Test: Sum & Difference Formula - JEE MCQ

sin 51° + cos 81°
= sin 51° + cos ( 90° − 9°)
= sin 51° + sin 9°

= 2sin 30° cos 21°

= cos(21°)

cosA - cos B = -2sin[(A+B)/2] sin [(A-B)/2]

So substituting, A = 9y and B = 5y, we get

cos9y - cos5y = -2sin7y sin2y

sinA - sinB = 2cos(A+B)/2 sin(A-B)/2
sin 35 – sin55 = 2cos(35+55)/2 sin(35-55)/2
= 2cos45 (-sin10)
= 2(√2/2) (-sin10)
= -√2 sin10

sin25° sin55°
Multiply and divide by '2'
1/2(2 sin25° sin55°)
cos(a-b) - cos(a+b) = 2sina sinb
1/2[cos(a-b) - cos(a+b)] = sina sinb
1/2[cos(25° - 55°) - cos(25° + 55°)] = sin25° sin55°
1/2[cos(-30°) - cos(80°)] = sin25° sin55°
1/2[cos(30°) - cos(80°)] = sin25° sin55°

Finding the value of sinA + sinB + sinC in a triangle.

In any triangle, the sum of all the interior angles is always 180°.

Therefore, In the triangle ABC, A + B + C = 180°

Now, solving for sinA + sinB + sinC:


Hence, Option (B) is correct.

CosA + Cos(120° - A) + Cos(120° + A)
 cosA + 2cos(120° - a + 120° + a)/(2cos(120° - a - 120° - a)
we know that formula
(cos C+ cosD = 2cos (C+D)/2.cos (C-D) /2)
⇒ cosA + 2cos120° cos(-A)
⇒ cosA+ 2cos (180° - 60°) cos(-A)
⇒ cosA + 2(-cos60°) cosA
⇒ cos A - 2 * 1/2cos A
⇒ cosA - cosA
⇒ 0

cos105° + cos75°
= cos(90º + 15) + cos(90° - 15) 
= - sin15 + sin15 = 0

The sum of the interior angles of any triangle is always 180°. Therefore, we can calculate angle C as:
C = 180° − (A + B)
Substitute the values of A and B:
C = 180° − (72° + 48°)
C = 60°
Thus, angle Ĉ (angle C) is 60°.

cos35^o cos45^o
Multiply and divide numerator and denominator by 2
1/2{2cos35^o cos45^o}
= 1/2{cos(35^o+45^o) + cos(35^o-45^o)}
= 1/2{cos80^o + cos(-10^o)}    { cos(-x) = cosx 
= {cos80^o + cos10^o}/2

We will use the cosine subtraction identity:

For cos(A) - cos(3A), substitute A and 3A into the identity:

Simplify:

cos(A) − cos(3A) = −2sin(2A) sin(−A)
Since sin(−A) = −sin(A), we get:
cos(A) − cos(3A) = −2sin(2A) (−sin(A))
cos(A) − cos(3A) = 2sin(2A) sin(A)

sin(A + B) - sin(A - B) = 2cosA sinB 
= 2cos(π/4)sinX
= 2 × 1/√2 sin X
= √2sinX

sin(n + 1)x cos(n + 2)x - cos(n + 1)x sin(n + 2)x
⇒ sin[(n + 1)x - (n + 2)x] 
As we know that sin(A - B) = sinA cosB - cosA sinB
⇒ sin(n + 1- n - 2)
sin (-x) 
= -sinx

cos 15° - sin 15°
= cos(45° - 30°) - sin(45° - 30°)
= (cos45°cos30° + sin45°sin30°) - (sin45°cos30° - cos45°sin30°)
= [√3/(2√2) + 1/(2√2)] - [√3/(2√2) - 1/(2√2)]
= 2/(2√2)
= 1/√2

sin36^º  = cos(90^º  - 36^º ) = cos54^º
So,cos18^º  + sin36^º = cos18^º  + cos54^º  = 2cos 36^º sin(-18^º ) = -2cos36^º sin18^º