# Proof this curious trigonometric identity

Proof that $$\cos^2{10^\circ} + \cos^2{50^\circ} - \sin{40^\circ}\sin{80^\circ} = \frac{3}{4}$$

I notice that $10^\circ + 80^\circ = 90^\circ$, and $50^\circ +40^\circ = 90^\circ$. I tried doing some manipulation but my efforts were futile. Any hints?

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$\displaystyle\cos^210^\circ+\cos^250^\circ=\cos^210^\circ-\sin^250^\circ+1=\cos60^\circ\cos40^\circ+1=\frac{\cos40^\circ}2+1$

Use Werner Formulas to find $\displaystyle2\sin40^\circ\sin80^\circ=\cos40^\circ-\cos120^\circ$

Now, $\displaystyle\cos120^\circ=\cos(180^\circ-60^\circ)=-\cos60^\circ=-\frac12$

Can you take it home from here?

Alternatively, $\displaystyle\cos^210^\circ+\cos^250^\circ-\sin40^\circ\sin80^\circ=\cos^210^\circ+\cos^250^\circ-\cos50^\circ\cos10^\circ$

$\displaystyle=\left(\cos10^\circ-\cos50^\circ\right)^2+\cos50^\circ\cos10^\circ$

Using Prosthaphaeresis Formulas, $\displaystyle\cos10^\circ-\cos50^\circ=2\sin30^\circ\sin20^\circ=\sin20^\circ$

Now use Double-Angle Formulas i.e., $\displaystyle\sin^2x=\frac{1-\cos2x}2$ for $x=20^\circ$

Using Werner Formulas, $\displaystyle2\cos50^\circ\cos10^\circ=\cos40^\circ+\cos60^\circ$

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Thanks for the reply! I am having troubles in proving it even after reading your observations. Could you please make it more explicit? – Yiyuan Lee Mar 8 '14 at 4:59
@YiyuanLee, First of all, are you familiar with the formulas used? If not, have you understood them? Finally please pinpoint the area(s) of confusion. – lab bhattacharjee Mar 8 '14 at 5:01
On second thoughts, after going through you answer again, I think I now get it. Thanks! – Yiyuan Lee Mar 8 '14 at 5:04

Convert into a polynomial of $\cos(40^\circ)$: \begin{align} &\cos^2(10^\circ)+\cos^2(50^\circ)-\sin(40^\circ)\sin(80^\circ)\\ &=\sin^2(80^\circ)+\sin^2(40^\circ)-\sin(40^\circ)\sin(80^\circ)\tag{1}\\ &=4\sin^2(40^\circ)\cos^2(40^\circ)+\sin^2(40^\circ)-2\sin^2(40^\circ)\cos(40^\circ)\tag{2}\\ &=1-2\cos(40^\circ)+3\cos^2(40^\circ)+2\cos^3(40^\circ)-4\cos^4(40^\circ)\tag{3} \end{align} Explanation:
$(1)$: $\cos(10^\circ)=\sin(80^\circ)$ and $\cos(50^\circ)=\sin(40^\circ)$
$(2)$: $\sin(80^\circ)=2\sin(40^\circ)\cos(40^\circ)$
$(3)$: $\sin^2(40^\circ)=1-\cos^2(40^\circ)$

Since $\cos(3x)=4\cos^3(x)-3\cos(x)$, we have $$4\cos^3(40^\circ)-3\cos(40^\circ)=\cos(120^\circ)=-\tfrac12\tag{4}$$ Thus, we write the polynomial from $(3)$, with $x=\cos(40^\circ)$, as $$\underbrace{1-2x+3x^2+2x^3-4x^4}_{\text{from }(3)}=\tfrac34+\left(\tfrac12-x\right)\underbrace{\left(\tfrac12-3x+4x^3\right)}_{0\text{ from }(4)}\tag{5}$$

Therefore, $(3)$, $(4)$ and $(5)$ say $$\cos^2(10^\circ)+\cos^2(50^\circ)-\sin(40^\circ)\sin(80^\circ)=\frac34\tag{6}$$

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Generalization :

$$I=\cos^2A+\cos^2B-\cos A\cos B$$

$$=\cos^2A-\sin^2B+1-\frac{\cos(A-B)+\cos(A-B)}2$$

$$=\cos(A-B)\cos(A+B)+1-\frac{\cos(A-B)+\cos(A+B)}2$$

$$I=\cos(A-B)\left(\cos(A+B)-\frac12\right)+1-\frac{\cos(A+B)}2$$

If $\displaystyle\cos(A+B)=\frac12=\cos60^\circ\iff A+B=2n\pi\pm60^\circ$ where $n$ is any integer

$I$ becomes $\displaystyle1-\frac{\dfrac12}2=\frac34$

$$\text{Again, }I=\cos(A+B)\left(\cos(A-B)-\frac12\right)+1-\frac{\cos(A-B)}2$$

What if $\displaystyle\cos(A-B)=\frac12 ?$

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