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Question: The trigonometric equation \(\cos 4x\cdot \cos 8x-\cos 5x\cdot \cos 9x=0\) if This question has m...

The trigonometric equation cos4xcos8xcos5xcos9x=0\cos 4x\cdot \cos 8x-\cos 5x\cdot \cos 9x=0 if
This question has multiple correct options
(a) cos(12x)=cos(14x)\cos (12x)=\cos (14x)
(b) sin(13x)=0\sin (13x)=0
(C) sinx=0\sin x=0
(d) cosx=0\cos x=0

Explanation

Solution

Hint: You can use the product-to-sum and sum-to-product formulas to rewrite the product and sum of the cosine, respectively for expanding or simplifying given trigonometric expressions.

Complete step-by-step solution -
The given trigonometric equation can be written as
cos4xcos8xcos5xcos9x=0\cos 4x\cdot \cos 8x-\cos 5x\cdot \cos 9x=0
cos4xcos8x=cos5xcos9x\cos 4x\cdot \cos 8x=\cos 5x\cdot \cos 9x
Multiplying both sides by 2, we get
2cos4xcos8x=2cos5xcos9x2\cos 4x\cdot \cos 8x=2\cos 5x\cdot \cos 9x
Applying the formula for the product of cosines 2cosAcosB=cos(AB)+cos(A+B)2\cos A\cos B=\cos (A-B)+\cos (A+B) , we get
We can then substitute the given angles into the formula and simplify.
cos(4x8x)+cos(4x+8x)=cos(5x9x)+cos(5x+9x)\cos \left( 4x-8x \right)+\cos \left( 4x+8x \right)=\cos \left( 5x-9x \right)+\cos \left( 5x+9x \right)
cos(4x)+cos(12x)=cos(4x)+cos(14x)\cos \left( -4x \right)+\cos \left( 12x \right)=\cos \left( -4x \right)+\cos \left( 14x \right)
We know that, cos(θ)=cosθ\cos (-\theta )=\cos \theta
cos(4x)+cos(12x)=cos(4x)+cos(14x)\cos \left( 4x \right)+\cos \left( 12x \right)=\cos \left( 4x \right)+\cos \left( 14x \right)
Cancelling the term cos(4x)\cos (4x) on both sides, we get
cos(12x)=cos(14x)............(1)\cos \left( 12x \right)=\cos \left( 14x \right)............(1)
Hence the correct option for the given trigonometric equation is option (a).
The equation (1) can be written as
cos(12x)cos(14x)=0\cos \left( 12x \right)-\cos \left( 14x \right)=0
Applying the formula for the sum of the cosine cosAcosB=2sin(A+B2)sin(AB2)\cos A-\cos B=-2\sin \left( \dfrac{A+B}{2} \right)sin\left( \dfrac{A-B}{2} \right) , we get
We can then substitute the given angles into the formula and simplify.
2sin(12x+14x2)sin(12x14x2)=0-2\sin \left( \dfrac{12x+14x}{2} \right)sin\left( \dfrac{12x-14x}{2} \right)=0
2sin(26x2)sin(2x2)=0-2\sin \left( \dfrac{26x}{2} \right)sin\left( \dfrac{-2x}{2} \right)=0
2sin(13x)sin(x)=0-2\sin \left( 13x \right)sin\left( -x \right)=0
We know that sin(θ)=sinθ\sin (-\theta )=-\sin \theta
2sin(13x)sin(x)=02\sin \left( 13x \right)sin\left( x \right)=0
Dividing both sides by 2, we get
sin(13x)sin(x)=0\sin \left( 13x \right)sin\left( x \right)=0
sin(13x)=0 or sin(x)=0\sin \left( 13x \right)\text{=0 or }sin\left( x \right)=0
Hence the correct options of the given trigonometric equation are option (b) and option(c).
Therefore, the correct options of the given question are option (a), option (b) and option(c).

Note: It is not true that cos (A) cos (B) is equal to cos (AB). There is no nice formula for cos (AB). You can use the product to sum formulas of the cosine for the trigonometric expression cos (A) cos (B).