Question

Find the value of ($cos{20}^{\circ }-sin{20}^{\circ }$) (1+4$sin{20}^{\circ }cos{20}^{\circ }$)

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

We don't know the values of cos20 or sin20.So we have to simplify the terms.

($cos{20}^{\circ }-sin{20}^{\circ }$) (1+4$cos{20}^{\circ }sin{20}^{\circ }$)

= $cos{20}^{\circ }+4co{s}^2{20}^{\circ }sin{20}^{\circ }-sin{20}^{\circ }-4si{n}^2{20}^{\circ }cos{20}^{\circ }$

= $cos{20}^{\circ }+4(1-si{n}^2{20}^{\circ })sin{20}^{\circ }-sin{20}^{\circ }-4(1-co{s}^2{20}^{\circ })cos{20}^{\circ }$

= $cos{20}^{\circ }+4sin{20}^{\circ }-4si{n}^3{20}^{\circ }-sin{20}^{\circ }-4cos{20}^{\circ }+4co{s}^3{20}^{\circ }$

= $3sin{20}^{\circ }-4si{n}^3{20}^{\circ }-3cos{20}^{\circ }+4co{s}^3{20}^{\circ }$

(sin3A = 3sin3A - 4sinA and cos3A = 4$co{s}^{3}A$ - 3cosA)

= sin$(3\times {20}^{\circ })$ + cos$(3\times {20}^{\circ })$

= sin${60}^{\circ }$ + cos${60}^{\circ }$

= $\frac{\sqrt{3}}{2}$ + $\frac{1}{2}$ = $\frac{(1+\sqrt{3})}{2}$