Prove that $\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4}$ I have a question that goes exactly like this:
By considering $(\sin^2\theta+\cos^2\theta)^2$ and $(\sin^2\theta+\cos^2\theta)^3$ prove that
$$
\text{a) }\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4},\qquad\text{b) }\sin^6\theta+\cos^6\theta=\frac{5+3\cos4\theta}{8}
$$
I have not idea how to do this. Please help.
 A: a) Since $\sin^2(\theta) + \cos^2(\theta) =1 $ we can write: $1=1^2= (\sin^2(\theta) + \cos^2(\theta))^2= \cos^4(\theta)+\sin^4(\theta)+2 \sin^2(\theta) \cos^2(\theta)$. Hence $\cos^4(\theta)+\sin^4(\theta) = 1-2\sin^2(\theta)\cos^2(\theta) = 1 - \frac{1}{2} \sin^2(2\theta) = 1 - \frac{1}{2} \frac{1-\cos(4\theta)}{2} $, the claim follows. b) is similar, even there you just have to remember basic goniometric ids (in particular bisection and dupliaction).
A: Hint: For $a$
$$L.H.S=\sin^4 A+ \cos^4 A$$
$$=(\sin^2 A+ \cos^2 A)^2-2\sin^2A.\cos^2A$$
$$=1-\frac {1}{4}(2\sin A.\cos A)^2$$
$$=1-\frac {1}{4} \sin^2 2A$$
$$=\frac {4-\sin^2 2A}{4}$$
$$=\frac {4-1+\cos 4A}{4}$$
$$=\frac {3+\cos 4A}{4}=R.H.S$$ 
Proved.
Just consider that $\theta=A$.
Follow similar steps above for $b$.
A: $\cos(m\theta)$ is a polynomial with degree $m$ in $\cos\theta$, in particular
$$ \cos(4\theta) = T_4(\cos\theta) = 8\cos^4\theta-8\cos^2\theta + 1\tag{1}$$
so the given identities are equivalent to the polynomial identities
$$ (1-x^2)^2+x^4 = \frac{3+T_4(x)}{4},\qquad (1-x^2)^3+x^6 = \frac{5+3 T_4(x)}{8}\tag{2}$$
that directly follow from $(1)$.
A: Write $c=\cos\theta,\,s=\sin\theta$ so $c^2+s^2=1$ and $$c^2s^2=\frac{1}{4}\sin^22\theta=\frac{1}{4}\left( 1-\cos^22\theta\right)=\frac{1}{8}\left( 1-\cos 4\theta\right).$$Hence $c^4+s^4=1-2c^2s^2=\frac{1}{4}\left( 3+\cos 4\theta\right)$.
Multiplying by $c^2+s^2$ gives $c^6+s^6+c^2s^2\left( c^2+s^2\right)$. Hence $$c^6+s^6=\frac{1}{4}\left( 3+\cos 4\theta\right)-c^2s^2=\frac{1}{8}\left( 5+3\cos 4\theta\right).$$
A: There is a general systematic method to prove this kind of linearisation identities. It's not as swift as the previous answers, bit it's purely automatic.
You use the Euler formulae
$$\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}2 \qquad \text{et}\qquad \sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$
and you expand the powers to get an expression with terms in $\cos(p\theta)$ and $\sin(q\theta)$.
\begin{align*}
\cos^4(\theta) &= \left(\frac{e^{i\theta}+e^{-i\theta}}2\right)^4\\
&= \frac{e^{i4\theta} + 4 e^{i2\theta} + 6 + 4 e^{-2i\theta} + e^{-i4\theta}}{16}\\
&= \frac{\cos 4\theta}8 + \frac{\cos 2\theta}2 + \frac 38.\\
\sin^4(\theta) &= \left(\frac{e^{i\theta}-e^{-i\theta}}{2i}\right)^4\\
&= \frac{e^{i4\theta} - 4 e^{i2\theta} + 6 - 4 e^{-2i\theta} + e^{-i4\theta}}{16}\\
&= \frac{\cos 4\theta}8 - \frac{\cos 2\theta}2 + \frac 38.\\
\text{therefore}\qquad \cos^4\theta+\sin^4\theta &= \frac{\cos 4\theta}4 + \frac 34.
\end{align*}
A: Hint:
$$\sin ^{4}\theta ={\frac {3-4\cos(2\theta )+\cos(4\theta )}{8}}\!$$
$$\cos ^{4}\theta ={\frac {3+4\cos(2\theta )+\cos(4\theta )}{8}}\!$$
A: From Euler's Identity and subsequent use of the binomial theorem, we have
$$\begin{align}
\cos(4\theta)&=\text{Re}(e^{i4\theta})\\\\
&=\text{Re}\left(\cos(\theta)+i\sin(\theta)\right)^4\\\\
&=\text{Re}\left(\sum_{k=0}^4\binom{4}{k}\cos^{4-k}(\theta)i^k\sin^k(\theta)\right)\\\\
&=\cos^4(\theta)-6\cos^2(\theta)\sin^2(\theta)+\sin^4(\theta)
\end{align}$$
Next, noting that $\sin^2(\theta)\cos^2(\theta)=\frac12-\frac12(\cos^4(\theta)+\sin^4(\theta))$ (this is a trivial consequence of $(\cos^2(\theta)+\sin^2(\theta))^2=1$), we obtain the coveted equality
$$\cos^4(\theta)+\sin^4(\theta)=\frac{3+\cos(4\theta)}{4}$$
A: Using  $\cos2A=2\cos^2A-1=1-2\sin^2A,$
$$(2\sin^2\theta)^2+(2\cos^2\theta)^2=(1-\cos2\theta)^2+(1+\cos2\theta)^2=2(1+\cos^22\theta)$$
Now $2\cos^22\theta=1+\cos2(2\theta)$
