Challenging O Level Trigonometry Question (A Maths)

A reader of our Mathtuition88.com blog asked the following Maths question:

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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Author: mathtuition88

http://mathtuition88.com

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