Review question

# Can we sketch the curve $y = \cos x - \sin x +2$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7581

## Solution

Give a sketch for $-\pi \leq x \leq \pi$ of the curve $y = \cos x - \sin x +2.$

Firstly we can rewrite the right hand side using one of the compound angle identities, such as $R\cos(x+\theta)=R\cos x\cos \theta- R\sin x \sin \theta.$

If we put $R\cos \theta = 1$ and $R \sin \theta = 1$, where $R > 0$ and $\theta$ is acute, we have (on dividing these equations) $\tan \theta = 1 \implies \theta = \dfrac{\pi}{4}$.

On squaring and adding these equations, we have $R^2 = 2 \implies R = \sqrt{2}.$

This gives us that $\sqrt{2}\cos\left(x+\dfrac{\pi}{4}\right) = \cos x - \sin x$, so the curve we are asked to sketch is $y = \sqrt{2}\cos\left(x+\dfrac{\pi}{4}\right) +2.$

So this is our standard cosine curve, stretched in the $y$-direction by a scale factor of $\sqrt{2}$, translated by $\dfrac{\pi}{4}$ to the left, and upwards by $2$.