Trignometry question

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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

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Other answers:

Yup. Keep going.
next wht
You now have \(cos^{2}(x)-2cos^{2}(x)+cos^{4}(x)+1\), right?
\(\large \color{black}{\begin{align} & \sin^{4}\theta +\cos^{4}\theta \hspace{.33em}\\~\\ &= (1-\cos^{2}\theta)^{2} +\cos^{4}\theta \hspace{.33em}\\~\\ &= 1-2\cos^{2}\theta+2\cos^{4}\theta \hspace{.33em}\\~\\ \end{align}}\)
Oh, yeah, I missed a 2 there.
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)
All right... We have to rewrite that. Nevermind what I did last post. 1-2cos²(θ)=-cos(2θ). We can work from here.
How about using the fact that\[1^2=(\sin^2 x+\cos^2 x)^2=\sin^4 x+\cos^4 x +2 \sin^2 x \cos^2 x\]
\(\sin^{4}x+\cos^{4}x=1-2\sin^{2}x\cos^{2}x\)
is 1 the answrer
Yes.
and\[\text{Your expression}=1-\frac{1}{2} (\sin(2x))^2\]yes the answer is 1
Notice \(\sin^4x\le \sin^2x\) and \(\cos^4x\le \cos^2x\). that implies \(\sin^4x+\cos^4x\le \sin^2x+\cos^2x=1\)
Nice observation gane +1
:) i answered this problem yesterday haha!
\(\large \color{black}{\begin{align} & \text{calculate} \hspace{.33em}\\~\\ & \tan 4^{\circ} \tan 43^{\circ} \tan 47^{\circ} \tan 86^{\circ} \hspace{.33em}\\~\\ \end{align}}\) i have to solve this within a minute
First notice that 4+86 = 43+47 = 90
next recall the indentity involving tan and cot : \[\tan(x) = \cot(90-x) = \dfrac{1}{\tan(90-x)}\]
1 is the answer
this was easy but looked hard at first sight
reason like this - you're supposed to solve the problem in minute, so they are "forced" to put only simple problems in the test
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..
\(\large \color{black}{\begin{align} & \text{If}\ \sin \theta +\sin^{2} \theta =1,\ \hspace{.33em}\\~\\ & \text{and}\ \cos^{2} \theta +\cos^{4} \theta =x,\ \hspace{.33em}\\~\\ & \text{then the value of }x= \hspace{.33em}\\~\\ &a.)\ \dfrac{\cos^{2} \theta}{\sin \theta} \hspace{.33em}\\~\\ &b.)\ \text{None} \hspace{.33em}\\~\\ & c.)\ 1 \hspace{.33em}\\~\\ &d.)\ \dfrac{\sin \theta}{\cos^{2} \theta} \hspace{.33em}\\~\\ \end{align}}\)
compare the first equation,\(\color{red}{\sin\theta}+\sin^2\theta=1\), with the identity : \(\color{red}{\cos^2\theta}+\sin^2\theta=1\)
\(\sin \theta =\cos^2 \theta \)
Yep, plug that in second equation
\(\large \color{black}{\begin{align} & x=\sin \theta +\sin ^{2} \theta \hspace{.33em}\\~\\ \end{align}}\)
\(\cos^2\theta + \cos^4\theta=x\) \(\sin\theta + \sin^2\theta=x\) \(1=x\)
\(\large \color{black}{\begin{align} & x\cos \theta -\sin \theta =1 \hspace{.33em}\\~\\ & x^{2}+(1+x^{2})\sin \theta \ \text{equals} \hspace{.33em}\\~\\ & a.)\ 0 \hspace{.33em}\\~\\ & b.)\ 2 \hspace{.33em}\\~\\ & c.)\ 1 \hspace{.33em}\\~\\ & d.)\ -1 \hspace{.33em}\\~\\ \end{align}}\)

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