Trigonometry

Solve the equation ( 2\cos^2 x + \sin x = 1 ) for ( 0 \le x < 2\pi ). Give your answers in terms of ( \pi ).",
"marks": 5,
"solution": "Step 1: The equation involves both cos and sin. We must use an identity to express it in terms of a single function. Using the identity ( \cos^2 x = 1 - \sin^2 x ):
( 2(1 - \sin^2 x) + \sin x = 1 )

Step 2: Expand and rearrange into a quadratic equation in ( \sin x ):
( 2 - 2\sin^2 x + \sin x = 1 )
( 0 = 2\sin^2 x - \sin x - 1 )

Step 3: Solve the quadratic. Let ( y = \sin x ). ( 2y^2 - y - 1 = 0 ). This factorises to ( (2y+1)(y-1) = 0 ).
So, ( y = -1/2 ) or ( y = 1 ).
This means ( \sin x = -1/2 ) or ( \sin x = 1 ).

Step 4: Find the principal values and all solutions in the range ( 0 \le x < 2\pi ).
For ( \sin x = 1 ), the principal value is ( x = \arcsin(1) = \pi/2 ). This is the only solution in the range.
For ( \sin x = -1/2 ), the principal value is ( x = \arcsin(-1/2) = -\pi/6 ). This is outside the range. We use the symmetry of the sine graph (or the unit circle) to find solutions in the range. The solutions are in the 3rd and 4th quadrants. ( x = \pi + \pi/6 = 7\pi/6 ) and ( x = 2\pi - \pi/6 = 11\pi/6 ).

Final answer: ( x = \pi/2, 7\pi/6, 11\pi/6 )",
"examiner_commentary": "This is a standard question style that earns full marks for a systematic approach. M1 is awarded for correctly substituting the Pythagorean identity. A1 for forming a correct quadratic equation. M1 for a valid attempt to solve the quadratic. A1 for finding at least two correct solutions. A1 for all three correct solutions and no extras. A common error is failing to find all secondary solutions for ( \sin x = -1/2 ). Sketching the sine wave is a good way to ensure all solutions are found."

Prove the identity ( \frac{\cos 2x}{\cos x - \sin x} \equiv \cos x + \sin x ).",
"marks": 3,
"solution": "Step 1: Start with the more complex side, the Left Hand Side (LHS).
LHS = ( \frac{\cos 2x}{\cos x - \sin x} )

Step 2: Choose an appropriate double angle formula for ( \cos 2x ). Since the denominator involves ( \cos x ) and ( \sin x ), the form ( \cos^2 x - \sin^2 x ) is the most promising.
LHS = ( \frac{\cos^2 x - \sin^2 x}{\cos x - \sin x} )

Step 3: The numerator is a difference of two squares, ( a^2 - b^2 = (a-b)(a+b) ).
LHS = ( \frac{(\cos x - \sin x)(\cos x + \sin x)}{\cos x - \sin x} )

Step 4: Cancel the common factor ( (\cos x - \sin x) ).
LHS = ( \cos x + \sin x ) = RHS.

Final answer: The identity is proven.",
"examiner_commentary": "Full credit is given for a clear, logical proof that starts on one side and manipulates it to match the other. M1 is awarded for selecting and using the correct ( \cos 2x ) identity. M1 for recognising and factorising the difference of two squares. A1 for the final correct result. Candidates who try to multiply both sides by the denominator will score no marks, as this assumes the identity is true from the start."

a) Express ( 5\cos heta - 12\sin heta ) in the form ( R\cos( heta + \alpha) ), where ( R > 0 ) and ( 0 < \alpha < \pi/2 ). [3 marks]
b) Hence, find the minimum value of the function ( f( heta) = 20 + 5\cos heta - 12\sin heta ) and the smallest positive value of ( heta ) at which it occurs. [3 marks]",
"marks": 6,
"solution": "a) Step 1: Find R. ( R = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 ).
Step 2: Expand the target form: ( R\cos( heta + \alpha) = R(\cos heta\cos\alpha - \sin heta\sin\alpha) ).
Step 3: Compare coefficients with ( 5\cos heta - 12\sin heta ):
( R\cos\alpha = 5 ) and ( R\sin\alpha = 12 ). So ( 13\cos\alpha = 5 ) and ( 13\sin\alpha = 12 ).
( an\alpha = (R\sin\alpha) / (R\cos\alpha) = 12/5 ).
( \alpha = \arctan(12/5) \approx 1.176 ) radians.
Final answer (a): ( 13\cos( heta + 1.176) )

b) Step 1: Substitute the result from part (a) into the function ( f( heta) ).
( f( heta) = 20 + 13\cos( heta + 1.176) ).
Step 2: The minimum value of the cosine function is -1. Therefore, the minimum value of ( f( heta) ) is ( 20 + 13(-1) = 7 ).
Step 3: This minimum occurs when ( \cos( heta + 1.176) = -1 ).
The principal value for ( \cos(X) = -1 ) is ( X = \pi ).
So, ( heta + 1.176 = \pi ).
( heta = \pi - 1.176 \approx 3.142 - 1.176 = 1.966 ) radians.
Final answer (b): Minimum value is 7, occurring at ( heta \approx 1.97 ).",
"examiner_commentary": "For part (a), M1 for finding R, M1 for finding ( an \alpha ), A1 for the correct final expression. For part (b), M1 is awarded for using the part (a) result to find the minimum value. A1 for the correct value of 7. M1 for setting ( \cos( heta + \alpha) = -1 ) and solving for ( heta ). A common mistake is to state the minimum value is -13, forgetting the '20+' part of the function."