Sina+cosa=root over cosA
To solve the equation sina + cosa = √(cosA), we need to use some trigonometric identities and algebraic manipulation.
Step 1: Recall the Pythagorean identity for sine and cosine:
sin^2(A) + cos^2(A) = 1
Step 2: Rearrange the equation sina + cosa = √(cosA) by squaring both sides:
(sinA + cosA)^2 = cosA
Step 3: Expand the left side of the equation using the binomial square formula:
sin^2(A) + 2sinAcosA + cos^2(A) = cosA
Step 4: Substitute sin^2(A) + cos^2(A) = 1 into the equation:
1 + 2sinAcosA = cosA
Step 5: Rearrange the equation to isolate sinAcosA on one side:
2sinAcosA - cosA + 1 = 0
Step 6: The equation is now in quadratic form. Treat sinAcosA as a quadratic term, and solve using factoring or the quadratic formula.
Let's use the quadratic formula:
sinAcosA = (cosA - 1) / 2
Step 7: Simplify the expression:
sinAcosA = cosA/2 - 1/2
Step 8: Apply trigonometric identity sin2A = 2sinAcosA:
sin2A = cosA - 1/2
Therefore, the equation sina + cosa = √(cosA) simplifies to sin2A = cosA - 1/2.