Prove the equation sinx+cosx=2 ,has no solution
Answers
Answer:
sinx+cosx=2—(1)sinx+cosx=2—(1)
Divide by 2–√2 on both LHSandRHS=>sinx∗12√+cosx∗12√=22√=2–√—(2)LHSandRHS=>sinx∗12+cosx∗12=22=2—(2)
Since sin45=cos45=12√sin45=cos45=12 , the equation (2)(2) can be written as sinxcos45+cosxsin45=2–√—(3)sinxcos45+cosxsin45=2—(3)
Since sin AcosB + cosAsinB = sin(A+B)sinAcosB + cosAsinB = sin(A+B) , equation (3)(3) can be written as sin(x+45)=2–√—(4)sin(x+45)=2—(4)
The known facts are:
The value for sinAsinA is always between −1−1 and 11
2–√=1.4142=1.414 , which is greater than 1
From the above facts, LHS of the equation (4) is always between -1 and 1 whereas RHS is > 1. This is not possible and hence sin + cos can never be equal to 2.
Answer:
Maximum value of sinx is 1 which occurs at π/2 + 2nπ and maximum value of cosx is 1 which occurs at 2nπ where n is any integer. ... There is no real solution to the equation since the maximum value which the given expression can attain is √2 at x=π/4.