Can we sketch the curve y = cos x − sin x + 2 y = \cos x - \sin x +2 y = cos x − sin x + 2
Question
Give a sketch for − π ≤ x ≤ π -\pi \leq x \leq \pi − π ≤ x ≤ π y = cos x − sin x + 2. y = \cos x - \sin x +2. y = cos x − sin x + 2.
Solution
Firstly we can rewrite the right hand side using one of the compound angle identities, such as R cos ( x + θ ) = R cos x cos θ − R sin x sin θ . R\cos(x+\theta)=R\cos x\cos \theta- R\sin x \sin \theta. R cos ( x + θ ) = R cos x cos θ − R sin x sin θ .
If we put R cos θ = 1 R\cos \theta = 1 R cos θ = 1 R sin θ = 1 R \sin \theta = 1 R sin θ = 1 R > 0 R > 0 R > 0 θ \theta θ tan θ = 1 ⟹ θ = π 4 \tan \theta = 1 \implies \theta = \dfrac{\pi}{4} tan θ = 1 ⟹ θ = 4 π
On squaring and adding these equations, we have R 2 = 2 ⟹ R = 2 . R^2 = 2 \implies R = \sqrt{2}. R 2 = 2 ⟹ R = 2 .
This gives us that 2 cos ( x + π 4 ) = cos x − sin x \sqrt{2}\cos\left(x+\dfrac{\pi}{4}\right) = \cos x - \sin x 2 cos ( x + 4 π ) = cos x − sin x y = 2 cos ( x + π 4 ) + 2. y = \sqrt{2}\cos\left(x+\dfrac{\pi}{4}\right) +2. y = 2 cos ( x + 4 π ) + 2.
So this is our standard cosine curve, stretched in the y y y 2 \sqrt{2} 2 π 4 \dfrac{\pi}{4} 4 π 2 2 2
