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

If 2sinx+5cosy+7sinz = 14, then find the value of 7tan⁡x2+4cos⁡y−6sin⁡z7\tan \dfrac{x}{2}+4\cos y-6\sin z.

Explanation

Solution

Hint: Use the fact that ∀x∈R,sin⁡x≤1\forall x\in \mathbb{R},\sin x\le 1 and ∀x∈R,cos⁡x≤1\forall x\in \mathbb{R},\cos x\le 1. Hence prove that if the solution of the equation exists then sin⁡z≥1\sin z\ge 1. Hence argue that sinz =1 and hence cosy = 1 and sinx = 1. Hence prove that x=π2,y=0x=\dfrac{\pi }{2},y=0 and z=π2z=\dfrac{\pi }{2}. Use the fact that tan⁡(π4)=1\tan \left( \dfrac{\pi }{4} \right)=1 and hence find the value of 7tan⁡(x2)+4cos⁡y−6sin⁡z7\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 ∀x∈R,sin⁡x≤1\forall x\in \mathbb{R},\sin x\le 1 and ∀x∈R,cos⁡x≤1\forall x\in \mathbb{R},\cos x\le 1
Hence, we have 2sin⁡x≤22\sin x\le 2 and 5cos⁡y≤55\cos y\le 5
Hence, we have 2sin⁡x+5cos⁡y≤5+2=72\sin x+5\cos y\le 5+2=7
Adding 7sinz on both sides, we get
2sin⁡x+5cos⁡y+7sin⁡z≤7+7sin⁡z2\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≤7+7sin⁡z14\le 7+7\sin z
Subtracting 7 from both sides, we get
7sin⁡z≥77\sin z\ge 7
Dividing both sides by 7, we get
sin⁡z≥1\sin z\ge 1
But, we know that sin⁡z≤1\sin z\le 1
Hence, we have sin⁡z=1\sin z=1
When sinz = 1, we have
2sin⁡x+5cos⁡y+7=142\sin x+5\cos y+7=14
Subtracting 7 from both sides, we get
2sin⁡x+5cos⁡y=7 (i)2\sin x+5\cos y=7\text{ }\left( i \right)
Now, we know that 5cos⁡y≤55\cos y\le 5
Adding 2 sinx from both sides, we get
2sin⁡x+5cos⁡y≤5+2sin⁡x2\sin x+5\cos y\le 5+2\sin x
From equation (i), we have 2sinx+5cosy = 7
Hence, we have
7≤5+2sin⁡x7\le 5+2\sin x
Subtracting 5 from both sides, we get
2sin⁡x≥22\sin x\ge 2
Dividing both sides by 2, we get
sin⁡x≥1\sin x\ge 1
But, we know that sin⁡x≤1\sin x\le 1
Hence, we have
sin⁡x=1\sin x=1
Hence equation (i) becomes
5cos⁡y+2=75\cos y+2=7
Subtracting 2 from both sides, we get
5cos⁡y=55\cos y=5
Dividing both sides by 5, we get
cos⁡y=1\cos y=1
Hence, we have x=π2,y=0x=\dfrac{\pi }{2},y=0 and z=π2z=\dfrac{\pi }{2}
Hence, we have 7tan⁡x2+4cos⁡y−6sin⁡z=7tan⁡(π4)+4cos⁡0−6sin⁡(π2)=7+4−6=57\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 7tan⁡x2+4cos⁡y−6sin⁡z7\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.