Question

If \[sinx+siny+sinz=3\]. Find the value of \[cosx+cosy+cosz\].

Answer 1

Hint: Considering the given equation , we know that we cannot find the value of three variables \[x\], \[y\] and \[z\] like we find in linear equation with three variables because for that we need three equations. So, we can see that it is not the same as a linear equation in three variables. Now we will see the properties and restrictions with sin function. So we know that the range of sin function is the closed interval \[\left[ -1,1 \right]\] i.e. for any value of \[x\], \[y\] and \[z\], the value of \[\sin x\], \[\sin y\] and \[\sin z\] will always lie in the interval \[\left[ -1,1 \right]\]. Using this, we can compute that the given equation will be possible only if all the three terms is equal to 1 since if one of the three terms is less than $1$, then there will be at least one term such that it is greater than 1, which is not true. Hence all three terms come out to be $1$.

Complete step by step solution :
We are given that
\[sinx+siny+sinz=3\]
We already know that the range of sin is the interval \[\left[ -1,1 \right]\].
And hence
\[sinx+siny+sinz=3\] is possible only when \[sinx=\sin y=\sin z=1\]
Also \[sin90{}^\circ =1\]
Therefore \[x=y=z=90{}^\circ \]
Thus we will use the value of \[x\], \[y\] and \[z\]as \[90{}^\circ \]
Now
\[cosx+cosy+cosz=cos90{}^\circ +cos90{}^\circ +cos90{}^\circ \]
We know that \[cos90{}^\circ =0\]
Using the value of \[cos\text{ }90{}^\circ \], we get
\[cosx+cosy+cosz=cos90{}^\circ +cos90{}^\circ +cos90{}^\circ =0+0+0=0\]

Thus, the value of \[cosx+cosy+cosz=0\]

Note :
It is important to note that we have memorized the value of trigonometric functions for various angles, basically \[0\text{ },\text{ }30\text{ },\text{ }45,\text{ }60\] and \[90\]. These values are often used in solving trigonometric equations.