Question

E = $\frac{cos6x+6cos4x+15cos2x+10}{(cos5x+5cos3x+10cosx)}$ equals to

• A. cos2x
• B. 2cosx
• C.
• D. 1 + cosx

Answer 1

The correct option is B

2cosx

We want to simplify numerator and denominator using the transformation formula cosA+cosB = $\frac{2cos(A+B)}{2}$ $\frac{2cos(A-B)}{2}$ . Important step here is finding the suitable pair to apply this formula in numerator and denominator. We will look at the last terms in numerator and denominator. They are 10 and 10 cosx. Second last term in numerator is 15 cos2x. If we combine 10 and 10cos2x of 15 cos2x, we will get 2 $\times$ 10 $co{s}^{2}x$ in numerator. In similar problems we solved, we saw that sometimes we take common factor from numerator or denominator. In this case, if the common factor is 2 cosx, we will be left with 10 cosx in the numerator. This gives us the hint to simplify numerator first before we do anything to denominator.

We will combine 10cos2x of 15 cos2x with 10, the remaining 5cos2x with 5cos4x and cos6x with cos4x.

E = $\frac{(cos6x+cos4x)+5(cos4x+cos2x)+10(1+cos2x)}{cos5x+5cos3x+10cosx}$

E = $\frac{2cos5xcosx+10cosx\times cos3x+20co{s}^{2}x}{cos5x+5cos3x+10cosx}$

= $\frac{2cosx(cos5x+5cos3x+10cosx)}{(cos5x+5cos3x+10cosx)}$

= 2cosx

As we expected, 2cosx was the common factor.

Key steps/concepts: (1) Simplifying numerator first

(2) Combining appropriate terms.