Trignometric Ratios, Functions and Equations Test - 5

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(90 º+ 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