Find the value of $A+B$?

Question

If

$A = 4\sin^2 10 + 4 \sin^2 50 \cos 20 + \cos 80 $

$B=\cos^2 (\pi/5) + \cos^2 (2\pi/15) + \cos^2(8\pi/15)$ then, $A+B$ =?
I tried simplifying the middle term of the first in this way:
$ A= 4\sin^2 10 +2\sin50(2\sin50\cos20) + \cos80$,
but this isn't helping even after applying the formula for $(2sinAcosB)$.Or maybe I am getting wrong somewhere. Please help.

Answer 1

1st case: Using the double angle formula we have that:

• $\sin^2 10^\circ = \frac{1-\cos 20^\circ}{2}\implies 4 \sin^2 10^\circ = 2 - 2\cos 20^\circ.$

• $\sin^2 50^\circ = \frac{1-\cos 100^\circ}{2}\implies 4\sin^2 50^\circ \cos 20^\circ = \cdots = 2\cos 20^\circ -2\cos 100^\circ \cos 20^\circ$

• $\cos 80^\circ = \cos(100^\circ - 20^\circ) = \cos 100^\circ \cos 20^\circ + \sin 100^\circ \sin 20^\circ$

In order to find $A$ just substitute and simplify.

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2nd case: Using again the double angle formula we have:

• $\cos^2\left(\frac{3\pi}{15}\right)= \frac{1+\cos(6\pi/15)}{2}$

• $\cos^2\left(\frac{2\pi}{15}\right) = \frac{1+\cos(4\pi/15)}{2}$

• $\cos^2\left(\frac{8\pi}{15}\right) = \frac{1+\cos(16\pi/15)}{2}$

Thus: $$ \cos^2\left(\frac{3\pi}{15}\right)+\cos^2\left(\frac{2\pi}{15}\right)+\cos^2\left(\frac{8\pi}{15}\right) =\frac{3+\cos(6\pi/15)+\cos(4\pi/15)+\cos(16\pi/15)}{2} $$

Using the identity: $$\cos x + \cos y = 2 \cos \left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right),$$

we have that $$\cos(6\pi/15) + \cos (4\pi/15) = 2\cos \left(\frac{10\pi}{30}\right)\cos \left(\frac{2\pi}{30}\right) = \cos\left(\frac{\pi}{15}\right).$$

$$\cos\left(\frac{16\pi}{15}\right) = \cos \left(\pi + \frac{\pi}{15} \right) = -\cos\left(\frac{\pi}{15}\right).$$

Hence $B = \frac 32.$

Answer 2

$$2S=\cos^236^\circ+\cos^224^\circ+\cos^296^\circ =1+\cos^236^\circ-\sin^224^\circ+\cos^296^\circ$$

Using $\cos2x=2\cos^2x-1,\cos(180^\circ+y)=-\cos y$ and Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,

$$2S=2+2\cos60^\circ\cos12^\circ+1+\cos(180^\circ+12^\circ)=3$$

Now for $A,$

$\sin10^\circ=\cos80^\circ=p$(say)

$2\sin^250^\circ=2\cos^240^\circ=\dfrac{1+\cos80^\circ}2;$

$\cos20^\circ=-\cos(180^\circ-20^\circ)=-(2\cos^280^\circ-1)$

$$\implies A=4p^2+2(1+p)(1-2p^2)+p=2-(4p^3-3p)=2-\cos(3\cdot80^\circ)$$

using $\cos3A$ formula