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Question: If \[\sin x+\sin y=a\] and \[\cos x+\cos y=b\] then find the values of \[x,y\]....

If sin⁡x+sin⁡y=a\sin x+\sin y=a and cos⁡x+cos⁡y=b\cos x+\cos y=b then find the values of x,yx,y.

Explanation

Solution

We can solve this question by first using the formulas of sum to product identities for both sine and cosine equations. Once we get the equations in products then we can divide both those equations. On dividing you can find the value of tan(x+y) and in this way you can solve the equations to find the value of x+y . And to find the value of x-y we need to square both the equations, add and simplify the values. From both those equations we can add and subtract them and get the value of x,y which we need.

Complete step by step answer:
To start solving this question let us start with what is given to us. Hence we can say that
sin⁡x+sin⁡y=a\sin x+\sin y=a
⇒cos⁡x+cos⁡y=b\Rightarrow \cos x+\cos y=b
Now simplify these equation using the formula of sum to product identity which gives us
2sin⁡(x+y2)cos⁡(x−y2)=a2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)=a
⇒2cos⁡(x+y2)cos⁡(x−y2)=b\Rightarrow 2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)=b
Dividing the two equations
2sin⁡(x+y2)cos⁡(x−y2)2cos⁡(x+y2)cos⁡(x−y2)=ab\dfrac{2\sin \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)}{2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)}=\dfrac{a}{b}
Simplifying
tan⁡(x+y2)=ab\tan \left( \dfrac{x+y}{2} \right)=\dfrac{a}{b}
Now we can find the inverse and get
x+y2=tan⁡−1(ab)\dfrac{x+y}{2}={{\tan }^{-1}}\left( \dfrac{a}{b} \right)
Now cross multiplying we get
x+y=2tan⁡−1(ab)x+y=2{{\tan }^{-1}}\left( \dfrac{a}{b} \right)

Now to find x,y we will next square the two equations given to us in the question and add them both giving
cos⁡2x+sin⁡2x+cos⁡2x+sin⁡2x+2cos⁡xcos⁡y+2sin⁡xsin⁡y=a2+b2{{\cos }^{2}}x+{{\sin }^{2}}x+{{\cos }^{2}}x+{{\sin }^{2}}x+2\cos x\cos y+2\sin x\sin y={{a}^{2}}+{{b}^{2}}
Simplifying
2+2cos⁡(x−y)=a2+b22+2\cos (x-y)={{a}^{2}}+{{b}^{2}}
From this
cos⁡(x−y)=a2+b2−22\cos (x-y)=\dfrac{{{a}^{2}}+{{b}^{2}}-2}{2}
Taking inverse
x−y=cos⁡−1(a2+b2−22)x-y={{\cos }^{-1}}\left( \dfrac{{{a}^{2}}+{{b}^{2}}-2}{2} \right)
Now we know the value of x+y and x-y. Adding and subtracting them we get the values of x and y which are
x=tan⁡−1ab+12cos⁡−1(a2+b2−22)x={{\tan }^{-1}}\dfrac{a}{b}+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{{{a}^{2}}+{{b}^{2}}-2}{2} \right)
∴y=tan⁡−1ab−12cos⁡−1(a2+b2−22)\therefore y={{\tan }^{-1}}\dfrac{a}{b}-\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{{{a}^{2}}+{{b}^{2}}-2}{2} \right)
Hence this is how we found the answer of both x and y.

Note: A problem that can be faced while trying to solve this question is not knowing how to start this question. If you see different variables in trigonometric questions in sum format, always try converting it in the form of products and singular function form so as to easily solve it. Trigonometric functions can be explained to be real functions which relate an angle of a right-angled triangle to ratios of two side lengths. The six trigonometric functions are sine,cosine,tangent,secant,cosecant and cotangent