WJEC Maths for A2 – Pure

1 Proof 14 2 Complete the following proof by contradiction to show that sin  θ + cos  θ ≤ √ 2 for all values of θ . Assume that there is a value of θ for which sin  θ + cos  θ > √ 2 Then squaring both sides, we have: … Answer 2 Assume that there is a value of θ for which sin  θ + cos  θ > √ 2 Then squaring both sides, we have: (sin  θ + cos  θ ) 2 > 2 sin 2 θ + 2 sin  θ  cos  θ + cos 2 θ > 2 Now sin 2 θ + cos 2 θ = 1 So 1 + 2 sin  θ  cos  θ > 2, giving 2 sin  θ  cos  θ > 1 Now 2 sin  θ  cos  θ = cos 2 θ So cos 2 θ > 1, which is a contradiction as the cosine of any angle is ≤ 1. 3 Prove by contradiction the following proposition: When x is real and x ≠ 0, | x + 1 x | ≥ 2 The ϐirst two lines of the proof are given below. Assume that there is a real value of x such that | x + 1 x | < 2 Then squaring both sides, we have: … Answer 3 Assume that there is a real value of x such that | x + 1 x | < 2 Then squaring both sides, we have: ( x + 1 x ) 2 < 4 x 2 + 2 + 1 x 2 < 4 Multiplying both sides by x 2 , we obtain x 4 + 2 x 2 + 1 < 4 x 2 x 4 − 2 x 2 + 1 < 0 ( x 2 − 1)( x 2 − 1) < 0 ( x 2 − 1) 2 < 0 , which is impossible since the square of a real number cannot be negative. As this is a contradiction, the assumption is incorrect, so | x + 1 x | ≥ 2. There is no doubt that proofs can be tricky. If you still feel a bit unsure about them why not try looking at YouTube videos on proof by contradiction. Try to choose ones that apply only to A-level work. You must be able to spot trig identities and use them in proofs such as this. Remember proofs can come up as part of any question. BOOST G ra d e ⇪⇪⇪⇪ The modulus of x is written as | x | and this means you take the numerical value of x ignoring the signs. Notice there is an x 2 . In equations such as this we can multiply both sides by x 2 to remove the fraction. Active Learning