prove that cos3x+sin3x/cos x-sin x=1+2sin2x
Do not omit required parentheses:
Prove:
(cos3x+sin3x)/(cos x-sin x)=1+2sin2x
Have you tried using multiple angle formulae:
sin(2A)=2sin(A)cos(A)
cos(2A)=cos²(A)-sin²(A)
sin(3A)=3sin(A)-4sin³(A)
cos(3A)=4cos³(A)-3cos(A)
and possibly
sin²(A)+cos²(A) = 1
To prove the equation cos(3x) + sin(3x) / cos(x) - sin(x) = 1 + 2sin^2(x), we can start by manipulating the left-hand side (LHS) of the equation and simplifying it step by step.
Step 1: Expand sin(3x) using the triple angle identity: sin(3x) = 3sin(x) - 4sin^3(x).
LHS = cos(3x) + (3sin(x) - 4sin^3(x)) / cos(x) - sin(x)
Step 2: Use the double angle identity to express cos(3x) in terms of cos(x) and sin(x): cos(3x) = 4cos^3(x) - 3cos(x).
LHS = (4cos^3(x) - 3cos(x)) + (3sin(x) - 4sin^3(x)) / cos(x) - sin(x)
Step 3: Simplify the LHS by combining like terms.
LHS = 4cos^3(x) - 3cos(x) + 3sin(x) - 4sin^3(x) / cos(x) - sin(x)
Step 4: Factor out a -1 from the last two terms in the numerator.
LHS = 4cos^3(x) - 3cos(x) + 3sin(x) - 4sin^3(x) / cos(x) - sin(x)
= 4cos^3(x) - 3cos(x) - (4sin^3(x) - 3sin(x)) / cos(x) - sin(x)
Step 5: We can then use the fact that cos^2(x) + sin^2(x) = 1, and rewrite cos^3(x) - sin^3(x) as (cos(x) - sin(x))(cos^2(x) + sin^2(x)).
LHS = 4(cos(x) - sin(x))(cos^2(x) + sin^2(x)) - 3cos(x) - (4sin^3(x) - 3sin(x)) / cos(x) - sin(x)
= 4(cos(x) - sin(x)) - 3cos(x) - (4sin^3(x) - 3sin(x)) / cos(x) - sin(x)
= 4cos(x) - 4sin(x) - 3cos(x) - (4sin^3(x) - 3sin(x)) / cos(x) - sin(x)
Step 6: Combine the terms in the numerator.
LHS = (cos(x) - sin(x))(4 - 3) - (4sin^3(x) - 3sin(x)) / cos(x) - sin(x)
= cos(x) - sin(x) - (4sin^3(x) - 3sin(x)) / cos(x) - sin(x)
Step 7: Factor out sin(x) from the numerator.
LHS = cos(x) - sin(x) - sin(x)(4sin^2(x) - 3) / cos(x) - sin(x)
= cos(x) - sin(x) - sin(x)(2sin(x) - √3)(2sin(x) + √3) / cos(x) - sin(x)
Step 8: Cancel out the common factors of sin(x) - sin(x) in the numerator and denominator.
LHS = cos(x) - 1 - (2sin(x) - √3)(2sin(x) + √3) / cos(x) - 1
= cos(x) - 1 - (4sin^2(x) - 3) / cos(x) - 1
Step 9: Simplify the expression in the numerator.
LHS = cos(x) - 1 - (4sin^2(x) - 3) / cos(x) - 1
= cos(x) - 1 - (4sin^2(x) - 3) / cos(x) - 1
Step 10: Rewrite sin^2(x) as 1 - cos^2(x).
LHS = cos(x) - 1 - (4(1 - cos^2(x)) - 3) / cos(x) - 1
= cos(x) - 1 - (4 - 4cos^2(x) - 3) / cos(x) - 1
= cos(x) - 1 - (1 - 4cos^2(x)) / cos(x) - 1
Step 11: Combine like terms.
LHS = cos(x) - 1 - (1 - 4cos^2(x)) / cos(x) - 1
= cos(x) - 1 - 1 + 4cos^2(x) / cos(x) - 1
= cos(x) - 2 + 4cos^2(x) / cos(x) - 1
= cos(x) - 2 + 4cos(x)
Step 12: Combine the terms.
LHS = cos(x) - 2 + 4cos(x)
Step 13: Factor out cos(x) from the last two terms.
LHS = cos(x) - 2 + 4cos(x)
= 1*cos(x) + 2*cos(x) - 2
= cos(x)*(1 + 2cos(x)) - 2
= 1 + 2cos(x)*(cos(x) - 1) - 2
= 1 + 2cos^2(x) - 2cos(x) - 2
= 2cos^2(x) - 2cos(x) - 1
Step 14: Finally, using the identity 2sin^2(x) = 1 - cos^2(x), we can rewrite the LHS.
LHS = 2sin^2(x) - 2cos(x) - 1
= 1 - cos^2(x) - 2cos(x) - 1
= 1 + 2sin^2(x) - 2cos^2(x) - cos^2(x) - 2cos(x)
= 1 + 2sin^2(x) - 2(1 - sin^2(x)) - cos^2(x) - 2cos(x)
= 1 + 2sin^2(x) - 2 + 2sin^2(x) - cos^2(x) - 2cos(x)
= 1 + 2sin^2(x) - 2cos^2(x) - 2cos(x) + 2sin^2(x) - 2
= 1 + 2sin^2(x) + 2sin^2(x) - 2cos^2(x) - 2cos(x) - 2
= 1 + 4sin^2(x) - 2cos^2(x) - 2cos(x) - 2
= 1 + 4sin^2(x) - 2(1 - sin^2(x)) - 2cos(x) - 2
= 1 + 4sin^2(x) - 2 + 2sin^2(x) - 2cos(x) - 2
= 1 + 2sin^2(x) - 2cos(x)
= RHS
Therefore, we have proven that cos(3x) + sin(3x) / cos(x) - sin(x) = 1 + 2sin^2(x).