Question

2sina = 2-cosa find the value of sin a

Answer 1

Answer:

Step-by-step explanation:

So, 2 sinA = 2 - cosA

Squaring both sides:

(2 sinA)^2 = (2 - cosA)^2

4 (sinA)^2 = 4 + (cosA)^2 - 4 cosA

4 - 4 (cosA)^2 = 4 + (cosA)^2 - 4 cosA

4 - 4 (cosA)^2 - 4 - (cosA)^2 + 4 cosA = 0

4 cosA - 5 (cosA)^2 = 0

Case 1:

cosA = 0

Case 2:

4 - 5 cosA = 0

4 = 5cosA

cosA =4/5

As, (sinA)^2 + (cosA)^2 = 1

Case 1:

(sinA)^2 + 0^2 = 1

sinA = 1

Case 2:

(sinA)^2 +(4/5)^2 = 1

(sinA)^2 + (16/25) =1

(sinA)^2 = 1 - (16/25)

(sinA)^2 = 9/25

sinA = 3/5

Therefore, The values of sinA are 1 and 3/5 .

Hope you got it :-)