Trigonometric functions of sum and difference

Six compound-angle formulae. For all real $x, y$ (with tangent/cotangent versions valid wherever the denominator is non-zero):

where: $x, y$ are the two angles being combined (in degrees or radians), and the tangent formulae require $1\mp\tan x\tan y \neq 0$. Notice the sign "swap" between the sine and cosine rules — $\sin(x+y)$ has a plus sign in the middle, but $\cos(x+y)$ has a minus — this is the single most common slip in the whole chapter.

Multiple-angle identities fall out immediately by setting $y=x$:

where: the three forms of $\cos 2x$ are algebraically identical (using $\sin^2x+\cos^2x=1$), and you pick whichever form matches the rest of a problem — all-$\sin$, all-$\cos$, or mixed. Applying $\sin(2x+x)$ and $\cos(2x+x)$ the same way gives $\sin 3x = 3\sin x - 4\sin^3x$ and $\cos 3x = 4\cos^3x-3\cos x$.

Sum-to-product identities come from a substitution trick: set $\alpha=x+y$, $\beta=x-y$, so $x=\tfrac{\alpha+\beta}{2}$ and $y=\tfrac{\alpha-\beta}{2}$, then feed these back into the compound-angle formulae:

where: $\alpha,\beta$ are any two angles whose sine or cosine is being added or subtracted; the trick of deriving these from the six basic formulae (rather than memorising them separately) is what keeps them straight under exam pressure.

Problem. Find the exact value of $\sin 75^\circ$, then use it to find $\cos 15^\circ$.

Step 1. Write $75^\circ$ as a sum of standard angles: $75^\circ = 45^\circ+30^\circ$.

Step 2. Apply $\sin(x+y)=\sin x\cos y+\cos x\sin y$ with $x=45^\circ$, $y=30^\circ$: $$\sin75^\circ = \sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ = \frac{1}{\sqrt2}\cdot\frac{\sqrt3}{2}+\frac{1}{\sqrt2}\cdot\frac12 = \frac{\sqrt3+1}{2\sqrt2}.$$

Step 3. Rationalise by multiplying by $\dfrac{\sqrt2}{\sqrt2}$: $$\sin75^\circ = \frac{\sqrt2(\sqrt3+1)}{4} = \frac{\sqrt6+\sqrt2}{4}.$$

Step 4. For $\cos15^\circ$, use the co-function identity $\cos15^\circ=\sin(90^\circ-15^\circ)=\sin75^\circ$, so $$\cos15^\circ = \frac{\sqrt6+\sqrt2}{4}.$$

Final answer: $\sin75^\circ = \dfrac{\sqrt6+\sqrt2}{4}$ and $\cos15^\circ = \dfrac{\sqrt6+\sqrt2}{4}$.

Competitive stretch

Prove $\cos\!\left(\dfrac{\pi}{4}+x\right)+\cos\!\left(\dfrac{\pi}{4}-x\right)=\sqrt2\cos x$. Expand both terms with the compound-angle cosine formula: $\cos\frac{\pi}{4}\cos x-\sin\frac{\pi}{4}\sin x$ and $\cos\frac{\pi}{4}\cos x+\sin\frac{\pi}{4}\sin x$; the $\sin\frac{\pi}{4}\sin x$ terms cancel on adding, leaving $2\cos\frac{\pi}{4}\cos x = 2\cdot\frac{1}{\sqrt2}\cos x = \sqrt2\cos x$, as required. The same result drops out instantly from the sum-to-product identity $\cos\alpha+\cos\beta=2\cos\tfrac{\alpha+\beta}{2}\cos\tfrac{\alpha-\beta}{2}$ with $\alpha=\tfrac{\pi}{4}+x$, $\beta=\tfrac{\pi}{4}-x$ — the two routes are the same identity used two ways.

• Mixing up the sign in the cosine formulae. $\cos(x+y)$ has a minus between the two products while $\sin(x+y)$ has a plus — the reverse of what intuition suggests.
• Forgetting the domain restriction on tangent addition. $\tan(x+y)$ is undefined whenever $1-\tan x\tan y=0$, independent of any restriction on $x$ or $y$ alone.
• Memorising sum-to-product formulae in isolation instead of deriving them from the compound-angle formulae via $\alpha=x+y,\beta=x-y$ — this makes it easy to mismatch which formula goes with $\sin$ vs $\cos$ and $+$ vs $-$.
• Picking the wrong form of $\cos2x$. All three forms ($\cos^2x-\sin^2x$, $1-2\sin^2x$, $2\cos^2x-1$) are equally correct; choosing the one matching the rest of the expression saves algebra.

Ask the tutor: want a walkthrough of deriving the compound-angle formulae from the unit circle, or more identity-proving practice with non-standard angles?