945o) = 3/2
iii)
Question 6: Simplify
Ans: 1
Solution Hint:
Question 7
Evaluate the following :
Question 8
Prove that : tan70o = tan20o + 2tan50o
Solution Hint:
Now cross-multiply and simplify the above fraction we get the required result.
Question 9
Question 10
If tan A = k tan B, then show that:
Solution Hint: tan A = k tan B
Now apply componendo and dividendo and using trigonometric formulas.
Question 11
If tan(A + B) = p, tan(A – B) = q, then show that :
Solution Hint:
Start from RHS and then putting the value of p and q then simplify and get the result.
Question 12
An angle α is divided into two parts such that the ratio of the tangents of the two parts = k and difference of two parts = x then show that:
Solution hint:
Let two parts of α are p and q, Then
ATQ : p + q = α, p – q = x,
Now applying componendo and dividendo and simplify we get the required result.
Question 13
Prove that : cos20o cos40o cos60o cos80o = 1/16
Question 14
Prove that : sin20o sin40o sin60o sin80o = 3/16
Question 15
Prove That :
Solution Hint :
Use tan θ = sin θ/cos θ, then using AB and CD formulas
Question 16
If cos(θ + 2 α) = n cos θ , then prove that :
Solution Hint:
Applying componendo and dividendo then applying CD formulas then simplify the fractions we get the required result.
Question 17
Prove that:
Solution Hint
Taking LHS and convert these into cosine functions.
Multiply and divide by 2 and the apply AB formulas
Question 18
Prove that:
Solution Hint:
Multiply and divide by 2 and then apply AB formulas
Question 19
Find the value of sin18o
Ans:
Solution:
θ = 18o ⇒ 5 θ = 90o ⇒ 2 θ + 3 θ = 90o ⇒ 2 θ = 90o - 3 θ
Now taking sin on both side we get
Sin (2 θ) = Sin (90o - 3 θ) Sin (2 θ) = Cos (3 θ)
2Sin θ cos θ = 4Cos3 θ – 3cos θ 2Sin θ = 4Cos2 θ – 3
2Sin θ = 4(1-sin2 θ) – 3
4 sin2 θ + 2sin θ – 1 = 0
Now using quadratic formula here and find the value of sin θ
Question 20
Prove that :
Solution Hint:
Question 21
i) Evaluate:
Solution Hint: Using:
ii) Evaluate:
Question 22
Prove that:
Solution Hint:
Multiply numerator and denominator by 2
Applying AB formulas we get
Now applying CD formulas we get the required result.
Question 23
Find ,
and
, when tan x =
, and x lie in II quadrant
Ans:
,
Question 25
If tan35o = α , then find the value of in terms of α
Ans:
Question 26
Prove that :
Question: 27
Prove that :
Solution Hint:
Use formula:
Now simplify and then apply CD formula we get the required result
Question 28
Prove that:
Question 29
If tan x + tan y + tan x tan y = 1, find (x + y).
Ans: x + y = 45o
Solution Hint:
tan x + tan y = 1-tan x tan y
Dividing on both side by 1-tanx tany we get
tan(x + y) = tan 45o ⇒ x + y = 45o
Question: 30
Prove that: Cos6θ = 32cos6 θ - 48cos4 θ + 18cos2 θ - 1
Solution Hint: Use cos6θ = cos3(2θ) = - 3cos2θ + 4cos3 2θ and then proceed.
Question: 31
Prove that: cos6x = 1 - 18sin2x + 48sin4x - 32sin6x
Solution Hint: Use cos6θ = cos3(2θ) = -3cos2θ + 4cos3 2θ .
Now putting cos2θ = 1 - 2sin2θ we get
cos6θ = -3(1 - 2sin2θ) + 4(1 - 2sin2θ)3
Solving this we get the required resultQuestion 32
Prove that:
Solution Hint
Using tanθ = sinθ/cosθ in numerator and in denominator
Taking LCM then using the formula sin(A+B) and sin (A-B) in numerator and in denominator respectively.
Now using sin2θ = 2sinθcosθ in the numerator for two times.
Question: 33
Prove that: sin3x cos3x + cos3xsin3x = sin4x
Solution Hint: Expand sin3x and cos3x
Now multiply and divide by 2
Applying 2sinxcosx = sin2x for two times we get the required result.
Question 34
Prove that :
Solution :
Question 35
Evaluate: and
Solution:
Prove that: sin2α
+ sin2(α - β) - 2sinα cosβ sin(α - β) = sin2β
Solution Hint:
LHS = sin2α + sin2(α - β) - [sin(α + β) + sin(α - β)]sin(α - β)
= sin2α + sin2(α - β) - sin(α + β)sin(α - β) - sin2(α - β)
= sin2α - [sin2α - sin2β] ..... [Using sin(A+B)sin(A-B) = sin2A - sin2B ]
= sin2β
Question 37
If , then find the value of xy + yz + zx
Answer 0
Solution Hint:
Let all equations = k
Now find the value of x, y, z in terms of k
Now find the value of , it should be '0'
Putting this value in xy + yz + zx =
=
= xyz ×0 = 0
Question 38
Find the value of :
Ans:
Solution Hint:
Taking "-" common from numerator
Using the formula:
Find the general solution of the equation: cos x + cos 2x + cos 3x = 0.
Solve:
Question 41
Solve the equation: sin 3x + sin 5x + sin 7x = 0,
Question 42
Solve:
Question 43
Solve:
Solution Hint:
Question 44
Solve:
Solution Hint:
Dividing on both side by
i.e. by 2
Now proceed this question as previous one.
Find the general solution of: 2 cos2x + 3 sin x = 0.
Question 46
Find general solution of cos2x cosec x + 3sinx + 3 = 0
Ans: ,
Solution Hint:
Convert all terms into sin x , we get a quadratic equation.
Solve the quadratic equations.
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