Question

Trignometry question

Answer 1

\(\large \color{black}{\begin{align}
& \normalsize \text{Find the maximum value of }\hspace{.33em}\\~\\
& \sin^{4}\theta +\cos^{4}\theta \hspace{.33em}\\~\\~\\
& a.)\ 3\hspace{.33em}\\~\\
& b.)\ \dfrac13\hspace{.33em}\\~\\
& c.)\ 1 \hspace{.33em}\\~\\
& d.)\ 2 \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

Rewrite either the sin or cos function. I.E \(sin^{2}(x)=1-cos^{2}(x)\)

Answer 3

\(\large \color{black}{\begin{align}

& \sin^{4}\theta +\cos^{4}\theta \hspace{.33em}\\~\\
& (1-\cos^{2}\theta)^{2} +\cos^{4}\theta \hspace{.33em}\\~\\
\end{align}}\)

Answer 4

Yup. Keep going.

Answer 5

next wht

Answer 6

You now have \(cos^{2}(x)-2cos^{2}(x)+cos^{4}(x)+1\), right?

Answer 7

\(\large \color{black}{\begin{align}

& \sin^{4}\theta +\cos^{4}\theta \hspace{.33em}\\~\\
&= (1-\cos^{2}\theta)^{2} +\cos^{4}\theta \hspace{.33em}\\~\\
&= 1-2\cos^{2}\theta+2\cos^{4}\theta \hspace{.33em}\\~\\
\end{align}}\)

Answer 8

Oh, yeah, I missed a 2 there.

Answer 9

Let's call \(x=1-2cos^{2}(\theta)+2cos^{4}(\theta)\), so \(\frac{x}{2}=\frac{1}{2}-cos^{2}(\theta)+cos^{4}(\theta)\). Well, I gotta think this for a bit (Assuming you can't use the obvious approach which is sin(x)+cos(x)=1)

Answer 10

All right... We have to rewrite that. Nevermind what I did last post. 1-2cos²(θ)=-cos(2θ).
We can work from here.

Answer 11

How about using the fact that\[1^2=(\sin^2 x+\cos^2 x)^2=\sin^4 x+\cos^4 x +2 \sin^2 x \cos^2 x\]

Answer 12

\(\sin^{4}x+\cos^{4}x=1-2\sin^{2}x\cos^{2}x\)

Answer 13

is 1 the answrer

Answer 14

Yes.

Answer 15

and\[\text{Your expression}=1-\frac{1}{2} (\sin(2x))^2\]yes the answer is 1

Answer 16

Notice \(\sin^4x\le \sin^2x\) and \(\cos^4x\le \cos^2x\).

that implies \(\sin^4x+\cos^4x\le \sin^2x+\cos^2x=1\)

Answer 17

Nice observation gane +1

Answer 18

:) i answered this problem yesterday haha!

Answer 19

\(\large \color{black}{\begin{align}
& \text{calculate} \hspace{.33em}\\~\\
& \tan 4^{\circ} \tan 43^{\circ} \tan 47^{\circ} \tan 86^{\circ} \hspace{.33em}\\~\\
\end{align}}\)

i have to solve this within a minute

Answer 20

First notice that 4+86 = 43+47 = 90

Answer 21

next recall the indentity involving tan and cot :
\[\tan(x) = \cot(90-x) = \dfrac{1}{\tan(90-x)}\]

Answer 22

1 is the answer

Answer 23

this was easy but looked hard at first sight

Answer 24

reason like this -
you're supposed to solve the problem in minute, so they are "forced" to put only simple problems in the test

Answer 25

that helps in approaching the problem because when you knw upfront that they wont be asking questions that cannot be solved within a minute, you wont get distracted trying all the fancy methods..

Answer 26

\(\large \color{black}{\begin{align}
& \text{If}\ \sin \theta +\sin^{2} \theta =1,\ \hspace{.33em}\\~\\
& \text{and}\ \cos^{2} \theta +\cos^{4} \theta =x,\ \hspace{.33em}\\~\\
& \text{then the value of }x= \hspace{.33em}\\~\\
&a.)\ \dfrac{\cos^{2} \theta}{\sin \theta} \hspace{.33em}\\~\\
&b.)\ \text{None} \hspace{.33em}\\~\\
& c.)\ 1 \hspace{.33em}\\~\\
&d.)\ \dfrac{\sin \theta}{\cos^{2} \theta} \hspace{.33em}\\~\\
\end{align}}\)

Answer 27

compare the first equation,\(\color{red}{\sin\theta}+\sin^2\theta=1\), with the identity :

\(\color{red}{\cos^2\theta}+\sin^2\theta=1\)

Answer 28

\(\sin \theta =\cos^2 \theta \)

Answer 29

Yep, plug that in second equation

Answer 30

\(\large \color{black}{\begin{align}
& x=\sin \theta +\sin ^{2} \theta \hspace{.33em}\\~\\
\end{align}}\)

Answer 31

\(\cos^2\theta + \cos^4\theta=x\)

\(\sin\theta + \sin^2\theta=x\)

\(1=x\)

Answer 32

\(\large \color{black}{\begin{align}
& x\cos \theta -\sin \theta =1 \hspace{.33em}\\~\\
& x^{2}+(1+x^{2})\sin \theta \ \text{equals} \hspace{.33em}\\~\\
& a.)\ 0 \hspace{.33em}\\~\\
& b.)\ 2 \hspace{.33em}\\~\\
& c.)\ 1 \hspace{.33em}\\~\\
& d.)\ -1 \hspace{.33em}\\~\\
\end{align}}\)