By first obtaining a quadratic equation in cos θ, solve the equation tan θ sin θ = 1 for 0° < θ < 360°.
- (a)By first obtaining a quadratic equation in cos θ, solve the equation tan θ sin θ = 1 for 0° < θ < 360°.[5]
- (b)Show that (tan θ / sin θ) - (sin θ / tan θ) = tan θ sin θ.[3]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2023). The mark or mark type is shown in brackets.
- [M1] tan θ sin θ = 1 leading to sin²θ = cos θ Use of tan θ = sin θ/cos θ and multiplication by cos θ.
- [M1] 1-cos²θ = cos θ or cos²θ + cos θ - 1 [=0] Use of trig identity to form a 3-term quadratic.
- [M1] [cos θ =] (-1±√5)/2 Use of formula or completion of the square must be seen on a 3-term quadratic. Expect 0.6180 .
- [A1] 51.8°, Both A marks dependent on the 2nd M1.
- [A1 FT] 308.2° FT for (360° – 1st soln), A0 if extra solutions in range. Radians 0.905 and 5.38, A1 only for both.
- [M1] (tan θ / sin θ) - (sin θ / tan θ) = (sin θ / (cos θ sin θ)) - (sin θ cos θ / sin θ) Use tan θ = sin θ/cos θ twice with correct use of fractions.
- [M1] 1/cos θ - cos θ = (1-cos²θ) / cos θ = sin²θ / cos θ Use 1-cos²θ = sin²θ with correct use of fractions.
- [A1] = tan θ sin θ WWW
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