Question

Solve $$1+\cos 2x+\cos 4x+\cos 6x=$$
A.$$2\cos x\cos 2x\cos 3x$$
B.$$4\sin x\cos 2x\cos 3x$$
C.$$4\cos x\cos 2x\cos 3x$$
D.None of these

Answer 1

Hint: In the given question the addition of trigonometric ratios are given , so we have to use the identity formula for compound angles for $$\cos x$$ . First solve for larger angles ($$\cos 4x$$and $$\cos 6x$$ ) than for smaller angles . Also , here direct $$1+\cos 2x$$ is given which is equal to $$1+\cos 2x=2{\cos }^{2}x$$

Complete step-by-step answer:
Given : $$1+\cos 2x+\cos 4x+\cos 6x$$ ,
Applying the identity formula $$\cos A+\cos B=2\cos {\displaystyle \frac{A+B}{2}}\cos {\displaystyle \frac{A-B}{2}}$$ for $$\cos 4x$$and $$\cos 6x$$ we get ,
$$=1+\cos 2x+2\cos ({\displaystyle \frac{6x-4x}{2}})\cos ({\displaystyle \frac{6x+4x}{2}})$$ , on solving we get ,
$$=1+\cos 2x+2\cos x\cos 5x$$
Now applying formula $$1+\cos 2x=2{\cos }^{2}x$$ we get ,
$$=2{\cos }^{2}x+2\cos x\cos 5x$$ taking $$2\cos x$$common we get ,
$$=2\cos x(\cos x+\cos 5x)$$
 Now again applying identity $$\cos A+\cos B=2\cos {\displaystyle \frac{A+B}{2}}\cos {\displaystyle \frac{A-B}{2}}$$ we get ,
$$=2\cos x\left[2\cos ({\displaystyle \frac{5x+x}{2}})\cos ({\displaystyle \frac{5x-x}{2}})\right]$$ , on solving we get ,
$$=2\cos x\left[2\cos 3x\cos 2x\right]$$
On solving further we get ,
$$=4\cos x\cos 2x\cos 3x$$ .
Therefore , option ( 3 ) is the correct answer .
So, the correct answer is “Option 3”.

Note: Alternative Method :
Given : $$1+\cos 2x+\cos 4x+\cos 6x$$
Now , this time taking the terms $$\cos 6x$$ and $$\cos 2x$$ together and applying the formula $$\cos A+\cos B=2\cos {\displaystyle \frac{A+B}{2}}\cos {\displaystyle \frac{A-B}{2}}$$ , again we get
$$=1+\cos 4x+2\cos ({\displaystyle \frac{6x-2x}{2}})\cos ({\displaystyle \frac{6x+2x}{2}})$$ , on solving we get ,
$$=1+\cos 4x+2\cos 2x\cos 4x$$
Now using the formula $$\cos 2x=2{\cos }^{2}x-1$$ we get ,
$$=2{\cos }^{2}2x+2\cos 2x\cos 4x$$, now taking the $$2\cos 2x$$ common we get
$$=2\cos 2x(\cos 2x+\cos 4x)$$
Now again applying the formula $$\cos A+\cos B=2\cos {\displaystyle \frac{A+B}{2}}\cos {\displaystyle \frac{A-B}{2}}$$ we get ,
$$=2\cos 2x\left[2\cos ({\displaystyle \frac{4x-2x}{2}})\cos ({\displaystyle \frac{4x+2x}{2}})\right]$$ on solving further we get ,
$$=2\cos 2x\left[2\cos x\cos 3x\right]$$, on simplifying we get
$$=4\cos x\cos 2x\cos 3x$$ .