# Challenging O Level Trigonometry Question (A Maths)

Given that $\sin x+\sin y=a$ and $\cos x+\cos y=a$, where $a\neq 0$, express $\sin x+\cos x$ in terms of $a$.

This is a rather challenging question, since there are many options to start. Which formula(s) should we use? Factor formula? R-formula? Give it a try first if you want to have a challenge.

Solution:

It turns out we can write:

$\sin y=a-\sin x$

#### $\cos y=a-\cos x$

Then, use $\sin^2 y+\cos ^2 y=1$

$(a-\sin x)^2+(a-\cos x)^2=1$

Expanding,

$a^2-2a\sin x+\sin^2 x+a^2-2a\cos x+\cos^2 x=1$

Rearranging,

$2a^2-2a(\sin x+\cos x)+1=1$

$2a(a-(\sin x+\cos x))=0$

Since $a\neq 0$, we have $a-(\sin x+\cos x)=0$.

Thus, $\boxed{\sin x+\cos x=a}$.

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