Question

Trigonometry question

Answer 1

ok,
lets start.

know the formula for \(a^3 + b^3\) ?

apply it for \(\Large (\sin^2x)^3 + (\cos^2 x)^3\)

where a = sin^2 x , b = cos^2 x

Answer 2

note: \(A^6 = (A^2)^3\)

Answer 3

why didnt u use formula \((a^2+b^2)^3\)

Answer 4

the K is in the form of a^3 + b^3 and not (a+b)^3 .. .

Answer 5

the final aim here is to represent K in the form of only one trigonometric function...like just sin2x..

Answer 6

\(\large \color{black}{\begin{align}
& \sin^{6} x+\cos^{6} x=(1)(\sin^{4}x-\sin^{2}x\cos^{2}x+\cos^{4}x) \hspace{.33em}\\~\\
\end{align}}\)

Answer 7

good!
now we use
\(a^2 + b^2 = (a+b)^2 - 2ab\)

Answer 8

a, b same

Answer 9

\(\large \color{black}{\begin{align}
& \sin^{6} x+\cos^{6} x=(1)(\sin^{4}x-\sin^{2}x\cos^{2}x+\cos^{4}x) \hspace{.33em}\\~\\
& =(1-3\sin^{2}x\cos^{2}x) \hspace{.33em}\\~\\
\end{align}}\)

Answer 10

excellent! we're very close now :)

recollect the double angle formula,
\(2\sin A \cos A = \sin 2A\)

Answer 11

\(\large \color{black}{\begin{align}
& \sin^{6} x+\cos^{6} x=(1)(\sin^{4}x-\sin^{2}x\cos^{2}x+\cos^{4}x) \hspace{.33em}\\~\\
& =(1-3\sin^{2}x\cos^{2}x) \hspace{.33em}\\~\\
& =(1-\dfrac{3}{4}\sin^{2}2x) \hspace{.33em}\\~\\
\end{align}}\)

Answer 12

now whats the range of sin^2 A ?

Answer 13

\(\large \color{black}{\begin{align}
& 0\leq \sin^{2}2x\leq 1 \hspace{.33em}\\~\\
\end{align}}\)

Answer 14

1. multiply both sides by -3/4
2. add 1 on both sides

Answer 15

**all sides :P

Answer 16

-3/4
its negative,
so signs will flip

Answer 17

\(\large \color{black}{\begin{align}
& 0\leq \sin^{2}2x\leq 1 \hspace{.33em}\\~\\
& 1\geq -\dfrac{3\sin^{2}2x}{4}+1\geq \dfrac{1}{4} \hspace{.33em}\\~\\
\end{align}}\)

Answer 18

\(\huge \checkmark \)

Answer 19

any doubts?

Answer 20

strange way to teach

Answer 21

didn't like it ?

Answer 22

joker likes it , cheers!

Answer 23

becuz joker is estrange.

Answer 24

lol joker jokes a joke :P

Answer 25

why so serious ?