Question

If 2sinx+5cosy+7sinz = 14, then find the value of $7\tan \dfrac{x}{2}+4\cos y-6\sin z$.

Answer 1

Hint: Use the fact that $\forall x\in \mathbb{R},\sin x\le 1$ and $\forall x\in \mathbb{R},\cos x\le 1$. Hence prove that if the solution of the equation exists then $\sin z\ge 1$. Hence argue that sinz =1 and hence cosy = 1 and sinx = 1. Hence prove that $x=\dfrac{\pi }{2},y=0$ and $z=\dfrac{\pi }{2}$. Use the fact that $\tan \left( \dfrac{\pi }{4} \right)=1$ and hence find the value of $7\tan \left( \dfrac{x}{2} \right)+4\cos y-6\sin z$.

Complete step-by-step answer:

We have 2sinx+5cosy+7sinz=14.
We know that $\forall x\in \mathbb{R},\sin x\le 1$ and $\forall x\in \mathbb{R},\cos x\le 1$
Hence, we have $2\sin x\le 2$ and $5\cos y\le 5$
Hence, we have $2\sin x+5\cos y\le 5+2=7$
Adding 7sinz on both sides, we get
$2\sin x+5\cos y+7\sin z\le 7+7\sin z$
From the equation, we have 2sinx+5cosy+7sinz = 14
Hence, we have $14\le 7+7\sin z$
Subtracting 7 from both sides, we get
$7\sin z\ge 7$
Dividing both sides by 7, we get
$\sin z\ge 1$
But, we know that $\sin z\le 1$
Hence, we have $\sin z=1$
When sinz = 1, we have
$2\sin x+5\cos y+7=14$
Subtracting 7 from both sides, we get
$2\sin x+5\cos y=7\text{ }\left( i \right)$
Now, we know that $5\cos y\le 5$
Adding 2 sinx from both sides, we get
$2\sin x+5\cos y\le 5+2\sin x$
From equation (i), we have 2sinx+5cosy = 7
Hence, we have
$7\le 5+2\sin x$
Subtracting 5 from both sides, we get
$2\sin x\ge 2$
Dividing both sides by 2, we get
$\sin x\ge 1$
But, we know that $\sin x\le 1$
Hence, we have
$\sin x=1$
Hence equation (i) becomes
$5\cos y+2=7$
Subtracting 2 from both sides, we get
$5\cos y=5$
Dividing both sides by 5, we get
$\cos y=1$
Hence, we have $x=\dfrac{\pi }{2},y=0$ and $z=\dfrac{\pi }{2}$
Hence, we have $7\tan \dfrac{x}{2}+4\cos y-6\sin z=7\tan \left( \dfrac{\pi }{4} \right)+4\cos 0-6\sin \left( \dfrac{\pi }{2} \right)=7+4-6=5$
Hence the value of $7\tan \dfrac{x}{2}+4\cos y-6\sin z$ is 5.

Note: [1] When we have an equation involving multiple variables in sines and cosines, we must check whether RHS is the maxima or minima of the expression or not. If RHS is the maxima or minima, then the sines and cosines also take extrema. This is the case in the above question.