If cos x + cos y = 1/3 and sin x + sin y = 1/4, Find the Values of cos(x+y), sin(x+y) and tan(x+y)

A standard trigonometric identities problem, not tied to a specific exam paper.

Problem

If $\cos x+\cos y=\dfrac13$ and $\sin x+\sin y=\dfrac14$, find the values of $\cos(x+y)$, $\sin(x+y)$ and $\tan(x+y)$.

Key insight. Turning both given sums into products via the sum-to-product formulas leaves a common factor of $\cos\left(\frac{x-y}2\right)$ in each — dividing one equation by the other cancels it and hands you $\tan\left(\frac{x+y}2\right)$ directly, from which every function of $x+y$ follows by the half-angle formulas.

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Approach

Rewrite both given sums using the sum-to-product identities, which each produce a factor of $\cos\left(\frac{x-y}2\right)$ alongside a factor of $\cos\left(\frac{x+y}2\right)$ or $\sin\left(\frac{x+y}2\right)$. Dividing the two equations eliminates the unknown $\cos\left(\frac{x-y}2\right)$ and isolates $\tan\left(\frac{x+y}2\right)$. From there, the tangent half-angle formulas convert this single value into $\cos(x+y)$, $\sin(x+y)$ and $\tan(x+y)$ directly.

Solution

Step 1 — Apply sum-to-product to both equations

$$\cos x+\cos y = 2\cos\left(\frac{x+y}2\right)\cos\left(\frac{x-y}2\right) = \frac13$$

$$\sin x+\sin y = 2\sin\left(\frac{x+y}2\right)\cos\left(\frac{x-y}2\right) = \frac14$$

Step 2 — Divide to eliminate the (x−y)/2 term

Dividing the second equation by the first cancels the common factor $2\cos\left(\frac{x-y}2\right)$, since it’s non-zero in both:

$$\frac{\sin\left(\frac{x+y}2\right)}{\cos\left(\frac{x+y}2\right)} = \frac{1/4}{1/3} = \frac34$$

$$\tan\left(\frac{x+y}2\right) = \frac34$$

This is the whole reason the two equations were set up as a sum of a cosine pair and a sum of a sine pair — dividing them removes $\cos\left(\frac{x-y}2\right)$ without ever needing to know its value.

Step 3 — Find cos(x+y) using the half-angle formula

With $t=\tan\left(\frac{x+y}2\right)=\dfrac34$, the tangent half-angle formula gives

$$\cos(x+y) = \frac{1-t^2}{1+t^2} = \frac{1-\frac9{16}}{1+\frac9{16}} = \frac{\frac{7}{16}}{\frac{25}{16}} = \frac{7}{25}$$

Step 4 — Find sin(x+y) using the half-angle formula

$$\sin(x+y) = \frac{2t}{1+t^2} = \frac{2\cdot\frac34}{\frac{25}{16}} = \frac{\frac32}{\frac{25}{16}} = \frac32\cdot\frac{16}{25} = \frac{24}{25}$$

Step 5 — Find tan(x+y)

$$\tan(x+y) = \frac{\sin(x+y)}{\cos(x+y)} = \frac{24/25}{7/25} = \frac{24}{7}$$

As a check, $\cos^2(x+y)+\sin^2(x+y) = \left(\frac{7}{25}\right)^2+\left(\frac{24}{25}\right)^2 = \frac{49+576}{625} = \frac{625}{625}=1$, confirming the pair $(7,24,25)$ satisfies the Pythagorean identity.

Answer

Common mistakes

• Adding or subtracting the two equations instead of dividing. Adding $\frac13$ and $\frac14$ doesn’t correspond to any useful trigonometric identity here — the division step is what makes the unknown $\cos\left(\frac{x-y}2\right)$ cancel out.
• Forgetting to square the half-angle tangent correctly. In $\dfrac{1-t^2}{1+t^2}$, a slip on $t^2=\left(\dfrac34\right)^2=\dfrac9{16}$ (rather than $\dfrac34$ itself) throws off both the numerator and denominator.
• Not verifying with the Pythagorean identity. Since $\cos(x+y)$ and $\sin(x+y)$ are computed from a common $t$, checking $\cos^2+\sin^2=1$ is a quick, free way to catch an arithmetic slip before reporting the answer.

Practise next

• If $\cos x-\cos y=\dfrac15$ and $\sin x-\sin y=\dfrac1{10}$, find $\cos(x+y)$ by the same division trick applied to the difference formulas.