Trigonometric Values?

Geometry Level 4

$\large \ \cos^{2} 73^{\circ} \ + \ \cos^{2} 47^{\circ} \ + \ \cos 73^{\circ} \cos 47^{\circ}$

If the value of the above expression is equal to $\dfrac {a}{b}$ , where $a$ and $b$ are coprime positive integers , find $a+b$ .

The answer is 7.

2 solutions

Rishabh Jain
Jun 17, 2016

$\mathcal K=\underbrace{ \cos^{2} 73^{\circ} + \cos^{2} 47^{\circ}}_{\mathcal P} + \underbrace{\cos 73^{\circ}\cdot\cos 47^{\circ}}_{\mathcal L}$ $\small{ \begin{aligned}\mathcal L=\cos 73^{\circ}\cdot\cos 47^{\circ}=&\dfrac 12\left(\underbrace{\cos 120^{\circ}}_{\frac{-1}2}+\cos 26^{\circ}\right)\\=&\dfrac{-1}{4}+\dfrac 12\cos 26^{\circ}\end{aligned}}$

$\small{\begin{aligned}\mathcal P=\cos^{2} 73^{\circ} + \cos^{2} 47^{\circ}=& 1+\color{#20A900}{\cos^{2} 73^{\circ} - \sin^{2} 47^{\circ}}~(**)\\=&1+\color{#20A900}{\underbrace{\cos (73^{\circ}+47^{\circ})}_{\color{#333333}{\dfrac{-1}2}}\cos (73^{\circ}-47^{\circ})}\\=&1-\dfrac 12 \cos 26^{\circ}\end{aligned}}$

Hence ,

$\large \mathcal{K=L+P}=1-\dfrac 14=\dfrac{3}{4}$

$\huge \therefore 3+4=\boxed{\color{#007fff}{7}}$

$\boxed{\color{#20A900}{\small{\cos^2 A-\sin^2 B=\cos(A+B)\cos(A-B)}}}$

$\small{\color{grey}{\text{PROOF:-}}}$

$\text{LHS:- }\small{\dfrac{1+\cos 2A}{2}-\dfrac{1-\cos 2B}{2}}$ $\small{=\cos 2A+\cos 2B=\cos(A+B)\cos(A-B)}$

Rishabh Jain - 4 years, 12 months ago

Why do you always go for a long solution ? :P ... Look at This one

$\cos^2{73^{\circ}}+\cos^2{47^{\circ}}+\cos{73^{\circ}} \cdot \cos{47^{\circ}} \implies \left(\cos{73^{\circ}}+\cos{47^{\circ}} \right)^2-\cos{73^{\circ}}\cos{47^{\circ}}\cdot\dfrac{2}{2} \\ \left( 2\cos{60^{\circ}}\cdot\cos{13^{\circ}}\right)^2-\dfrac{1}{2}\left( \underbrace{\cos{120^{\circ}}}_{-\frac12}+\underbrace{\cos{26^{\circ}}}_{2\cos^2{13^{\circ}}-1}\right) \\ \cos^2{13^{\circ}}+\dfrac{1}{4}-\cos^2{13^{\circ}}+\dfrac{1}{2} = \boxed{\dfrac34} \\ 3+4=\boxed{7}$

Sabhrant Sachan - 4 years, 12 months ago

Lol one line for $\mathfrak L$ and two lines for $\mathcal P$ doesnt make the solution long.... :-) . I always do these type of questions without pen and paper but its always hard to explain it to others... And I well could have combined without evaluating $\mathcal P$ and $\mathcal L$ separately but I did it to provide better clarity but if that makes the solution long no problem and BTW that proof was added just for fun ... :-)

Rishabh Jain - 4 years, 12 months ago

i agree with you :-) ,but think about the others who are weak in trigonometry . you gave Jee-Advanced ?

Sabhrant Sachan - 4 years, 12 months ago

@Sabhrant Sachan – Said many times on brilliant... Leave it (Jee Adv) ....

Rishabh Jain - 4 years, 12 months ago

How did you evaluated $\mathcal {P}$ ? @Rishabh Cool

Rishabh Tiwari - 4 years, 12 months ago

I've added the proof :-)

Rishabh Jain - 4 years, 12 months ago

Ok thanx , you mistook and wrote $\cos^{2}47^{\circ}$ instead of $\sin^{2} 47^{\circ}$ , that's what confused me. Btw perfect solution ! (+1)! What was your rank in Jee-Advanced?

Rishabh Tiwari - 4 years, 12 months ago

@Rishabh Tiwari – Leave it..

Rishabh Jain - 4 years, 12 months ago

Even

$cos^{2}(\theta) + cos^{2}(120- \theta) + cos(\theta).cos(120-\theta)$ = $0.75$ .

Aakash Khandelwal - 4 years, 12 months ago

Rajen Kapur
Apr 22, 2017

Use: $c^2 = a^2 + b^2 + ab$ where side c is given in terms of sides a and b having an enclosed angle of $120^{\circ}$ . Now given that, $a= cos73^{\circ}=sin17^{\circ}$ and $b=cos47^{\circ}=sin43^{\circ}$ of a triangle which has $c=sin120^{\circ}=\dfrac{\sqrt{3}}{2}$ by Law of sines. Hence $c^2=\dfrac{3}{4}$