Question

sinA+2sinA=1 then cosA+2cosA=?​

Answer 1

Answer:

From the definition of sine:

−1≤sinx≤1 for all x∈R

Our starting condition sinA+sinB=2 therefore implies that sinA=sinB=1 .

This happens when A and B are π2 (plus or minus a multiple of 2π ):

A=π2+2kπ

B=π2+2hπ

where k,h∈Z.

Thus cosA+cosB=cos(π2+2kπ)+cos(π2+2hπ)=0+0=0.

or

CosA+SinA=√2CosA

SinA=√2CosA-CosA take it eq.1

SO, CosA-SinA

Putting the value of eq. 1

CosA-√2CosA+CosA

2CosA-√2CosA

√2(√2CosA-CosA)

Putting the value of eq.1

√2SinA

Answer 2

Step-by-step explanation:

Hello!!!!!!! I can answer this question:

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 :-)