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Question: How do you prove \( \left[ {\dfrac{{\sin 3x}}{{\sin x}}} \right] - \left[ {\dfrac{{\cos 3x}}{{\cos x...

How do you prove [sin3xsinx][cos3xcosx]=2?\left[ {\dfrac{{\sin 3x}}{{\sin x}}} \right] - \left[ {\dfrac{{\cos 3x}}{{\cos x}}} \right] = 2?

Explanation

Solution

Hint : As we know that there are three basic trigonometric identities that involve the sums of angles, the functions which are involved in these identities are sine, cosine and tangent. A trigonometric identity is an equation based on trigonometry which is always true. The angle sum identity like sin(AB)=sinAcosBcosAsinB\sin (A - B) = \sin A\cos B - \cos A\sin B . Whether it is sum or difference of angles in trigonometric functions, they are used to find out the most functional values or exact value of any angles. Some of the most familiar values of all trigonometric ratios are 30,45,90,60{30^ \circ },{45^ \circ },{90^ \circ },{60^ \circ } and so on.

Complete step-by-step answer :
As per the given question we have: [sin3xsinx][cos3xcosx]=2\left[ {\dfrac{{\sin 3x}}{{\sin x}}} \right] - \left[ {\dfrac{{\cos 3x}}{{\cos x}}} \right] = 2 . We have to prove that the left hand side expression is equal to the right hand expression by using the double angle formula.
We know that double angle formula which states that sin2a=2sinacosa\sin 2a = 2\sin a\cos a .
We will first take the left hand side of the equation and solve it like normal fraction by taking the L.C.M i.e.
[sin3xsinx][cos3xcosx]=sin3xcosxcos3xsinxsinxcosx\left[ {\dfrac{{\sin 3x}}{{\sin x}}} \right] - \left[ {\dfrac{{\cos 3x}}{{\cos x}}} \right] = \dfrac{{\sin 3x\cos x - \cos 3x\sin x}}{{\sin x\cos x}} .
We will recall the identity here which states that : sin(AB)=sinAcosBcosAsinB\sin (A - B) = \sin A\cos B - \cos A\sin B ,here A=3x,B=xA = 3x,B = x .
Therefore according to the identity we can write: sin(3xx)sinxcosx\dfrac{{\sin (3x - x)}}{{\sin x\cos x}} .
It gives sin2xsinxcosx\dfrac{{\sin 2x}}{{\sin x\cos x}} . We know that the value of sin2x\sin 2x can be written as 2sinxcosx2\sin x\cos x . So we will put the values and solve
2sinxcosxsinxcosx=2\dfrac{{2\sin x\cos x}}{{\sin x\cos x}} = 2 .
So the left hand side is now equal to the right hand side value.
Hence it is proved that [sin3xsinx][cos3xcosx]=2\left[ {\dfrac{{\sin 3x}}{{\sin x}}} \right] - \left[ {\dfrac{{\cos 3x}}{{\cos x}}} \right] = 2 .
So, the correct answer is “2.

Note : We should always remember the angle sum identity of every trigonometric function before solving it. Trigonometric functions are also called circular functions and these basic functions are also known as trigonometric ratios. There are multiple trigonometric formulas and identities which represent the relation between the functions and enable to find the value of unknown angles. We should always remember to determine in which quadrant the angle will lie as it will say about the positive and negative value of cosine. There is both angle sum identity and angle difference formula to calculate the values of angles.