Test: Basic Trigonometric Formula


15 Questions MCQ Test Mathematics (Maths) Class 11 | Test: Basic Trigonometric Formula


Description
This mock test of Test: Basic Trigonometric Formula for JEE helps you for every JEE entrance exam. This contains 15 Multiple Choice Questions for JEE Test: Basic Trigonometric Formula (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Basic Trigonometric Formula quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Basic Trigonometric Formula exercise for a better result in the exam. You can find other Test: Basic Trigonometric Formula extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

cos 68° cos 8° + sin 68° sin 8° = ?

Solution:

Cos68°cos8°+sin68°sin8°

We know, 

cosAcosB+sinAsinB=cos(A-B)

Cos(68-8)

Cos60°

=1/2

QUESTION: 2

cos130° cos40°+ sin130° sin40° is

Solution:

cos(a-b) = Cos a Cos b + Sin a Sin b

Cos(130-40)=cos90=0    

QUESTION: 3

Which of the following is correct:

Solution:

QUESTION: 4

The value of Cos75°is equal to

Solution:

    

QUESTION: 5

Sin A = 1/√10 , Sin B = 1/√5 If A and B are both acute angles, then, A+B =?

Solution:

If sin A = 1 / ( 10 ) ( 1 / 2 )and sin B = 1 / ( 5 ) ( 1 / 2 ) then cos A = ( 1 - sin A ) ( 1 / 2 )

= cos A = 3 / ( 10 ) ( 1 / 2 ) ,similarly ,cos B = ( 1 - sinB )(1 / 2 )

= cos B = 2 / ( 5 ) ( 1 / 2 )

= sin ( A + B ) =  sin ( pi / 4 )

= sin A cos B  +  sin B cos A  = sin ( pi / 4 ) 

QUESTION: 6

The value of sin 120° cos 60° + cos 120° sin 60° is equal to

Solution:

Sin A Cos B + Cos A SinB =Sin( A + B) 

 = Sin(120+60)  = Sin (180) = 0

                   

                    

QUESTION: 7

The values of sin105° is equal to

Solution:

sin (105º) = sin (90º + 15º) = cos 150

QUESTION: 8

Chose which of the following expressions equals sinA + cosA.

Solution:

SinA+cosA=√2(sinA/√2+cosA/√2)
=√2(sinAcosπ/4+cosAsinπ/4)
=√2sin(A+π/4)
by using sin(A+B) formula

QUESTION: 9

The value of cos45°cos15° + sin 45° sin15° is

Solution:



QUESTION: 10

sin(60° + A) cos(30° – B) + cos(60° + A) sin(30° – B) is equal to

Solution:

Sin(60+a)cos(30-b)+ cos(60+a)sin(30-b) =sin(60+a+30-b) sin(a)cos(b) + cos(a)sin(b)= sin(a+b) =sin(90+《a-b》) =cos(a-b)

QUESTION: 11

sin(45° – A) cos(45° – B) +  cos(45° – A) sin(45° – B)

Solution:

It is equal to SinACosB + CosASinB = cos(A+B)

                  hence, = Sin(45-A+45-B)

                             = Sin(90-(A+B))

                             = Cos(A+B)

QUESTION: 12

Sin(n+1)A sin(n+2)A + cos(n+1)A cos(n+2)A =

Solution:

Sin(n+1)ASin(n+2)A + Cos(n+1)ACos(n+2)A =Cos (n+1)ACos(n+2)A + Sin(n+1)ASin(n+2)A =Cos{A(n+2-n-1)} =cos (A.1) =cos A

QUESTION: 13

cos(π/4 - x) cos (π/4 - y) -sin(π/4 - x) sin(π/4 - y) =

Solution:

Cos(π/4-x)cos (π/4-y) - sin (π/4-x) sin(π/4-y)

= CosA*Cos B - Sin A*Sin B =

                                      Cos (A+B)

=cos(π/4-x+π/4-y)

=cos(π/2-x-y)

=cos{π/2 - (x+y)}

= sin(x+y)

QUESTION: 14

cos130° cos40° + sin130° sin40° is equal to

Solution:

cos 130cos 40+sin 130sin 40 = cos (130-40) = cos 90 =0

 

QUESTION: 15

Which of the following is correct:

Solution:

Related tests